A145460 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where sequence a_k of column k is the exponential transform of C(n,k).
1, 1, 1, 1, 1, 2, 1, 0, 3, 5, 1, 0, 1, 10, 15, 1, 0, 0, 3, 41, 52, 1, 0, 0, 1, 9, 196, 203, 1, 0, 0, 0, 4, 40, 1057, 877, 1, 0, 0, 0, 1, 10, 210, 6322, 4140, 1, 0, 0, 0, 0, 5, 30, 1176, 41393, 21147, 1, 0, 0, 0, 0, 1, 15, 175, 7273, 293608, 115975, 1, 0, 0, 0, 0, 0, 6, 35, 1176, 49932, 2237921, 678570
Offset: 0
Examples
Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, ... 1, 1, 0, 0, 0, 0, ... 2, 3, 1, 0, 0, 0, ... 5, 10, 3, 1, 0, 0, ... 15, 41, 9, 4, 1, 0, ... 52, 196, 40, 10, 5, 1, ...
Links
- Alois P. Heinz, Antidiagonals n = 0..140, flattened
- N. J. A. Sloane, Transforms
Crossrefs
Programs
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Maple
exptr:= proc(p) local g; g:= proc(n) option remember; `if`(n=0, 1, add(binomial(n-1, j-1) *p(j) *g(n-j), j=1..n)) end: end: A:= (n,k)-> exptr(i-> binomial(i, k))(n): seq(seq(A(n, d-n), n=0..d), d=0..12);
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Mathematica
Exptr[p_] := Module[{g}, g[n_] := g[n] = If[n == 0, 1, Sum[Binomial[n-1, j-1] *p[j]*g[n-j], {j, 1, n}]]; g]; A[n_, k_] := Exptr[Function[i, Binomial[i, k]]][n]; Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Jan 15 2014, translated from Maple *)
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Ruby
def ncr(n, r) return 1 if r == 0 (n - r + 1..n).inject(:*) / (1..r).inject(:*) end def A(k, n) ary = [1] (1..n).each{|i| ary << (0..i - 1).inject(0){|s, j| s + ncr(i - 1, j) * ncr(j + 1, k) * ary[i - 1 - j]}} ary end def A145460(n) a = [] (0..n).each{|i| a << A(i, n - i)} ary = [] (0..n).each{|i| (0..i).each{|j| ary << a[i - j][j] } } ary end p A145460(20) # Seiichi Manyama, Sep 28 2017
Formula
A(0,k) = 1 and A(n,k) = Sum_{i=0..n-1} binomial(n-1,i) * binomial(i+1,k) * A(n-1-i,k) for n > 0. - Seiichi Manyama, Sep 28 2017
Comments