Search: a016730 -id:a016730
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A120754
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Records in continued fraction expansion of 1/log(2) (cf. A016730).
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+20
3
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1, 2, 3, 6, 10, 13, 14, 18, 32, 52, 113, 485, 3377, 4714, 963664, 10467647, 14779710, 15407967, 70919074, 73672410, 363144903, 409121736, 628298429, 2803904265, 4054561652, 53155160769
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OFFSET
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1,2
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COMMENTS
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a(n) are also the incrementally largest terms in the continued fraction of log(2) = [a_0; a_1, a_2, ...] = [0; 1, 2, 3, 1, 6, 3, 1, 1, 2] (excluding the a_0 term). - Eric W. Weisstein, Aug 20 2013
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LINKS
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MAPLE
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with(numtheory); read transforms; t1:= cfrac(1/log(2), 2000, 'quotients'); RECORDS(t1);
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CROSSREFS
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Cf. A016730 (continued fraction of log(2)).
Cf. A120755 (positions of records in the continued fraction of 1/log(2) and log(2)).
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KEYWORD
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nonn,cofr
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A120755
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Where records occur in continued fraction expansion of 1/log(2) (cf. A016730).
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+20
3
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1, 2, 3, 5, 15, 28, 41, 47, 74, 109, 123, 166, 501, 6528, 9168, 465506, 3456790, 28568688, 47066208, 146052963, 201331652, 415612810, 1047079803, 1464289355, 2294768489, 2565310827
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OFFSET
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1,2
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COMMENTS
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a(n) are also the positions of incrementally highest terms in the continued fraction of log(2) = [a_0; a_1, a_2, ...] = [0; 1, 2, 3, 1, 6, 3, 1, 1, 2] (excluding the term a_0) - Eric W. Weisstein, Aug 20 2013
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LINKS
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CROSSREFS
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Cf. A016730 (continued fraction of log(2)).
Cf. A120754 (records in the continued fraction expansion of 1/log(2) and log(2).
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KEYWORD
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nonn,cofr
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A002162
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Decimal expansion of the natural logarithm of 2.
(Formerly M4074 N1689)
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+10
218
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6, 9, 3, 1, 4, 7, 1, 8, 0, 5, 5, 9, 9, 4, 5, 3, 0, 9, 4, 1, 7, 2, 3, 2, 1, 2, 1, 4, 5, 8, 1, 7, 6, 5, 6, 8, 0, 7, 5, 5, 0, 0, 1, 3, 4, 3, 6, 0, 2, 5, 5, 2, 5, 4, 1, 2, 0, 6, 8, 0, 0, 0, 9, 4, 9, 3, 3, 9, 3, 6, 2, 1, 9, 6, 9, 6, 9, 4, 7, 1, 5, 6, 0, 5, 8, 6, 3, 3, 2, 6, 9, 9, 6, 4, 1, 8, 6, 8, 7
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OFFSET
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0,1
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COMMENTS
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Newton calculated the first 16 terms of this sequence.
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REFERENCES
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G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 227.
S. R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3 and 6.2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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M. Kontsevich and D. Zagier, Periods, pp. 4-5.
Albert Stadler, Problem 3567, Crux Mathematicorum, Vol. 36 (Oct. 2010), p. 396; Oliver Geupel, Solution, Crux Mathematicorum, Vol. 37 (Oct. 2011), pp. 400-401.
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FORMULA
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log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
log(2) = Integral_{t=0..1} dt/(1+t).
log(2) = (2/3) * (1 + Sum_{k>=1} 2/((4*k)^3-4*k)) (Ramanujan).
log(2) = 4*Sum_{k>=0} (3-2*sqrt(2))^(2*k+1)/(2*k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = (1/4)*(3 - Sum_{n>=1} 1/(n*(n+1)*(2*n+1))).
log(2) = (230166911/9240 - Sum_{k>=1} (1/2)^k*(11/k + 10/(k+1) + 9/(k+2) + 8/(k+3) + 7/(k+4) + 6/(k+5) - 6/(k+7) - 7/(k+8) - 8/(k+9) - 9/(k+10) - 10/(k+11)))/35917. (End)
From log(1-x-x^2) at x=1/2, log(2) = (1/2)*Sum_{k>=1} L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). - Jaume Oliver Lafont, Oct 24 2009
log(2) = 105*(Sum_{n>=1} 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7)))-319/44100).
log(2) = 319/420 - (3/2)*Sum_{n>=1} 1/(6*n^2+39*n+63)). (End)
log(2) = lim_{s->1} zeta(s)*(1-1/2^(s-1)). - Mats Granvik, Jun 18 2013
log(2) = 2*Sum_{n>=1} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
log(2) = (1/3)*Sum_{n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = (1/12)*Sum_{n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
log(2) = (3/16)*Sum_{n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = (1/12)*Sum_{n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = Sum_{n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
log(2) = (2/3)*Sum_{k >= 0} 1/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second-order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = (2/3)*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
Asymptotic expansions:
for N even, log(2) - Sum_{k = 1..N/2} (-1)^(k-1)/k ~ (-1)^(N/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1. See Borwein et al., Theorem 1 (b);
for N odd, log(2) - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/k ~ (-1)^((N-1)/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), by Borwein et al., Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then set x = (N - 1)/2, where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. (End)
log(2) = lim_{n->oo} Sum_{k=1..n} sin(1/(n+k)). See Mathematical Reflections link. - Michel Marcus, Jan 07 2017
Equals Sum_{k>=2} zeta(k)/2^k.
Equals -Sum_{k>=2} log(1 - 1/k^2).
Equals Integral_{x=0..Pi/3} tan(x) dx. (End)
log(2) = Integral_{x=0..Pi/2} (sec(x) - tan(x)) dx. - Clark Kimberling, Jul 08 2020
log(2) = Integral_{x = 0..1} (x - 1)/log(x) dx (Boros and Moll, p. 97).
log(2) = (1/2)*Integral_{x = 0..1} (x + 2)*(x - 1)^2/log(x)^2 dx.
log(2) = (1/4)*Integral_{x = 0..1} (x^2 + 3*x + 4)*(x - 1)^3/log(x)^3 dx. (End)
log(2) = 2*arcsinh(sqrt(2)/4) = 2*sqrt(2)*Sum_{n >= 0} (-1)^n*C(2*n,n)/ ((8*n+4)*32^n) = 3*Sum_{n >= 0} (-1)^n/((8*n+4)*(2^n)*C(2*n,n)). - Peter Bala, Jan 14 2022
log(2) = Integral_{x=0..oo} ( e^(-x) * (1-e^(-2x)) * (1-e^(-4x)) * (1-e^(-6x)) ) / ( x * (1-e^(-14x)) ) dx (see Crux Mathematicorum link). - Bernard Schott, Jul 11 2022
log(2) = 23/32 + 2!^3/16 * Sum_{n >= 1} (-1)^n * (n + 1)/(n*(n + 1)*(n + 2))^2 = 707/1024 - 4!^3/(16^2 * 2!^2) * Sum_{n >= 1} (-1)^n * (n + 2)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^2 = 42611/61440 + 6!^3/(16^3 * 3!^2) * Sum_{n >= 1} (-1)^n * (n + 3)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6))^2.
More generally, it appears that for k >= 0, log(2) = c(k) + (2*k)!^3/(16^k * k!^2) * Sum_{n >= 1} (-1)^(n+k+1) * (n + k)/(n*(n + 1)*...*(n + 2*k))^2 , where c(k) is a rational approximation to log(2). The first few values of c(k) are [0, 23/32, 707/1024, 42611/61440, 38154331/55050240, 76317139/110100480, 26863086823/38755368960, ...].
Let P(n,k) = n*(n + 1)*...*(n + k).
Conjecture: for k >= 0 and r even with r - 1 <= k, the series Sum_{n >= 1} (-1)^n * (d/dn)^r (P(n,k)) / (P(n,k)^2 = A(r,k)*log(2) + B(r,k), where A(r,k) and B(r,k) are both rational numbers. (End)
log(2) = 5/8 + (1/8)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)^2 / ( k*(k + 1) )^4
= 257/384 + (3!^5/2^9)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)^2*(2*k + 5) / ( k*(k + 1)*(k + 2)*(k + 3) )^4
= 267515/393216 + (5!^5/2^19)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)*(2*k + 5)^2*(2*k + 7)*(2*k + 9) / ( k*(k + 1)*(k + 2)*(k + 3)*(k + 4)*(k + 5) )^4
log(2) = 3/4 - 1/128 * Sum_{k >= 0} (-1/16)^k * (10*k + 12)*binomial(2*k+2,k+1)/ ((k + 1)*(2*k + 3)). The terms of the series are O(1/(k^(3/2)*4^n)). (End)
log(2) = eta(1) is a period, where eta(x) is the Dirichlet eta function. - Andrea Pinos, Mar 19 2024
log(2) = K_{n>=0} (n^2 + [n=0])/1, where K is the Gauss notation for an infinite continued fraction. In the expanded form, log(2) = 1/(1 + 1/(1 + 4/(1 + 9/1 + 16/(1 + 25/(1 + ... (see Clawson at p. 227). - Stefano Spezia, Jul 01 2024
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EXAMPLE
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0.693147180559945309417232121458176568075500134360255254120680009493393...
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MATHEMATICA
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PROG
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(PARI) { default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } \\ Harry J. Smith, Apr 21 2009
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CROSSREFS
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KEYWORD
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STATUS
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approved
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A067882
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Factorial expansion of log(2) = Sum_{n>=1} a(n)/n!.
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+10
5
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0, 1, 1, 0, 3, 1, 0, 3, 6, 2, 5, 4, 6, 11, 4, 11, 5, 12, 3, 5, 13, 2, 22, 6, 22, 13, 20, 7, 1, 0, 1, 20, 2, 6, 4, 1, 18, 14, 35, 2, 11, 31, 16, 19, 42, 36, 41, 0, 14, 31, 25, 43, 4, 13, 34, 53, 50, 57, 2, 30, 12, 25, 45, 24, 2, 39, 57, 51, 30, 41, 65, 15, 9, 55, 23, 4, 35, 18, 77, 43
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OFFSET
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1,5
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LINKS
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FORMULA
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a(n) = floor(n!*log(2)) - n*floor((n-1)!*log(2)).
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EXAMPLE
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log(2) = 0 + 1/2! + 1/3! + 0/4! + 3/5! + 1/6! + 0/7! + 3/8! + 6/9! + ...
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MATHEMATICA
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With[{b = Log[2]}, Table[If[n == 1, Floor[b], Floor[n!*b] - n*Floor[(n - 1)!*b]], {n, 1, 100}]] (* G. C. Greubel, Nov 26 2018 *)
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PROG
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(PARI) default(realprecision, 250); b = log(2); for(n=1, 80, print1(if(n==1, floor(b), floor(n!*b) - n*floor((n-1)!*b)), ", ")) \\ G. C. Greubel, Nov 26 2018
(Magma) SetDefaultRealField(RealField(250)); [Floor(Log(2))] cat [Floor(Factorial(n)*Log(2)) - n*Floor(Factorial((n-1))*Log(2)) : n in [2..80]]; // G. C. Greubel, Nov 26 2018
(Sage)
if (n==1): return floor(log(2))
else: return expand(floor(factorial(n)*log(2)) - n*floor(factorial(n-1)*log(2)))
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CROSSREFS
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KEYWORD
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easy,nonn
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STATUS
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approved
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A129935
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Numbers n such that ceiling( 2/(2^(1/n)-1) ) is not equal to floor( 2n/log(2) ).
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+10
5
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777451915729368, 140894092055857794, 1526223088619171207, 3052446177238342414, 54545811706258836911039145, 624965662836733496131286135873807507, 1667672249427111806462471627630318921648499, 36465374036664559522628534720215805439659141
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OFFSET
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1,1
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COMMENTS
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If n belongs to this sequence and m = ceiling(2/(2^(1/n)-1)), then 0 < m/(2n) - 1/log(2) < (log(2)/3) * (1/(2n)^2) implying that m/(2n) is a convergent of 1/log(2) (note that m and 2n are not necessarily coprime). - Max Alekseyev, Jun 06 2007
"Some background to Max Alekseyev's comments: The key point is that the Laurent series for 2/(2^(1/n)-1) about n=infinity is 2/log(2)*n - 1 + (1/6)*log(2)/n + O(1/n^3).
"Also, since 2/log(2) is irrational, 2n/log(2) is never integral, so floor(2n/log(2)) = ceiling(2n/log(2)-1).
"So the question becomes: when is 2n/log(2)-1 so close to an integer that 2/(2^(1/n)-1) is on the other side of the integer? That is why the continued fraction expansion of 2/log(2) is relevant." (End)
The appropriate generalization of ceiling(2/(2^(1/n)-1)) = ? floor(2n/log(2)) is floor(a/(b^(1/n)-1)+a/2) = ceiling(an/log(b)). When a=2, the a/2 can be hidden in floor() + 1 = ceiling(). - David Applegate, Jun 08 2007 [edited Jun 11 2007]
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REFERENCES
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S. W. Golomb and A. W. Hales, "Hypercube Tic-Tac-Toe", in "More Games of No Chance", ed. R. J. Nowakowski, MSRI Publications 42, Cambridge University Press, 2002, pp. 167-182. Here it is stated that the first counterexample is at n=6847196937, an error due to faulty multiprecision arithmetic. The correct value was found by J. Buhler in 2004 and is reported in S. Golomb, "Martin Gardner and Tictacktoe," in Demaine, Demaine, and Rodgers, eds., A Lifetime of Puzzles, A K Peters, 2008, pp. 293-301.
Dean Hickerson, Email to Jon Perry and N. J. A. Sloane, Dec 16 2002. Gives first three terms: 777451915729368, 140894092055857794, 1526223088619171207, as well as five later terms. - N. J. A. Sloane, Apr 30 2014
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LINKS
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S. W. Golomb and A. W. Hales, Hypercube Tic-Tac-Toe, More Games of No Chance, MSRI Publications, Vol. 42, 2002.
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MATHEMATICA
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(* Mma 9.0.1 code from Bill Gosper, Mar 15 2013. He comments: "This reproduces the hundred values in the b-file, and probably works up to around half a billion digits. When Mathematica gets fixed, change 999999999 to infinity." *)
$MaxExtraPrecision = 999999999; For[{lo = {0, 1}, hi = {1, 0}, nu = {0, 0}, n = 0}, nu[[2]] < 10^386, nu = lo + hi; For[{k = nu[[2]]}, Floor[k*2/Log[2]] != Ceiling[2/(2^(1/k) - 1)], k += nu[[2]], Print[{++n, k}]];
If[nu[[1]]*Log[2] > 2*nu[[2]], hi = nu, lo = nu]]
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PROG
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(PARI) prec=1500; default(realprecision, prec); c=contfrac(log(2)/2); default(realprecision, prec*2+50); i=0; for(n=2, #c-1, cand=contfracpnqn(vecextract(c, 2^n-1))[1, 1]; forstep(m=cand, c[n+1]*cand, cand, if(ceil(2/(2^(1/m)-1)) != floor(2*m/log(2)), i++; print(i" "m), break))) /* Phil Carmody, Mar 20 2013 */
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CROSSREFS
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Cf. A078608 for the sequence ceiling( 2/(2^(1/n)-1) ).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A317557
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Number of binary digits to which the n-th convergent of the continued fraction expansion of log(2) matches the correct value.
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+10
3
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0, -1, 3, 6, 9, 13, 14, 17, 19, 20, 23, 20, 25, 20, 33, 37, 35, 38, 41, 43, 45, 43, 47, 48, 52, 54, 58, 61, 68, 70, 74, 77, 78, 81, 86, 89, 92, 93, 92, 99, 105, 109, 113, 116, 118, 121, 127, 133, 136, 135, 139, 141, 145, 149, 154, 159, 161, 165, 171, 173, 172, 180
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OFFSET
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1,3
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COMMENTS
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Binary expansion of log(2) in A068426.
For number of correct decimal digits see A317558.
For the similar case of number of correct binary digits of Pi see A305879.
The denominator of the k-th convergent obtained from a continued fraction satisfying the Gauss-Kuzmin distribution will tend to exp(k*A100199), A100199 being the inverse of Lévy's constant; the error between the k-th convergent and the constant itself tends to exp(-2*k*A100199), or in binary digits 2*k*A100199/log(2) bits after the binary point.
The sequence for quaternary digits is obtained by floor(a(n)/2), the sequence for octal digits is obtained by floor(a(n)/3), and the sequence for hexadecimal digits is obtained by floor(a(n)/4).
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LINKS
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FORMULA
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EXAMPLE
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n convergent binary expansion a(n)
== ============ ============================= ====
1 0 / 1 0.0 0
2 1 / 1 1.0 -1
3 2 / 3 0.1010... 3
4 7 / 10 0.1011001... 6
5 9 / 13 0.1011000100... 9
6 61 / 88 0.10110001011101... 13
7 192 / 277 0.101100010111000... 14
8 253 / 365 0.101100010111001001... 17
9 445 / 642 0.10110001011100100000... 19
10 1143 / 1649 0.101100010111001000011... 20
oo lim = log(2) 0.101100010111001000010111... --
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MATHEMATICA
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a[n_] := Block[{k = 1, a = RealDigits[ Log@2, 2, 4 + 10][[1]], b = RealDigits[ FromContinuedFraction@ ContinuedFraction[Log@2, n + 1], 2, 4n + 10][[1]]}, While[ a[[k]] == b[[k]], k++]; k - 1]; a[1] = 0; a[2] = -1; Array[a, 61] (* Robert G. Wilson v, Aug 09 2018 *)
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CROSSREFS
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KEYWORD
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sign,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A317558
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Number of decimal digits to which the n-th convergent of the continued fraction expansion of log(2) matches the correct value.
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+10
3
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0, -1, 1, 0, 2, 4, 5, 4, 5, 6, 6, 6, 7, 8, 9, 10, 11, 10, 12, 13, 13, 13, 14, 15, 15, 16, 17, 18, 20, 22, 22, 23, 23, 24, 25, 26, 27, 27, 28, 29, 31, 32, 33, 34, 35, 36, 38, 40, 39, 41, 39, 43, 44, 45, 46, 48, 48, 49, 51, 52, 52, 54, 54, 55, 55, 56, 57, 57, 58
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OFFSET
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1,5
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COMMENTS
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Decimal expansion of log(2) in A002162.
For the number of correct binary digits see A317557.
For the similar case of number of correct decimal digits of Pi see A084407.
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LINKS
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FORMULA
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EXAMPLE
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n convergent decimal expansion a(n)
== ============ ==================== ====
1 0 / 1 0.0 0
2 1 / 1 1.0 -1
3 2 / 3 0.66... 1
4 7 / 10 0.7... 0
5 9 / 13 0.692... 2
6 61 / 88 0.69318... 4
7 192 / 277 0.693140... 5
8 253 / 365 0.69315... 4
9 445 / 642 0.693146... 5
10 1143 / 1649 0.6931473... 6
oo lim = log(2) 0.693147180559945... --
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MATHEMATICA
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a[n_] := Block[{k = 1, a = RealDigits[Log@2, 10, n + 10][[1]], b = RealDigits[ FromContinuedFraction@ ContinuedFraction[ Log@2, n], 10, n + 10][[1]]}, While[a[[k]] == b[[k]], k++]; k - 1]; a[1] = 0; a[2] = -1; Array[a, 69] (* Robert G. Wilson v, Aug 09 2018 *)
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CROSSREFS
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KEYWORD
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sign,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A228269
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First position of n in the continued fraction of log(2).
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+10
1
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1, 2, 3, 30, 40, 5, 29, 89, 88, 15, 187, 125, 28, 41, 364, 394, 70, 47, 105, 378, 483, 2096, 520, 1239, 390, 1207, 299, 117, 687, 295, 179, 74, 1842, 1531, 1546, 1302, 1720, 1544, 119, 916, 880, 2081, 110, 614, 865, 310, 951, 2094, 1292, 1064, 6139
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OFFSET
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1,2
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COMMENTS
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Smallest positive integers not appearing in the first 9702699208 terms of the c.f. are 42112, 42387, 43072, 45089, ... - Eric W. Weisstein, Aug 21 2013
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LINKS
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CROSSREFS
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Cf. A016730 (continued fraction of log(2)).
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KEYWORD
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nonn,cofr
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AUTHOR
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STATUS
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approved
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A228270
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First position of n in the continued fraction of log(10).
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+10
1
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4, 0, 1, 11, 18, 7, 44, 159, 74, 212, 260, 182, 43, 152, 59, 84, 40, 186, 27, 89, 927, 38, 20, 83, 277, 17, 101, 65, 194, 2244, 492, 779, 88, 632, 411, 634, 1090, 1624, 177, 228, 2358, 1720, 1502, 2809, 2933, 897, 1452, 6833, 5467, 1860, 126, 1010, 1908, 1789
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OFFSET
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1,1
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COMMENTS
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Smallest positive integers not appearing in the first 9702786891 terms of the c.f. are 40230, 45952, 46178, 46530, ... - Eric W. Weisstein, Aug 28 2013
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LINKS
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MATHEMATICA
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Flatten[With[{cf=ContinuedFraction[Log[10], 7000]}, Table[SequencePosition[cf, {n}, 1][[All, 1]], {n, 60}]]]-1 (* Harvey P. Dale, Oct 09 2022 *)
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CROSSREFS
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Cf. A016730 (continued fraction of log(10)).
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KEYWORD
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nonn,cofr
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AUTHOR
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STATUS
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approved
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A113160
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Table read by antidiagonals of continued fractions for log(n).
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+10
0
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0, 1, 1, 2, 10, 1, 3, 7, 2, 1, 1, 9, 1, 1, 1, 6, 2, 1, 1, 1, 1, 3, 2, 2, 1, 3, 1, 2, 1, 1, 3, 1, 1, 17, 12, 2, 1, 3, 7, 3, 4, 2, 1, 5, 2, 2, 1, 6, 1, 18, 19, 1, 14, 3, 2, 1, 32, 4, 1, 2, 1, 2, 4, 3, 2, 2, 1, 2, 1, 1, 330, 11, 2, 1, 3, 1, 2, 2, 1, 17, 1, 3, 3, 2, 1, 2, 1, 1, 16, 1, 2, 1, 1, 21, 4, 1, 1, 8, 2
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OFFSET
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2,4
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LINKS
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EXAMPLE
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The table starts:
0,1,2,3,1,6,...
1,10,7,9,2,2,...
1,2,1,1,2,3,...
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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