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%I #17 Sep 10 2021 22:51:15
%S 1,1,1,1,1,2,5,10,17,26,38,59,101,182,326,564,945,1566,2622,4476,7750,
%T 13455,23231,39837,68101,116611,200526,346137,598438,1034227,1785400,
%U 3080418,5317009,9187567,15893830,27515434,47647774,82513447,142902640,247553410,429020710,743846284
%N G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * x^(4*k).
%C Limit a(n)/a(n+1) = t^2 = 0.569840290998053... where t = A075778 is the positive real root of 1 - x - x^5 = 0.
%H Michael De Vlieger, <a href="/A246884/b246884.txt">Table of n, a(n) for n = 0..1000</a>
%H Hacène Belbachir and Abdelghani Mehdaoui, <a href="https://doi.org/10.2989/16073606.2020.1729269">Recurrence relation associated with the sums of square binomial coefficients</a>, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
%F G.f.: Sum_{n>=0} (2*n)!/(n!)^2 * x^(5*n) / (1 - x + x^5)^(2*n+1). - _Paul D. Hanna_, Oct 15 2014
%F G.f.: Sum_{n>=0} x^n * [Sum_{k>=0} C(n+k,k)^2 * x^(4*k)] * (1-x^4)^(2*n+1).
%F G.f.: Sum_{n>=0} x^(5*n) * [Sum_{k>=0} C(n+k,k)^2 * x^k].
%F G.f.: Sum_{n>=0} x^(5*n) * [Sum_{k=0..n} C(n,k)^2 * x^k] /(1-x)^(2n+1).
%F G.f.: exp( Sum_{n>=1} (x^n/n) * Sum_{k=0..n} C(2*n,2*k) * x^(4*k) ).
%F G.f.: exp( Sum_{n>=1} (x^n/n) * ((1+x^2)^(2*n) + (1-x^2)^(2*n))/2 ).
%F G.f.: 1 / sqrt((1 - x - 2*x^3 - x^5)*(1 - x + 2*x^3 - x^5)).
%F G.f.: 1 / sqrt((1 - x - x^5)^2 - 4*x^6).
%F G.f.: 1 / sqrt((1 - x + x^5)^2 - 4*x^5).
%F a(n) = Sum_{k=0..[n/4]} C(n-4*k, k)^2.
%e G.f.: A(x) = 1 + x + x^2 + x^3 + x^4 + 2*x^5 + 5*x^6 + 10*x^7 + 17*x^8 +...
%e where, by definition,
%e A(x) = 1 + x*(1 + x^4) + x^2*(1 + 2^2*x^4 + x^8)
%e + x^3*(1 + 3^2*x^4 + 3^2*x^8 + x^12)
%e + x^4*(1 + 4^2*x^4 + 6^2*x^8 + 4^2*x^12 + x^16)
%e + x^5*(1 + 5^2*x^4 + 10^2*x^8 + 10^2*x^12 + 5^2*x^16 + x^20) +...
%e which is also given by the series identity:
%e A(x) = 1/(1-x+x^5) + 2*x^5/(1-x+x^5)^3 + 6*x^10/(1-x+x^5)^5 + 20*x^15/(1-x+x^5)^7 + 70*x^20/(1-x+x^5)^9 + 252*x^25/(1-x+x^5)^11 + 924*x^30/(1-x+x^5)^13 +...
%e The logarithm of the g.f. begins:
%e log(A(x)) = x*(1 + x^4) + x^2*(1 + 6*x^4 + x^8)/2
%e + x^3*(1 + 15*x^4 + 15*x^8 + x^12)/3
%e + x^4*(1 + 28*x^4 + 70*x^8 + 28*x^12 + x^16)/4
%e + x^5*(1 + 45*x^4 + 210*x^8 + 210*x^12 + 45*x^16 + x^20)/5 +...
%e more explicitly,
%e log(A(x)) = x + x^2/2 + x^3/3 + x^4/4 + 6*x^5/5 + 19*x^6/6 + 36*x^7/7 + 57*x^8/8 + 82*x^9/9 + 116*x^10/10 + 199*x^11/11 + 391*x^12/12 +...
%e where the logarithmic derivative equals
%e A'(x)/A(x) = (1-x+5*x^4+6*x^5-5*x^9)/((1+x+x^2)*(1-2*x+x^2-x^3)*(1-x+2*x^3-x^5)).
%t CoefficientList[Series[1/Sqrt[(1 - x + x^5)^2 - 4 x^5], {x, 0, 41}], x] (* _Michael De Vlieger_, Sep 10 2021 *)
%o (PARI) /* By definition: */
%o {a(n)=local(A=1); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2*x^(4*k)) +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 40, print1(a(n), ", "))
%o (PARI) /* From closed formula: */
%o {a(n)=local(A=1); A= 1/sqrt((1 - x - x^5)^2 - 4*x^6 +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 40, print1(a(n), ", "))
%o (PARI) /* From a series identity: */
%o {a(n)=local(A=1+x); A=sum(m=0, n, (2*m)!/(m!)^2 * x^(5*m) / (1 - x + x^5 +x*O(x^n))^(2*m+1)); polcoeff(A, n)}
%o for(n=0, 40, print1(a(n), ", "))
%o (PARI) /* From a binomial series identity: */
%o {a(n)=local(A=1+x); A=sum(m=0, n, x^m*(1-x^4)^(2*m+1)*sum(k=0, n, binomial(m+k, k)^2*x^(4*k)) +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 40, print1(a(n), ", "))
%o (PARI) /* From a binomial series identity: */
%o {a(n)=local(A=1+x); A=sum(m=0, n\5, x^(5*m)*sum(k=0, n-4*m, binomial(m+k, k)^2*x^k) +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 40, print1(a(n), ", "))
%o (PARI) /* From a binomial series identity: */
%o {a(n)=local(A=1+x); A=sum(m=0, n\5, x^(5*m) * sum(k=0, m, binomial(m, k)^2*x^k) / (1-x +x*O(x^n))^(2*m+1) ); polcoeff(A, n)}
%o for(n=0, 40, print1(a(n), ", "))
%o (PARI) /* From exponential formula: */
%o {a(n)=local(A=1); A=exp(sum(m=1, n, sum(k=0, m, binomial(2*m, 2*k)*x^(4*k)) * x^m/m) +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 40, print1(a(n), ", "))
%o (PARI) /* From exponential formula: */
%o {a(n)=local(A=1); A=exp(sum(m=1, n, ((1+x^2)^(2*m) + (1-x^2)^(2*m))/2 * x^m/m) +x*O(x^n)); polcoeff(A, n)}
%o for(n=0, 40, print1(a(n), ", "))
%o (PARI) /* From formula for a(n): */
%o {a(n)=sum(k=0, n\4, binomial(n-4*k, k)^2)}
%o for(n=0, 40, print1(a(n), ", "))
%Y Cf. A075778, A181665, A246840, A246883, A248193.
%K nonn
%O 0,6
%A _Paul D. Hanna_, Sep 06 2014
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