cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A108962 Number of partitions that are "3-close" to being self-conjugate.

Original entry on oeis.org

1, 1, 2, 3, 5, 5, 9, 11, 16, 20, 28, 34, 47, 57, 75, 92, 119, 143, 183, 220, 277, 332, 412, 491, 605, 718, 874, 1036, 1252, 1475, 1772, 2082, 2483, 2909, 3450, 4027, 4755, 5533, 6499, 7545, 8826, 10213, 11904, 13741, 15955, 18372, 21262, 24422
Offset: 0

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Author

John McKay (mckay(AT)cs.concordia.ca), Jul 22 2005

Keywords

Comments

Let (a1,a2,a3,...ad:b1,b2,b3,...bd) be the Frobenius symbol for a partition pi of n. Then pi is m-close to being self-conjugate if for all k, |ak-bk| <= m.

Examples

			1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 5*x^5 + 9*x^6 + 11*x^7 + 16*x^8 + 20*x^9 + 28*x^10 + ...
1/q + q^23 + 2*q^47 + 3*q^71 + 5*q^95 + 5*q^119 + 9*q^143 + 11*q^167 + 16*q^191 + ...
		

References

  • D. M. Bressoud, Analytic and combinatorial generalizations of the Rogers-Ramanujan identities, Mem. Amer. Math. Soc. 24 (1980), no. 227, 54 pp.

Crossrefs

Cf. A000700 for m=0 (self-conjugate), A070047 for m=1, A108961 for m=2, A271661 for m=4, A280937 for m=5, A280938 for m=6.

Programs

  • Mathematica
    nmax = 50; CoefficientList[Series[Product[(1 - x^(5*k))^2 / ((1 - x^k) * (1 - x^(10*k))), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Nov 13 2016 *)
  • PARI
    {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^5 + A)^2 / (eta(x + A) * eta(x^10 + A)), n))} /* Michael Somos, Jun 08 2012 */

Formula

Define the Dedekind eta function = q^1/24. Product(1-q^k), k >=1. Then the number of m-close partitions is q^(1/24).(m+2)^2/(1.(2m+4)) (where m denotes eta(q^m)).
From Vaclav Kotesovec, Nov 13 2016, extended Jan 11 2017: (Start)
a(n, m) ~ exp(Pi*sqrt((2*m+1)*n/(3*(m+2)))) * (2*m+1)^(1/4) / (2*3^(1/4)*(m+2)^(3/4)*n^(3/4)).
For m=3, a(n) ~ 7^(1/4) * exp(sqrt(7*n/15)*Pi) / (2*3^(1/4)*5^(3/4)*n^(3/4)) * (1 -(3*sqrt(15)/(8*Pi*sqrt(7)) + Pi*sqrt(7)/(48*sqrt(15)))/sqrt(n) + (7*Pi^2/69120 - 225/(896*Pi^2) + 5/128)/n).
(End)
a(n) ~ 2*Pi * BesselI(1, Pi/6 * sqrt((24*n-1)*7/10)) / (5*sqrt((24*n-1)/7)). - Vaclav Kotesovec, Jan 11 2017