OFFSET
1,1
COMMENTS
Comments from Alexander Adamchuk, Jul 02 2006: (Start)
Every a(n) is divisible by prime 2, a(n)/2 = A079309(n).
a(n) is divisible by prime 3 only for n=12,30,36,84,90,108,120,... A083096.
a(p) is divisible by p^2 for primes p=5,11,17,23,29,41,47,... Primes of form 6n-1. A007528.
a(p-1) is divisible by p^2 for primes p=7,13,19,31,37,43,... Primes of form 6n+1. A002476.
Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p=7,13,19,31,37,43,... Primes of form 6n+1. A002476.
Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p>3.
a(p^2-1), a(p^2-2) and a(p^2-3) are divisible by p^2 for prime p>3.
a(p^2-4) is divisible by p^2 for prime p>5.
a(p^2-5) is divisible by p^2 for prime p>7.
a(p^2-6) is divisible by p^2 for prime p>7.
a(p^2-7) is divisible by p^2 for prime p>11.
a(p^2-8) is divisible by p^2 for prime p>13.
a(p^3) is divisible by p^2 for prime 2 and prime p=5,11,... Primes of form 6n-1. A007528.
a(p^3-1) is divisible by p^2 for prime p=7,13,... Primes of form 6n+1. A002476.
a(p^4-1) is divisible by p^2 for prime p>3. (End)
Mod[ a(3^k), 9 ] = 1 for integer k>0. Smallest number k such that 2^n divides a(k) is k(n) = {1,2,2,11,11,46,46,707,707,707,...}. Smallest number k such that 3^n divides a(k) is k(n) = (12,822,2466,...}. a(2(p-1)/3) is divisible by p^2 for prime p = {7,13,19,31,37,43,61,...} = A002476 Primes of form 6n+1. Every a(n) from a(p^2-(p+1)/2) to a(p^2-1) is divisible by p^2 for prime p>3. Every a(n) from a((4p+3)(p-1)/6) to a((2p+3)(p-1)/3) is divisible by p^2 for prime p = {7,13,19,31,37,43,61,...} = A002476 Primes of form 6n+1. - Alexander Adamchuk, Jan 04 2007
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000 (Terms 1 to 200 computed by Harry J. Smith; terms 201 to 1000 computed by G. C. Greubel, Jan 15 2017)
Guo-Shuai Mao, Proof of a conjecture of Adamchuk, arXiv:2003.09810 [math.NT], 2020.
Guo-Shuai Mao, On a supercongruence conjecture of Z.-W. Sun, arXiv:2003.14221 [math.NT], 2020.
Guo-Shuai Mao, On some supercongruence conjectures of Z.-W. Sun, Nanjing Univ. Info. Sci. Tech. (China, 2023).
Guo-Shuai Mao, Proof of some congruences via the hypergeometric identities, Nanjing Univ. Info. Sci. Tech. (China, 2023).
Guo-Shuai Mao and Roberto Tauraso, Three pairs of congruences concerning sums of central binomial coefficients, arXiv:2004.09155 [math.NT], 2020.
Z.-W. Sun, Fibonacci numbers modulo cubes of primes, arXiv:0911.3060 [math.NT], 2009-2013; Taiwanese J. Math., to appear 2013. - From N. J. A. Sloane, Mar 01 2013
Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
Eric Weisstein's World of Mathematics, Binomial Sums.
FORMULA
a(n) = A006134(n) - 1; generating function: (sqrt(1-4*x)-1)/(sqrt(1-4*x)*(x-1)) - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 11 2003, corrected by Vaclav Kotesovec, Nov 06 2012
a(n) = Sum_{k=1..n}(2k)!/(k!)^2. - Alexander Adamchuk, Jul 02 2006
a(n) = Sum_{k=1..n}binomial(2k,k). - Alexander Adamchuk, Jan 04 2007
a(n) ~ 2^(2*n+2)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Nov 06 2012
MATHEMATICA
Table[Sum[(2k)!/(k!)^2, {k, 1, n}], {n, 1, 50}] (* Alexander Adamchuk, Jul 02 2006 *)
Table[Sum[Binomial[2k, k], {k, 1, n}], {n, 1, 30}] (* Alexander Adamchuk, Jan 04 2007 *)
PROG
(PARI) { a=0; for (n=1, 200, write("b066796.txt", n, " ", a+=binomial(2*n, n)) ) } \\ Harry J. Smith, Mar 27 2010
(PARI) a(n) = sum(i=1, n, binomial(2*i, i)); \\ Michel Marcus, Jan 04 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Jan 18 2002
STATUS
approved