A099173 Array, A(k,n), read by diagonals: g.f. of k-th row x/(1-2*x-(k-1)*x^2).
0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 4, 4, 0, 1, 2, 5, 8, 5, 0, 1, 2, 6, 12, 16, 6, 0, 1, 2, 7, 16, 29, 32, 7, 0, 1, 2, 8, 20, 44, 70, 64, 8, 0, 1, 2, 9, 24, 61, 120, 169, 128, 9, 0, 1, 2, 10, 28, 80, 182, 328, 408, 256, 10, 0, 1, 2, 11, 32, 101, 256, 547, 896, 985, 512, 11
Offset: 0
Examples
Square array, A(n, k), begins as: 0, 1, 2, 3, 4, 5, 6, 7, 8, ... A001477; 0, 1, 2, 4, 8, 16, 32, 64, 128, ... A000079; 0, 1, 2, 5, 12, 29, 70, 169, 408, ... A000129; 0, 1, 2, 6, 16, 44, 120, 328, 896, ... A002605; 0, 1, 2, 7, 20, 61, 182, 547, 1640, ... A015518; 0, 1, 2, 8, 24, 80, 256, 832, 2688, ... A063727; 0, 1, 2, 9, 28, 101, 342, 1189, 4088, ... A002532; 0, 1, 2, 10, 32, 124, 440, 1624, 5888, ... A083099; 0, 1, 2, 11, 36, 149, 550, 2143, 8136, ... A015519; 0, 1, 2, 12, 40, 176, 672, 2752, 10880, ... A003683; 0, 1, 2, 13, 44, 205, 806, 3457, 14168, ... A002534; 0, 1, 2, 14, 48, 236, 952, 4264, 18048, ... A083102; 0, 1, 2, 15, 52, 269, 1110, 5179, 22568, ... A015520; 0, 1, 2, 16, 56, 304, 1280, 6208, 27776, ... A091914; Antidiagonal triangle, T(n, k), begins as: 0; 0, 1; 0, 1, 2; 0, 1, 2, 3; 0, 1, 2, 4, 4; 0, 1, 2, 5, 8, 5; 0, 1, 2, 6, 12, 16, 6; 0, 1, 2, 7, 16, 29, 32, 7; 0, 1, 2, 8, 20, 44, 70, 64, 8; 0, 1, 2, 9, 24, 61, 120, 169, 128, 9; 0, 1, 2, 10, 28, 80, 182, 328, 408, 256, 10;
Links
- G. C. Greubel, Antidiagonals n = 0..50, flattened
- Ralf Stephan, Prove or disprove. 100 Conjectures from the OEIS, #16, arXiv:math/0409509 [math.CO], 2004.
Crossrefs
Programs
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Magma
A099173:= func< n,k | (&+[n^j*Binomial(k,2*j+1): j in [0..Floor(k/2)]]) >; [A099173(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 17 2023
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Mathematica
A[k_, n_]:= Which[k==0, n, n==0, 0, True, ((1+Sqrt[k])^n - (1-Sqrt[k])^n)/(2 Sqrt[k])]; Table[A[k-n, n]//Simplify, {k, 0, 12}, {n, 0, k}]//Flatten (* Jean-François Alcover, Jan 21 2019 *)
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PARI
A(k,n)=sum(i=0,n\2,k^i*binomial(n,2*i+1))
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SageMath
def A099173(n,k): return sum( n^j*binomial(k, 2*j+1) for j in range((k//2)+1) ) flatten([[A099173(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 17 2023
Formula
A(n, k) = Sum_{i=0..floor(k/2)} n^i * C(k, 2*i+1) (array).
Recurrence: A(n, k) = 2*A(n, k-1) + (n-1)*A(n, k-2), with A(n, 0) = 0, A(n, 1) = 1.
T(n, k) = A(n-k, k) (antidiagonal triangle).
T(2*n, n) = A357502(n).
A(n, k) = ((1+sqrt(n))^k - (1-sqrt(n))^k)/(2*sqrt(n)). - Jean-François Alcover, Jan 21 2019