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a(n) = 2^n*P_n(5), 2^n times the Legendre polynomial of order n at 5.
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%I #44 May 21 2023 22:09:32

%S 1,10,148,2440,42256,752800,13660480,251113600,4660568320,87140108800,

%T 1638884021248,30970912737280,587599919386624,11185644310405120,

%U 213540626285805568,4086692369433395200,78378887309200261120

%N a(n) = 2^n*P_n(5), 2^n times the Legendre polynomial of order n at 5.

%C Central coefficients of (1 + 10*x + 24*x^2)^n. 2^n*LegendreP(n,k) yields the central coefficients of (1 + 2*k*x + (k^2-1)*x^2)^n, with g.f. 1/sqrt(1 - 4*k*x + 4*x^2).

%H Vincenzo Librandi, <a href="/A098270/b098270.txt">Table of n, a(n) for n = 0..200</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.

%H Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Szalay/szalay42.html">Diagonal Sums in the Pascal Pyramid, II: Applications</a>, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LegendrePolynomial.html">Legendre Polynomial.</a>

%F G.f.: 1/sqrt(1-20*x+4*x^2).

%F a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n,k)*binomial(2*(n-k), n)*5^(n-2*k).

%F D-finite with recurrence: n*a(n) +10*(1-2*n)*a(n-1) +4*(n-1)*a(n-2) = 0. - _R. J. Mathar_, Sep 26 2012

%F a(n) ~ sqrt(72+30*sqrt(6))*(10+4*sqrt(6))^n/(12*sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 14 2012

%F a(n) = A059473(n,n). - _Alois P. Heinz_, Oct 05 2017

%F From _Peter Bala_, Nov 28 2021: (Start)

%F a(n) = (1/3)*Sum_{k >= n} binomial(k,n)^2*(2/3)^k.

%F a(n) = (4^n)*Sum_{k = 0..n} binomial(n,k)^2*(3/2)^k.

%F a(n) = (1/3)*(2/3)^n*hypergeometric2F1([n+1, n+1], [1], 2/3).

%F a(n) = (4^n)*hypergeometric2F1([-n, -n], [1], 3/2)

%F a(n) = [x^n] ((2*x - 2)*(3 - 2*x))^n.

%F a(n) = (2^n)*A006442(n). (End)

%t Table[SeriesCoefficient[1/Sqrt[1-20*x+4*x^2],{x,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 14 2012 *)

%t Table[2^n*LegendreP[n, 5], {n,0,40}] (* _G. C. Greubel_, May 21 2023 *)

%o (PARI) a(n)=pollegendre(n,5)<<n \\ _Charles R Greathouse IV_, Oct 25 2011

%o (Sage)

%o def A098270(n): return 2^n*gen_legendre_P(n, 0, 5)

%o [A098270(n) for n in (0..16)] # _Peter Luschny_, Oct 14 2012

%o (Magma) [2^n*Evaluate(LegendrePolynomial(n), 5): n in [0..40]]; // _G. C. Greubel_, May 21 2023

%Y Sequences of the form 2^n*LegendreP(n, 2*m+1): A000079 (m=0), A084773 (m=1), this sequence (m=2).

%Y Cf. A006442, A059473, A069835.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Sep 01 2004