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A107026 Row sums of inverse of Riordan array (1/(1+x),x/(1+x)^4). 4
1, 2, 10, 62, 426, 3112, 23686, 185684, 1488554, 12144248, 100489320, 841268078, 7112138790, 60629940152, 520591221412, 4498091003272, 39079909924522, 341193986978008, 2991881019936760, 26338436818801496, 232688056611178216 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
The Riordan array (1/(1+x),x/(1+x)^4) has general term (-1)^(n-k)*binomial(n+3k,4k).
LINKS
FORMULA
G.f.: A(x)=y satisfies (2y)^4*x-(y+1)^3*(y-1)=0; a(n)=3*binomial(4n, n)-2*sum{k=0..n, binomial(4n, 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
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
CROSSREFS
Cf. A047098.
Sequence in context: A370275 A304443 A243034 * A107841 A175936 A175937
KEYWORD
easy,nonn
AUTHOR
Paul Barry, May 09 2005
STATUS
approved

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Last modified March 1 17:48 EST 2024. Contains 370442 sequences. (Running on oeis4.)