A309010 Square array A(n, k) = Sum_{j=0..n} binomial(n,j)^k, n >= 0, k >= 0, read by antidiagonals.
1, 1, 2, 1, 2, 3, 1, 2, 4, 4, 1, 2, 6, 8, 5, 1, 2, 10, 20, 16, 6, 1, 2, 18, 56, 70, 32, 7, 1, 2, 34, 164, 346, 252, 64, 8, 1, 2, 66, 488, 1810, 2252, 924, 128, 9, 1, 2, 130, 1460, 9826, 21252, 15184, 3432, 256, 10, 1, 2, 258, 4376, 54850, 206252, 263844, 104960, 12870, 512, 11
Offset: 0
Examples
Square array, A(n, k), begins: 1, 1, 1, 1, 1, 1, ... A000012; 2, 2, 2, 2, 2, 2, ... A007395; 3, 4, 6, 10, 18, 34, ... A052548; 4, 8, 20, 56, 164, 488, ... A115099; 5, 16, 70, 346, 1810, 9826, ... 6, 32, 252, 2252, 21252, 206252, ... Antidiagonals, T(n, k), begin: 1; 1, 2; 1, 2, 3; 1, 2, 4, 4; 1, 2, 6, 8, 5; 1, 2, 10, 20, 16, 6; 1, 2, 18, 56, 70, 32, 7; 1, 2, 34, 164, 346, 252, 64, 8; 1, 2, 66, 488, 1810, 2252, 924, 128, 9; 1, 2, 130, 1460, 9826, 21252, 15184, 3432, 256, 10;
References
- R. P. Stanley, Enumerative Combinatorics Vol I, Second Edition, Cambridge, 2011, Example 3.18.3 d, page 366.
Links
- Seiichi Manyama, Antidiagonals n = 0..100, flattened
Crossrefs
Programs
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Magma
[(&+[Binomial(k,j)^(n-k): j in [0..k]]): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 26 2022
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Mathematica
nn = 8; Table[ek[x_] := Sum[x^n/n!^k, {n, 0, nn}];Range[0, nn]!^k CoefficientList[Series[ek[x]^2, {x, 0, nn}],x], {k, 0, nn}] // Transpose // Grid (* Geoffrey Critzer, Apr 17 2020 *)
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PARI
A(n, k) = sum(j=0, n, binomial(n, j)^k); \\ Seiichi Manyama, Jan 08 2022
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SageMath
flatten([[sum(binomial(k,j)^(n-k) for j in (0..k)) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 26 2022
Formula
A(n, k) = Sum_{j=0..n} binomial(n,j)^k (array).
A(n, n+1) = A328812(n).
A(n, n) = A167010(n).
T(n, k) = A(k, n-k) (antidiagonals).
T(n, n) = A000027(n+1).
T(n, n-1) = A000079(n-1).
T(n, n-2) = A000984(n-2).
T(n, n-3) = A000172(n-3).
T(n, n-4) = A005260(n-4).
T(n, n-5) = A005261(n-5).
T(n, n-6) = A069865(n-6).
T(n, n-7) = A182421(n-7).
T(n, n-8) = A182422(n-8).
T(n, n-9) = A182446(n-9).
T(n, n-10) = A182447(n-10).
T(n, n-11) = A342294(n-11).
T(n, n-12) = A342295(n-12).
Sum_{n>=0} A(n,k) x^n/(n!^k) = (Sum_{n>=0} x^n/(n!^k))^2. - Geoffrey Critzer, Apr 17 2020
Comments