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Convolution of triangular numbers with every third number of Narayana's Cows sequence.
0

%I #33 Aug 17 2022 07:10:44

%S 0,1,7,31,114,385,1250,3987,12619,39810,125425,394955,1243433,3914383,

%T 12322293,38789576,122105944,384377494,1209981891,3808901216,

%U 11990036895,37743426054,118812495000,374009739009,1177344897390,3706162867858,11666626518622,36725362368682,115607732787126,363921470561515

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

%C 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.

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

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (7,-18,23,-16,6,-1).

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

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

%F a(n) = A052529(n)-A000217(n), n>0. - _R. J. Mathar_, Aug 17 2022

%e 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).

%t CoefficientList[

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

%Y Cf. A000217, A000930.

%K nonn,easy

%O 1,3

%A _Greg Dresden_, Jan 04 2022