OFFSET
0,3
COMMENTS
The signs of the terms form a seemingly unpredictable pattern.
The sequence begins [1, 0, -1, 0, -11, 0, 59, 0, ...]. As is standard practice in the database, the zeros have been omitted.
Even order terms of the sequence (i.e., the nonzero ones) gives the diagonal of rational function 1/(1 - (x + y + x*z - y*z - x*y*z)). - Gheorghe Coserea, Aug 29 2018
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..300
FORMULA
G.f.: Sum_{n>=0} (-x)^n * Sum_{k=0..n} (-1)^k * C(n,k)^2 * Sum_{j=0..k} C(k,j)^2 * x^j = 1 - x^2 - 11*x^4 + 59*x^6 + 241*x^8 - ....
a(n) = Sum_{k=0..n} (-1)^k * Sum_{j=0..2*n-2*k} C(2*n-k, k+j)^2 * C(k+j, j)^2.
From Peter Bala, Mar 13 2018: (Start)
a(n) = Sum_{k = 0..n} (-1)^k*C(2*k,k)*C(n+k,n-k)^2.
n^2*(2*n-5)*a(n) = -( (2*n-5)*(2*n-1)^2*a(n-1) + (2*n-3)*(70*n^2-210*n+87)*a(n-2) + (2*n-1)*(2*n-5)^2*a(n-3) + (2*n-1)*(n-3)^2*a(n-4) ). (End)
a(n) = hypergeom([-n, -n, n + 1, n + 1], [1/2, 1, 1], -1/4). - Peter Luschny, Mar 14 2018
y = Sum_{n>=0} a(n)*x^n satisfies: 0 = 2*x*(x^4 + 4*x^3 + 70*x^2 + 4*x + 1)*y''' - (3*x^6 + 12*x^5 + 190*x^4 - 52*x^3 - 837*x^2 - 32*x - 4)*y'' - (9*x^5 + 24*x^4 + 166*x^3 - 106*x^2 - 947*x - 18)*y' - (3*x^4 + 3*x^3 - 67*x^2 - 29*x - 106)*y. - Gheorghe Coserea, Aug 29 2018
From Mark van Hoeij, Nov 04 2023: (Start)
a(n) = LegendreP(n, 2+sqrt(5)) * LegendreP(n, 2-sqrt(5)).
G.f.: hypergeom([1/4, 3/4],[1],-64*x^2/(x+1)^4)/(x+1). (End)
EXAMPLE
G.f.: A(x) = 1 - x^2 - 11*x^4 + 59*x^6 + 241*x^8 - 3361*x^10 - 419*x^12 + ...
where the g.f. is given by the series:
A(x) = 1/(1+x) + x/(1+x)^3 * (1+x) * (1-x)
+ x^2/(1+x)^5 * (1 + 2^2*x + x^2) * (1 - 2^2*x + x^2)
+ x^3/(1+x)^7 * (1 + 3^2*x + 3^2*x^2 + x^3) * (1 - 3^2*x + 3^2*x^2 - x^3)
+ x^4/(1+x)^9 * (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4) * (1 - 4^2*x + 6^2*x^2 - 4^2*x^3 + x^4)
+ x^5/(1+x)^11 * (1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5) * (1 - 5^2*x + 10^2*x^2 - 10^2*x^3 + 5^2*x^4 - x^5) + ...
in which the coefficients of the odd powers of x vanish.
We can also express the g.f. by the series:
A(x) = 1 - x*(1 - (1+x)) + x^2*(1 - 2^2*(1+x) + (1+2^2*x+x^2))
- x^3*(1 - 3^2*(1+x) + 3^2*(1+2^2*x+x^2) - (1+3^2*x+3^2*x^2+x^3))
+ x^4*(1 - 4^2*(1+x) + 6^2*(1+2^2*x+x^2) - 4^2*(1+3^2*x+3^2*x^2+x^3) + (1+4^2*x+6^2*x^2+4^2*x^3+x^4))
- x^5*(1 - 5^2*(1+x) + 10^2*(1+2^2*x+x^2) - 10^2*(1+3^2*x+3^2*x^2+x^3) + 5^2*(1+4^2*x+6^2*x^2+4^2*x^3+x^4) - (1+5^2*x+10^2*x^2+10^2*x^3+5^2*x^4+x^5)) + ...
MAPLE
seq(add((-1)^k*binomial(2*k, k)*binomial(n+k, n-k)^2, k = 0..n), n = 0..20); # Peter Bala, Mar 13 2018
MATHEMATICA
a[n_] := HypergeometricPFQ[{-n, -n, n + 1, n + 1}, {1/2, 1, 1}, -1/4]; Table[a[n], {n, 0, 22}] (* Peter Luschny, Mar 14 2018 *)
PROG
(PARI) /* By definition: */
{a(n)=local(A=1); A=sum(m=0, n, x^m/(1+x)^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * x^k) * sum(k=0, m, binomial(m, k)^2 * (-x)^k) +x*O(x^n)); polcoeff(A, n)}
for(n=0, 25, print1(a(2*n), ", "))
(PARI) /* By a binomial identity: */
{a(n)=polcoeff(sum(m=0, n, (-x)^m*sum(k=0, m, (-1)^k * binomial(m, k)^2 * sum(j=0, k, binomial(k, j)^2 * x^j) +x*O(x^n))), n)}
for(n=0, 25, print1(a(2*n), ", "))
(PARI) /* Formula for a(2*n): */
{a(n)=sum(k=0, n\2, sum(j=0, n-2*k, (-1)^j * binomial(n-k, k+j)^2 * binomial(k+j, j)^2 ))}
for(n=0, 25, print1(a(2*n), ", "))
(PARI)
seq(N) = {
my(a=vector(N), t1, t2); a[1]=-1; a[2]=-11; a[3]=59; a[4]=241;
for (n=5, N,
t1 = (2*n-5)*(2*n-1)^2*a[n-1] + (2*n-3)*(70*n^2-210*n+87)*a[n-2];
t2 = (2*n-1)*(2*n-5)^2*a[n-3] + (2*n-1)*(n-3)^2*a[n-4];
a[n] = -(t1 + t2)/(n^2*(2*n-5)));
concat(1, a);
};
seq(22)
\\ test: y=Ser(seq(303), 'x); 0 == 2*x*(x^4 + 4*x^3 + 70*x^2 + 4*x + 1)*y''' - (3*x^6 + 12*x^5 + 190*x^4 - 52*x^3 - 837*x^2 - 32*x - 4)*y'' - (9*x^5 + 24*x^4 + 166*x^3 - 106*x^2 - 947*x - 18)*y' - (3*x^4 + 3*x^3 - 67*x^2 - 29*x - 106)*y
\\ Gheorghe Coserea, Aug 29 2018
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Paul D. Hanna, Nov 07 2014
STATUS
approved