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%I #37 Jan 08 2022 03:43:37
%S 1,14,786,61340,5562130,549676764,57440496036,6242164112184,
%T 698300344311570,79881547652046140,9301427008157320036,
%U 1098786921802152516024,131361675994216221116836,15863471168011822803270200,1932252897656224864335299400
%N de Bruijn's S(4,n).
%C a(n) is divisible by (n+1). Prime p divides a(p-1). Prime p>2 divides all a(n) from a((p+1)/2) to a(p-1). - _Alexander Adamchuk_, Jul 05 2006
%D N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.
%H Seiichi Manyama, <a href="/A050983/b050983.txt">Table of n, a(n) for n = 0..471</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BinomialSums.html">Binomial Sums</a>
%F a(n) = Sum_{k=-n..+n} (-1)^k*C(2*n,n+k)^4. - _Benoit Cloitre_, Mar 02 2005
%F a(n) = (-1)^n * HypergeometricPFQ[ {-2n, -2n, -2n, -2n}, {1, 1, 1}, -1]. - _Alexander Adamchuk_, Jul 05 2006
%F E.g.f.: Sum(n>=0,I^n*x^n/n!^4) * Sum(n>=0,(-I)^n*x^n/n!^4) = Sum(n>=0,a(n)*x^(2*n)/n!^4) where I^2=-1. - _Paul D. Hanna_, Dec 21 2011
%F a(n) ~ 0.125 k^(8n+3)/(Pi*n)^(3/2) where k = 2 cos(Pi/8) = A179260. This formula is due to de Bruijn 1958. - _Charles R Greathouse IV_, Dec 28 2011
%F Recurrence: a(0) = 1, a(1) = 14, 4 * (n + 1) * (2*n + 1)^3 * (48*n^2 + 162*n + 137) * a(n) + (n + 2)^3 * (2*n + 3) * (48*n^2 + 66*n + 23) * a(n+2) = 2 * (4 * (n + 1)^2 * (2*n + 3)^2 * (408*n^2 + 969*n + 431) - (n + 1) * (2*n + 3) * (69*n + 31) + 57*n + 92) * a(n+1). - _Vladimir Reshetnikov_, Sep 26 2016
%t Sum[ (-1)^(k+n)Binomial[ 2n, k ]^4, {k, 0, 2n} ]
%t RecurrenceTable[{a[0] == 1, a[1] == 14, 4 (n + 1) (2 n + 1)^3 (48 n^2 + 162 n + 137) a[n] + (n + 2)^3 (2 n + 3) (48 n^2 + 66 n + 23) a[n + 2] == 2 (4 (n + 1)^2 (2 n + 3)^2 (408 n^2 + 969 n + 431) - (n + 1) (2 n + 3) (69 n + 31) + 57 n + 92) a[n + 1]}, a[n], {n, 0, 20}] (* _Vladimir Reshetnikov_, Sep 26 2016 *)
%o (PARI) a(n)=sum(k=0,2*n,(-1)^(k+n)*binomial(2*n,k)^4) \\ _Charles R Greathouse IV_, Dec 28 2011
%Y Cf. A000984, A006480, A050984, A227357, A249332.
%K nonn
%O 0,2
%A _Eric W. Weisstein_
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