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A227845
G.f.: Sum_{n>=0} x^n / (1-x)^(2*n+1) * [ Sum_{k=0..n} C(n,k)^2*x^k ]^2.
15
1, 2, 7, 28, 125, 590, 2891, 14536, 74497, 387450, 2038743, 10830148, 57986773, 312542678, 1694166275, 9228580464, 50486521785, 277239830210, 1527533993871, 8441627856300, 46776754474709, 259830443968046, 1446468759734131, 8068688342238328, 45091854560015025, 252423540736438890
OFFSET
0,2
COMMENTS
Equals antidiagonal sums of table A143007.
LINKS
FORMULA
G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^2 * Sum_{j=0..k} binomial(k,j)^2 * x^j.
a(n) = Sum_{k=0..[n/2]} Sum_{j=k..n-k} binomial(n-k,j)^2 * binomial(j,k)^2.
Recurrence: n^2*a(n) = 2*(3*n^2 - 3*n + 1)*a(n-1) - 2*(3*n^2 - 9*n + 7)*a(n-3) + (n-2)^2*a(n-4). - Vaclav Kotesovec, Jul 05 2014
a(n) ~ (3+2*sqrt(2))^(n+1) / (4*Pi*n). - Vaclav Kotesovec, Jul 05 2014
G.f.: 1 / AGM((1+x)^2, 1 - 6*x + x^2), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) denotes the arithmetic-geometric mean. - Paul D. Hanna, Jul 31 2014
G.f. satisfies: A(x) = F(x*A(x))^2, where F(x) is the g.f. of A258053. - Paul D. Hanna, May 17 2015
G.f.: hypergeom([1/2, 1/2], [1], -16*x^2/((x+1)^2*(x^2-6*x+1)))/((x+1)*sqrt(x^2-6*x+1)). - Mark van Hoeij, Jul 08 2024
a(n) = U(n)*U(n-1) where the sequences U(-1),U(1),U(3),... and U(0),U(2),U(4),... satisfy a second order recurrence n^2*U(n) = 2*(3*n^2-3*n+1)*U(n-2) - (n-1)^2*U(n-4) with initial terms U(-1), U(1)=2 and U(0)=1, U(2)=7/2. - Mark van Hoeij, Jul 10 2024
EXAMPLE
G.f.: A(x) = 1 + 2*x + 7*x^2 + 28*x^3 + 125*x^4 + 590*x^5 + 2891*x^6 +...
where
A(x) = 1/(1-x) + x/(1-x)^3 * (1+x)^2 + x^2/(1-x)^5*(1 + 2^2*x + x^2)^2
+ x^3/(1-x)^7 * (1 + 3^2*x + 3^2*x^2 + x^3)^2
+ x^4/(1-x)^9 * (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4)^2
+ x^5/(1-x)^11 * (1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5)^2
+ x^6/(1-x)^13 * (1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)^2 +...
We can also express the g.f. by the binomial series identity:
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)) +...
The square-root of the g.f. is an integer series:
A(x)^(1/2) = 1 + x + 3*x^2 + 11*x^3 + 47*x^4 + 215*x^5 + 1029*x^6 +...+ A227846(n)*x^n +...
The g.f. also satisfies A(x) = F(x*A(x)^2) and F(x)^2 = A(x/F(x)^2)) where
F(x) = 1 + x + x^2 + x^4 - 2*x^5 - 4*x^6 - 7*x^8 + 20*x^9 + 42*x^10 + 84*x^12 - 272*x^13 - 584*x^14 - 1239*x^16 +...+ A258053(n)*x^n +...
such that A258053(4*n+3) = 0 for n>=0.
MAPLE
U := proc(n) options remember;
if n < 1 then 1
elif n = 1 then 2
elif n = 2 then 7/2
else
(2*(3*n^2-3*n+1)*U(n-2) - (n-1)^2*U(n-4))/n^2
fi
end:
seq(U(n)*U(n-1), n=0..25); # Mark van Hoeij, Jul 10 2024
MATHEMATICA
Table[Sum[Sum[Binomial[n-k, j]^2*Binomial[j, k]^2, {j, k, n-k}], {k, 0, Floor[n/2]}], {n, 0, 20}] (* Vaclav Kotesovec, Jul 05 2014 *)
PROG
(PARI) /* From 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)^2+x*O(x^n)); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) /* From alternate g.f.: */
{a(n)=polcoeff(sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2*sum(j=0, k, binomial(k, j)^2*x^j)+x*O(x^n))), n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) /* From formula for a(n): */
{a(n)=sum(k=0, n\2, sum(j=k, n-k, binomial(n-k, j)^2*binomial(j, k)^2))}
for(n=0, 30, print1(a(n), ", "))
(PARI) /* From g.f.: 1/AGM((1+x)^2, 1-6*x+x^2) */
{a(n)=local(A); A = 1 / agm((1+x)^2, 1-6*x+x^2 +x*O(x^n)); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Aug 01 2013
EXTENSIONS
Name changed by Paul D. Hanna, Sep 07 2014
STATUS
approved