A107026 Row sums of inverse of Riordan array (1/(1+x),x/(1+x)^4).

1, 2, 10, 62, 426, 3112, 23686, 185684, 1488554, 12144248, 100489320, 841268078, 7112138790, 60629940152, 520591221412, 4498091003272, 39079909924522, 341193986978008, 2991881019936760, 26338436818801496, 232688056611178216

Offset: 0

Author

Paul Barry, May 09 2005

Keywords

• easy
• nonn

Comments

The Riordan array (1/(1+x),x/(1+x)^4) has general term (-1)^(n-k)*binomial(n+3k,4k).

Crossrefs

Cf. A047098, A107027, A107030.

Programs

Maple

A107026 := proc(n)
    3*binomial(4*n,n)-2*add(binomial(4*n,k),k=0..n) ;
end proc: # R. J. Mathar, Feb 20 2015

Formula

G.f.: A(x)=y satisfies (2y)^4*x-(y+1)^3*(y-1)=0.

a(n) = 3*binomial(4*n, n) - 2*Sum_{k=0..n} binomial(4*n, k).

Conjecture: +189*n*(3*n-1)*(3*n-2)*a(n) +72*(-1034*n^3+3098*n^2-3754*n+1655)*a(n

-1) +384*(2700*n^3-12828*n^2+20426*n-10785)*a(n-2) +4096*(-1066*n^3+6666*n^2-129

50*n+7365)*a(n-3) -65536*(4*n-15)*(2*n-7)*(4*n-13)*a(n-4)=0. - R. J. Mathar, Feb 20 2015

Conjecture: 3*n*(3*n-1)*(3*n-2)*(22*n^2-62*n+43)*a(n) +8*(-1892*n^5+8280*n^4-13330*n^3+9660*n^2-3048*n+315)*a(n-1) +128*(4*n-7)*(2*n-3)*(4*n-5)*(22*n^2-18*n+3)*a(n-2)=0. - R. J. Mathar, Feb 20 2015