A050983 de Bruijn's S(4,n).

1, 14, 786, 61340, 5562130, 549676764, 57440496036, 6242164112184, 698300344311570, 79881547652046140, 9301427008157320036, 1098786921802152516024, 131361675994216221116836, 15863471168011822803270200, 1932252897656224864335299400

Offset: 0

Author

Eric W. Weisstein

Keywords

• nonn
• easy

Comments

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

References

• N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..471
• T. Amdeberhan, De Bruijn's sequence is odd iff n = 2m - 1: Part I, MathOverflow 383035.
• Eric Weisstein's World of Mathematics, Binomial Sums

Crossrefs

Cf. A000984, A006480, A050984, A099601, A177322, A227357, A249332.

Programs

Mathematica

Sum[ (-1)^(k+n)Binomial[ 2n, k ]^4, {k, 0, 2n} ]
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 *)

PARI

a(n)=sum(k=0,2*n,(-1)^(k+n)*binomial(2*n,k)^4) \\ Charles R Greathouse IV, Dec 28 2011

Formula

a(n) = Sum_{k=-n..+n} (-1)^k*C(2*n,n+k)^4. - Benoit Cloitre, Mar 02 2005

a(n) = (-1)^n * HypergeometricPFQ[ {-2n, -2n, -2n, -2n}, {1, 1, 1}, -1]. - Alexander Adamchuk, Jul 05 2006

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

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

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

From Peter Bala, Nov 02 2024; (Start)

a(n) = 1/n * Sum_{k = 0..2*n} (-1)^(n+k) * k * binomial(2*n, k)^4 for n >= 1.

a(n) = binomial(2*n, n) * Sum_{k = 0..n} binomial(2*n, n+k)^2 * binomial(2*n+k,k) = binomial(2*n, n) * Sum_{k = 0..n} (-1)^(n+k) * binomial(2*n, n+k) * binomial(2*n+k, k)^2. (End)