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A081297
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Array T(k,n), read by antidiagonals: T(k,n) = ((k+1)^(n+1)-(-k)^(n+1))/(2k+1).
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8
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1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 7, 5, 1, 1, 1, 13, 13, 11, 1, 1, 1, 21, 25, 55, 21, 1, 1, 1, 31, 41, 181, 133, 43, 1, 1, 1, 43, 61, 461, 481, 463, 85, 1, 1, 1, 57, 85, 991, 1281, 2653, 1261, 171, 1, 1, 1, 73, 113, 1891, 2821, 10501, 8425, 4039, 341, 1, 1, 1, 91, 145, 3305
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OFFSET
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0,9
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COMMENTS
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Square array of solutions of a family of recurrences.
Rows of the array give solutions to the recurrences a(n)=a(n-1)+k(k-1)a(n-2), a(0)=a(1)=1.
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LINKS
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FORMULA
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T(k, n) = ((k+1)^(n+1)-(-k)^(n+1))/(2k+1).
Rows of the array have g.f. 1/((1+kx)(1-(k+1)x)).
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EXAMPLE
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Rows begin
1, 1, 1, 1, 1, 1, ...
1, 1, 3, 5, 11, 21, ...
1, 1, 7, 13, 55, 133, ...
1, 1, 13, 25, 181, 481, ...
1, 1, 21, 41, 461, 1281, ...
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MATHEMATICA
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T[n_, k_]:=((n + 1)^(k + 1) - (-n)^(k + 1)) / (2n + 1); Flatten[Table[T[n - k, k], {n, 0, 10}, {k, 0, n}]] (* Indranil Ghosh, Mar 27 2017 *)
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PROG
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(PARI)
for(k=0, 10, for(n=0, 9, print1(((k+1)^(n+1)-(-k)^(n+1))/(2*k+1), ", "); ); print(); ) \\ Andrew Howroyd, Mar 26 2017
(Python)
def T(n, k): return ((n + 1)**(k + 1) - (-n)**(k + 1)) // (2*n + 1)
for n in range(11):
print([T(n - k, k) for k in range(n + 1)]) # Indranil Ghosh, Mar 27 2017
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CROSSREFS
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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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