A350498 Convolution of triangular numbers with every third number of Narayana's Cows sequence.

0, 1, 7, 31, 114, 385, 1250, 3987, 12619, 39810, 125425, 394955, 1243433, 3914383, 12322293, 38789576, 122105944, 384377494, 1209981891, 3808901216, 11990036895, 37743426054, 118812495000, 374009739009, 1177344897390, 3706162867858, 11666626518622, 36725362368682, 115607732787126, 363921470561515

Offset: 1

Author

Greg Dresden, Jan 04 2022

Keywords

• nonn
• easy

Comments

This is the convolution of N(3*n-1) with t(n); in other words, a(n) = Sum_{i=1..n} N(3*i-1)*t(n-i) where N(k)=A000930(k) is the k-th number in Narayana's Cows sequence and t(k)=A000217(k) is the k-th triangular number.

Examples

			For n=4, a(4) = N(2)*t(3) + N(5)*t(2) + N(8)*t(1) + N(11)*t(0) = 1*6 + 4*3 + 13*1 + 41*0 = 31, where N(k)=A000930(k) and t(k)=A000217(k).
		

References

• G. Dresden and M. Tulskikh, "Convolutions of Sequences with Single-Term Signature Differences", preprint.

Links

• Index entries for linear recurrences with constant coefficients, signature (7,-18,23,-16,6,-1).

Crossrefs

Cf. A000217, A000930.

Programs

Mathematica

CoefficientList[
Series[x/((-1 + x)^3 (-1 + 4 x - 3 x^2 + x^3)), {x, 0, 30}], x]

Formula

a(n) = N(3*n-1) - A000217(n) where N(k)=A000930(k).

G.f.: x^2/((1 - x)^3 * (1 - 4*x + 3*x^2 - x^3)).

a(n) = A052529(n)-A000217(n), n>0. - R. J. Mathar, Aug 17 2022