OFFSET
0,5
COMMENTS
LINKS
Robert Israel, Table of n, a(n) for n = 0..2484 (antidiagonals 0 to 69, flattened)
FORMULA
T(n,k) = T(n,k-1) + T(n-1,3*k) for n > 0, k > 0, with T(0,n)=T(n,0)=1 for n >= 0.
G.f. of row n is g_n(z) where g_{n+1}(z) = (1-z)^(-1)*Sum_{w^3=1} g_n(w*z^(1/3)) (the sum being over the cube roots of unity). - Robert Israel, Jun 02 2019
EXAMPLE
Recurrence T(n,k) = T(n,k-1) + T(n-1,3*k) is illustrated by:
T(3,3) = T(3,2) + T(2,9) = 93 + 145 = 238;
T(4,3) = T(4,2) + T(3,9) = 1632 + 4195 = 5827;
T(5,3) = T(5,2) + T(4,9) = 68457 + 273925 = 342382.
Rows of this table begin:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...;
1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, ...;
1, 23, 93, 238, 485, 861, 1393, 2108, 3033, 4195, 5621, ...;
1, 239, 1632, 5827, 15200, 32856, 62629, 109082, 177507, 273925,...;
1, 5828, 68457, 342382, 1144664, 3013980, 6769672, 13570796, ...;
1, 342383, 7112055, 50110483, 215155493, 690729981, 1828979530, ...;
1, 50110484, 1879090014, 18757984045, 103674882878, 406279238154,..;
1, 18757984046, 1287814075131, 18318289003447, 130648799730635, ...;
Triangle A078122 begins:
1;
1, 1;
1, 3, 1;
1, 12, 9, 1;
1, 93, 117, 27, 1;
1, 1632, 3033, 1080, 81, 1;
1, 68457, 177507, 86373, 9801, 243, 1; ...
where row sums form column 1 of this table A125790,
Matrix square A078122^2 begins:
1;
2, 1;
5, 6, 1;
23, 51, 18, 1;
239, 861, 477, 54, 1;
5828, 32856, 25263, 4347, 162, 1; ...
where row sums form column 2 of this table A125790,
MAPLE
f[0]:= 1/(1-z):
S[0]:= series(f[0], z, 21):
for n from 1 to 20 do
ff:= unapply(f[n-1], z);
f[n]:= simplify(1/3*sum(ff(w*z^(1/3)), w=RootOf(Z^3-1, Z)))/(1-z);
S[n]:= series(f[n], z, 21-n)
od:
seq(seq(coeff(S[s-i], z, i), i=0..s), s=0..20); # Robert Israel, Jun 02 2019
MATHEMATICA
T[0, _] = T[_, 0] = 1; T[n_, k_] := T[n, k] = T[n, k-1] + T[n-1, 3 k]; Table[T[n-k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 08 2016 *)
PROG
(PARI) T(n, k, p=0, q=3)=local(A=Mat(1), B); if(n<p || p<0, 0, for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i || j==1, B[i, j]=1, B[i, j]=(A^q)[i-1, j-1]); )); A=B); return((A^(k+1))[n+1, p+1]))
CROSSREFS
AUTHOR
Paul D. Hanna, Dec 10 2006
STATUS
proposed