OFFSET
0,3
COMMENTS
Row n of triangle A130457 has 3n+1 terms, where row n+1 is generated by taking partial sums of row n and then appending 2 zeros followed by the final term in the partial sums of row n, for n>=0.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
FORMULA
a(n) = Sum_{k=1..n}((k*binomial(4*n-3*k-1,n-k))/(3*n-2*k)), n>0, a(0)=1. - Vladimir Kruchinin, Jun 27 2016
Recurrence: 3*(n-1)*(3*n - 4)*(3*n - 2)*(5491*n^4 - 57274*n^3 + 222641*n^2 - 382514*n + 245280)*a(n) = 8*(2*n - 5)*(4*n - 13)*(4*n - 11)*(5491*n^4 - 35310*n^3 + 83765*n^2 - 87090*n + 33624)*a(n-1) + 3*(n-1)*(3*n - 4)*(3*n - 2)*(5491*n^4 - 57274*n^3 + 222641*n^2 - 382514*n + 245280)*a(n-2) - (1553953*n^7 - 22978945*n^6 + 143006047*n^5 - 485316835*n^4 + 969747472*n^3 - 1141036300*n^2 + 732514848*n - 198216000)*a(n-3) + 8*(2*n - 5)*(4*n - 13)*(4*n - 11)*(5491*n^4 - 35310*n^3 + 83765*n^2 - 87090*n + 33624)*a(n-4). - Vaclav Kotesovec, Jun 27 2016
a(n) ~ 2^(8*n + 9/2) / (3025 * sqrt(Pi) * 3^(3*n - 3/2) * n^(3/2)). - Vaclav Kotesovec, Jun 27 2016
EXAMPLE
Triangle A130457 begins:
1;
1, 0, 0, 1;
1, 1, 1, 2, 0, 0, 2;
1, 2, 3, 5, 5, 5, 7, 0, 0, 7;
1, 3, 6, 11, 16, 21, 28, 28, 28, 35, 0, 0, 35;
1, 4, 10, 21, 37, 58, 86, 114, 142, 177, 177, 177, 212, 0, 0, 212; ...
MATHEMATICA
Flatten[{1, Table[Binomial[4*n-4, n-1] * HypergeometricPFQ[{2, 1 - 3*n/2, 3/2 - 3*n/2, 1 - n}, {4/3 - 4*n/3, 5/3 - 4*n/3, 2 - 4*n/3}, 4/27] / (3*n-2), {n, 1, 20}]}] (* Vaclav Kotesovec, Jun 27 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, May 26 2007
STATUS
approved