OFFSET
1,1
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..10000
Robert Israel, Proof of recurrence for column k
FORMULA
Empirical: T(n,k) = sum{i=0..floor(n/(k+1))} ( (-1)^i * (k+1)^(n-(k+1)*i) * binomial(n-k*i,i) ) (after A076264)
Empirical for column k: a(n) = (k+1)*a(n-1) - a(n-(k+1)).
Formula for column k verified by Robert Israel, Dec 17 2017 (see link).
EXAMPLE
Table starts
2 3 4 5 6 7 8 9 10 11 ...
3 9 16 25 36 49 64 81 100 121 ...
4 26 64 125 216 343 512 729 1000 1331 ...
5 75 255 625 1296 2401 4096 6561 10000 14641 ...
6 216 1016 3124 7776 16807 32768 59049 100000 161051 ...
7 622 4048 15615 46655 117649 262144 531441 1000000 1771561 ...
8 1791 16128 78050 279924 823542 2097152 4782969 10000000 19487171 ...
9 5157 64257 390125 1679508 5764787 16777215 43046721 100000000 214358881 ...
...
MAPLE
N:= 20: # for the first N antidiagonals
for k from 1 to N-1 do
F[k]:= gfun:-rectoproc({a(n)=(k+1)*a(n-1) - a(n-k-1), seq(a(j)=(k+1)^j, j=1..k), a(k+1)=(k+1)^(k+1)-1}, a(n), remember)
od:
seq(seq(F[m-j](j), j=1..m-1), m=1..N); # Robert Israel, Dec 17 2017
MATHEMATICA
nmax = 20;
col[k_] := col[k] = Module[{a}, a[n_ /; n>2] := a[n] = (k+1)*a[n-1]-a[n-k-1]; a[0]=1; a[1]=k+1; a[2]=(k+1)^2; a[_?Negative]=0; Array[a, nmax]];
T[n_, k_] := If[k == 1, n+1, col[k][[n]]];
Table[T[n-k+1, k], {n, 1, nmax}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jul 22 2022 *)
CROSSREFS
AUTHOR
R. H. Hardin, Feb 07 2012
STATUS
approved