A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)*binomial(k-floor((j+1)/2), floor(j/2))*n^(k-j), n >= 1, k >= 0.
1, 1, 2, 1, 3, 1, 1, 4, 5, -1, 1, 5, 11, 7, -2, 1, 6, 19, 29, 9, -1, 1, 7, 29, 71, 76, 11, 1, 1, 8, 41, 139, 265, 199, 13, 2, 1, 9, 55, 239, 666, 989, 521, 15, 1, 1, 10, 71, 377, 1393, 3191, 3691, 1364, 17, -1, 1, 11, 89, 559, 2584, 8119, 15289, 13775, 3571, 19, -2
Offset: 1
Examples
Array begins: 1 2 1 -1 -2 -1 1 2 1 -1 1 3 5 7 9 11 13 15 17 19 1 4 11 29 76 199 521 1364 3571 9349 1 5 19 71 265 989 3691 13775 51409 191861 1 6 29 139 666 3191 15289 73254 350981 1681651 1 7 41 239 1393 8119 47321 275807 1607521 9369319 1 8 55 377 2584 17711 121393 832040 5702887 39088169 1 9 71 559 4401 34649 272791 2147679 16908641 133121449 1 10 89 791 7030 62479 555281 4935050 43860169 389806471 1 11 109 1079 10681 105731 1046629 10360559 102558961 1015229051
Links
- Robert Price, Table of n, a(n) for n = 1..5050
- Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.
Crossrefs
Programs
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Mathematica
(* Array: *) Grid[Table[LinearRecurrence[{n, -1}, {1, 1 + n}, 10], {n, 10}]] (* Array antidiagonals flattened (gives this sequence): *) A294099[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] n^(k - j), {j, 0, k}]; Flatten[Table[A294099[n - k, k], {n, 11}, {k, 0, n - 1}]]
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PARI
{A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*n^(k-j))}; /* Michael Somos, Jun 19 2023 */
Formula
A(n,0) = 1, A(n,1) = n + 1, A(n,k) = n*A(n,k-1) - A(n,k-2), n >= 1, k >= 2.
G.f. for row n: (1 + x)/(1 - n*x + x^2), n >= 1.
A(n, k) = B(-n, k) where B = A299045. - Michael Somos, Jun 19 2023
Comments