A227845 - OEIS

%I #55 Jul 12 2024 10:18:00

%S 1,2,7,28,125,590,2891,14536,74497,387450,2038743,10830148,57986773,

%T 312542678,1694166275,9228580464,50486521785,277239830210,

%U 1527533993871,8441627856300,46776754474709,259830443968046,1446468759734131,8068688342238328,45091854560015025,252423540736438890

%N G.f.: Sum_{n>=0} x^n / (1-x)^(2*n+1) * [ Sum_{k=0..n} C(n,k)^2*x^k ]^2.

%C Equals antidiagonal sums of table A143007.

%H Vincenzo Librandi, <a href="/A227845/b227845.txt">Table of n, a(n) for n = 0..200</a>

%F 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.

%F a(n) = Sum_{k=0..[n/2]} Sum_{j=k..n-k} binomial(n-k,j)^2 * binomial(j,k)^2.

%F 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

%F a(n) ~ (3+2*sqrt(2))^(n+1) / (4*Pi*n). - _Vaclav Kotesovec_, Jul 05 2014

%F 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

%F 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

%F 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

%F 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

%e G.f.: A(x) = 1 + 2*x + 7*x^2 + 28*x^3 + 125*x^4 + 590*x^5 + 2891*x^6 +...

%e where

%e A(x) = 1/(1-x) + x/(1-x)^3 * (1+x)^2 + x^2/(1-x)^5*(1 + 2^2*x + x^2)^2

%e + x^3/(1-x)^7 * (1 + 3^2*x + 3^2*x^2 + x^3)^2

%e + x^4/(1-x)^9 * (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4)^2

%e + x^5/(1-x)^11 * (1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5)^2

%e + 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 +...

%e We can also express the g.f. by the binomial series identity:

%e A(x) = 1 + x*(1 + (1+x)) + x^2*(1 + 2^2*(1+x) + (1+2^2*x+x^2))

%e + 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))

%e + 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))

%e + 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)) +...

%e The square-root of the g.f. is an integer series:

%e 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 +...

%e The g.f. also satisfies A(x) = F(x*A(x)^2) and F(x)^2 = A(x/F(x)^2)) where

%e 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 +...

%e such that A258053(4*n+3) = 0 for n>=0.

%p U := proc(n) options remember;

%p         if n < 1 then 1

%p         elif n = 1 then 2

%p         elif n = 2 then 7/2

%p         else

%p                 (2*(3*n^2-3*n+1)*U(n-2) - (n-1)^2*U(n-4))/n^2

%p         fi

%p end:

%p seq(U(n)*U(n-1), n=0..25); # _Mark van Hoeij_, Jul 10 2024

%t 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 *)

%o (PARI) /* From definition: */

%o {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)}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) /* From alternate g.f.: */

%o {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)}

%o for(n=0, 30, print1(a(n), ", "))

%o (PARI) /* From formula for a(n): */

%o {a(n)=sum(k=0,n\2,sum(j=k,n-k,binomial(n-k,j)^2*binomial(j,k)^2))}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) /* From g.f.: 1/AGM((1+x)^2, 1-6*x+x^2) */

%o {a(n)=local(A);A = 1 / agm((1+x)^2, 1-6*x+x^2 +x*O(x^n));polcoeff(A,n)}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A227846, A245930, A143007.

%Y Cf. A258053.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Aug 01 2013

%E Name changed by _Paul D. Hanna_, Sep 07 2014