Search: a084171 -id:a084171
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A090888
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Matrix defined by a(n,k) = 3^n*Fibonacci(k) - 2^n*Fibonacci(k-2), read by antidiagonals.
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+10
12
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1, 2, 0, 4, 1, 1, 8, 5, 3, 1, 16, 19, 9, 4, 2, 32, 65, 27, 14, 7, 3, 64, 211, 81, 46, 23, 11, 5, 128, 665, 243, 146, 73, 37, 18, 8, 256, 2059, 729, 454, 227, 119, 60, 29, 13, 512, 6305, 2187, 1394, 697, 373, 192, 97, 47, 21, 1024, 19171, 6561, 4246, 2123, 1151, 600, 311
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OFFSET
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0,2
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COMMENTS
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a(n,1) = a(n-1,1) + a(n-1,3) for n > 0; a(n,1) = A001047(n) = 2^(2n) - A083324(n); a(n,2) = A000244(n) = 2^(2n) - A005061(n); a(n,3) = 2a(n-1,4) for n > 0; a(n,3) = A027649(n); a(n,4) = A083313(n+1); a(n,5) = A084171(n+1).
Sum[a(n-k,k), {k,0,n}] = A098703(n+1), antidiagonal sums.
Let R, S and T be binary relations on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xRy if x is a subset of y or y is a subset of x, xSy if x is a subset of y and xTy if x is a proper subset of y. Then a(n,3) = |R|, a(n,2) = |S| and a(n,1) = |T|. Note that a binary relation W on P(A) can be defined also such that for every element x, y of P(A) xWy if x is a proper subset of y and there are no z in P(A) such that x is a proper subset of z and z is a proper subset of y. A090802(n,1) = |W|. Also, a(n,0) = |P(A)|.
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LINKS
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FORMULA
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a(n, k) = 3^n*Fibonacci(k) - 2^n*Fibonacci(k-2).
a(n, 0) = 2^n, a(n, 1) = 3^n - 2^n, a(n, k) = a(n, k-1) + a(n, k-2) for k > 1.
a(0, k) = Fibonacci(k-1), a(1, k) = Lucas(k), a(n, k) = 5a(n-1, k) - 6a(n-2, k) for n > 1.
O.g.f. (by rows) = (-2^n + (2^(n+1) - 3^n)x)/(-1+x+x^2). - Ross La Haye, Mar 30 2006
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EXAMPLE
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1 0 1 1 2 3 5 8 13 21 34
2 1 3 4 7 11 18 29 47 76 123
4 5 9 14 23 37 60 97 157 254 411
8 19 27 46 73 119 192 311 503 814 1317
16 65 81 146 227 373 600 973 1573 2546 4119
32 211 243 454 697 1151 1848 2999 4847 7846 12693
64 665 729 1394 2123 3517 5640 9157 14797 23954 38751
a(5,3) = 454 because Fibonacci(3) = 2, Fibonacci(1) = 1 and (2 * 3^5) - (1 * 2^5) = 454.
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MATHEMATICA
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Table[3^(n - k) Fibonacci@ k - 2^(n - k) Fibonacci[k - 2], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 28 2015 *)
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A250167
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T(n,k)=Number of length n+1 0..k arrays with the sum of adjacent differences multiplied by some arrangement of +-1 equal to zero
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+10
11
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2, 3, 4, 4, 11, 8, 5, 20, 37, 16, 6, 33, 96, 119, 32, 7, 48, 211, 436, 373, 64, 8, 67, 380, 1269, 1880, 1151, 128, 9, 88, 639, 2860, 7109, 7836, 3517, 256, 10, 113, 976, 5831, 19896, 37881, 32032, 10679, 512, 11, 140, 1437, 10460, 49037, 129648, 195927
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OFFSET
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1,1
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COMMENTS
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Table starts
....2.....3.......4........5.........6..........7..........8...........9
....4....11......20.......33........48.........67.........88.........113
....8....37......96......211.......380........639........976........1437
...16...119.....436.....1269......2860.......5831......10460.......17765
...32...373....1880.....7109.....19896......49037.....103556......203615
...64..1151....7836....37881....129648.....380939.....938128.....2121089
..128..3517...32032...195927....810964....2810751....7989940....20567199
..256.10679..129572...996933...4962056...20169871...65768448...191480917
..512.32293..521256..5029417..30034672..142786013..532548628..1748028901
.1024.97391.2091052.25262121.180893724.1004527983.4281269376.15822382297
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LINKS
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FORMULA
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Empirical for column k:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = 5*a(n-1) -6*a(n-2)
k=3: a(n) = 8*a(n-1) -21*a(n-2) +22*a(n-3) -8*a(n-4)
k=4: [order 8]
Empirical for row n:
n=1: a(n) = n + 1
n=2: a(n) = 2*a(n-1) -2*a(n-3) +a(n-4); also a quadratic polynomial plus a constant quasipolynomial with period 2
n=3: a(n) = 2*a(n-1) +a(n-2) -4*a(n-3) +a(n-4) +2*a(n-5) -a(n-6); also a cubic polynomial plus a linear quasipolynomial with period 2
n=4: [order 12; also a quartic polynomial plus a quadratic quasipolynomial with period 12]
n=5: [order 24; also a polynomial of degree 5 plus a cubic quasipolynomialwith period 60]
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EXAMPLE
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Some solutions for n=5 k=4
..3....0....3....4....0....3....4....4....2....4....4....2....0....4....3....1
..2....0....4....2....0....4....1....4....4....1....3....1....1....2....1....1
..4....4....0....4....4....2....2....2....4....3....4....3....1....2....3....3
..0....2....0....1....2....1....3....2....1....0....3....2....3....1....0....4
..1....2....4....1....3....1....3....3....3....0....0....2....0....4....3....3
..1....0....3....2....0....1....2....4....0....4....0....2....0....2....3....1
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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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