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a(n) = A006442(n)^2.
5

%I #27 May 19 2023 15:19:00

%S 1,25,1369,93025,6974881,553425625,45558768025,3848757330625,

%T 331434586569025,28966516730025625,2561512789823546329,

%U 228690489716580520225,20579914168308199841761,1864413002713001259355225,169871744046114667846619929,15554069096581207471331850625

%N a(n) = A006442(n)^2.

%C In general, we have the binomial identity:

%C if b(n) = Sum_{k=0..n} t^k * C(2*k, k) * C(n+k, n-k),

%C then b(n)^2 = Sum_{k=0..n} (t^2+t)^k * C(2*k, k)^2 * C(n+k, n-k),

%C where the g.f. of b(n) is 1/sqrt(1 - (4*t+2)*x + x^2),

%C and the g.f. of b(n)^2 is 1 / AGM(1-x, sqrt((1+x)^2 - (4*t+2)^2*x)), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.

%C Note that the g.f. of A006442 is 1/sqrt(1 - 10*x + x^2).

%C Limit_{n -> oo} a(n+1)/a(n) = (5 + 2*sqrt(6))^2 = 49 + 20*sqrt(6).

%H Seiichi Manyama, <a href="/A243943/b243943.txt">Table of n, a(n) for n = 0..500</a>

%F G.f.: 1 / AGM(1-x, sqrt(1-98*x+x^2)). - _Paul D. Hanna_, Aug 30 2014

%F a(n) = Sum_{k=0..n} 6^k * C(2*k, k)^2 * C(n+k, n-k).

%F a(n)^(1/2) = Sum_{k=0..n} 2^k * C(2*k, k) * C(n+k, n-k).

%F a(n) ~ (5+2*sqrt(6))^(2*n+1) / (4*Pi*sqrt(6)*n). - _Vaclav Kotesovec_, Sep 28 2019

%e G.f.: A(x) = 1 + 9*x + 169*x^2 + 3969*x^3 + 103041*x^4 + 2832489*x^5 +...

%t Table[Sum[6^k * Binomial[2*k, k]^2 * Binomial[n+k, n-k], {k,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Sep 28 2019 *)

%t LegendreP[Range[0,40], 5]^2 (* _G. C. Greubel_, May 17 2023 *)

%o (PARI) {a(n) = sum(k=0, n, 6^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}

%o for(n=0, 20, print1(a(n), ", "))

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

%o for(n=0, 20, print1(a(n), ", "))

%o (PARI) /* Using AGM: */

%o {a(n)=polcoeff( 1 / agm(1-x, sqrt((1+x)^2 - 10^2*x +x*O(x^n))), n)}

%o for(n=0, 20, print1(a(n), ", ")) \\ _Paul D. Hanna_, Aug 30 2014

%o (Magma) [Evaluate(LegendrePolynomial(n), 5)^2 : n in [0..40]]; // _G. C. Greubel_, May 17 2023

%o (SageMath) [gen_legendre_P(n,0,5)^2 for n in range(41)] # _G. C. Greubel_, May 17 2023

%Y Sequences of the form LegendreP(n, 2*m+1)^2: A000012 (m=0), A243949 (m=1), this sequence (m=2), A243944 (m=3), A243007 (m=4).

%Y Cf. A006442.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Aug 17 2014