A087659 - OEIS

A087659

a(n) = hypergeom([ -n, (n+4)/2, (n+5)/2], [3, 2], -4).

1, 6, 57, 701, 10147, 164317, 2888282, 54047434, 1062530119, 21739192762, 459685114665, 9993072855135, 222421656113435, 5052215132332492, 116808526607319823, 2742986603349411311, 65306671610636210891, 1574090246599071243962, 38361262640988126803839

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OFFSET

0,2

COMMENTS

From Bill Gosper, Feb 04 2004: (Start)

A few weeks ago I conjectured that 2 binomial(n,i) (n+2i+3)! / ((i+1)!(i+2)!(n+3)!) is always an integer (summed on i, this gives the current sequence).

This is the special case C(3,i,n-i) of C(m,k,n) := (n+k)!(n+m)!/(n!(n+m+k)!) * Product_{j=1..k} (j - 1)! (n + j m + m)!/((m + j - 1)! (n + j m)!)

which I also conjecture integral.

(End)

From Alec Mihailovs, Feb 04 2004: (Start)

These conjectures are true. Consider the partition p(m,k,n)=(n+m,m,...,m) of n+m*(k+1), where m is repeated k times. It is easy to see that C(m,k,n) equals the dimension of the irreducible representation of S_(n+m*(k+1)) corresponding to p(m,k,n) calculated using hook length formula.

Another formula for C(m,k,n) is ((n+mk+m)!/n!) * Product_{i=0..m-1} i!/((k+i)!(n+k+i+1)!).

(End)

Cloitre has characterized the sequence mods 2 and 3. Remarkably, a(9k+6) mod 3 = 2*A014578(k+1), the binary expansion of the "Thue constant", 110110111110110111110110110..., wherein the 3n-th bit is the complement of the n-th. - Bill Gosper, Mar 19 2004

LINKS

Table of n, a(n) for n=0..18.

FORMULA

a(n) = Sum _{i=0..n} 2*C(n,i) * (n+2*i+3)! / ( (i+1)! * (i+2)! * (n+3)! ).

From Vaclav Kotesovec, Jul 05 2018: (Start)

D-finite with Recurrence: (n+2)^2*(n+3)*(3*n - 1)*a(n) = 2*(n+2)*(3*n + 1)*(15*n^2 + 5*n - 2)*a(n-1) - (n-1)*(9*n^3 - 3*n^2 - 4*n - 4)*a(n-2) + (n-3)*(n-2)*(n-1)*(3*n + 2)*a(n-3).

a(n) ~ sqrt(957 + 1/3*(5/2*(9465769685 - 18403*sqrt(5)))^(1/3) + 1/3*(5/2*(9465769685 + 18403*sqrt(5)))^(1/3)) * (66 + 10*2^(2/3)*(73 + sqrt(5))^(1/3) + 3*2^(1/3)*(73 + sqrt(5))^(2/3))^n / (Pi * n^4 * 2^(2*n/3) * (73 + sqrt(5))^(n/3)). (End)

MATHEMATICA

a[n_] := HypergeometricPFQ[{-n, (n + 4)/2, (n + 5)/2}, {3, 2}, -4];

Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Feb 19 2018 *)

PROG

(PARI) a(n)= sum(i=0, n, 2*binomial(n, i)*(n+2*i+3)!/((i+1)!*(i+2)!*(n+3)!)) \\ Benoit Cloitre

(Sage)

def A087659():

x, y, z, n = 1, 6, 57, 2

while True:

yield x

n += 1

x, y, z = y, z, ((n-1)*(n-2)*(n-3)*(3*n+2)*x-(n-1)*(9*n^3-3*n^2-4*n-4)*y+(2*(3*n+1))*(n+2)*(15*n^2+5*n-2)*z)/((n+3)*(3*n-1)*(n+2)^2)