A264592 Let G[1](q) denote the g.f. for A003114 and G[2](q) the g.f. for A003106 (the two Rogers-Ramanujan identities). For i>=3, let G[i](q) = (G[i-1](q)-G[i-2](q))/q^(i-2). Sequence gives coefficients of G[5](q).

1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 7, 8, 9, 11, 12, 14, 16, 18, 20, 23, 25, 29, 32, 36, 40, 46, 50, 57, 63, 71, 78, 88, 96, 108, 119, 132, 145, 162, 177, 197, 216, 239, 262, 290, 317, 350, 383, 421, 460, 507, 552, 606, 661, 724, 789, 864, 939, 1027, 1117

Offset: 0

Author

N. J. A. Sloane, Nov 19 2015

Keywords

• nonn

Comments

It is conjectured that G[i](q) = 1 + O(q^i) for all i.

From Wolfdieter Lang, Nov 03 2016: (Start)

The generalized Rogers-Ramanujan [R-R] series G[i](q) of Andrews and Baxter [A-B] have a standard combinatorial interpretation of the Schur and MacMahon type (see Hardy [H] and Hardy-Wright [H-W] for the original [R-R] case) inferred from the formula G[i](q) = Sum_{m>=0} q^(m*(m+i-1))/Product_{j=1..m} (1 - q^j) ([A-B], eq. (5.1)). Define GI_k(q) = G[2*k+1](q) and GII_k(q) = G[2*k](q), for k = 0, 1,..., and prove the two formulas I(m,k): m*(m+2*k) = Sum_{j = 1..2*m-1} (2*k + j), and II(m,k): m*(m+2*k+1) = Sum_{j = 1..m} (2*(k + j)) for fixed positive m by induction over k = 0, 1, ... . For GI_k(q) define the special m-part partition SPI(m,k) = [2*k+2*m-1,2*k+2*m-3,...,2*k+1] of m*(m+2*k), and for GII_k(q) the special m-part partition SPII(m,k) [2*(k+1),2*(k+2),...,2*(k+1))] of m*(m+2*k+1).

Then GI_k(q) = 1 + Sum_{n >=1} aI(k,n)*q^n with aI(k,n) the number of partitions of n without parts 1, 2, ..., 2*k, and the parts differ by at least 2. GII_k(q) = 1 + Sum_{n >=1} aII(k,n)*q^n with aII(k,n) the number of partitions of n without parts 1, 2, ..., 2*k+1, and the parts differ by at least 2. The proof can be directly adapted from the one given in [H] or [H-W] for k=1.

For the partitions of n generated by GI_k(q) one needs the maximal part number MmaxI(k,n) = floor(-k + sqrt(k^2 + n)). For the GII_k(q) case MmaxII(k,n) = floor(-(2*k+1) + sqrt((2*k+1)^2 + 4*n)).

The present sequence is aI(2,n), in [A-B] notation generated by G[5](q), giving the number of partitions of n without parts 1, 2, 3 and 4, and the parts differ by at least 2.

(End)

Examples

			From _Wolfdieter Lang_, Nov 03 2016: (Start)
a(5) = 1 because the only partition of n = 5 without parts 1, 2, 3 and 4, and parts differing by at least 2 is [5].
a(12) = 2 from the two partitions [12] and [7,5] of n = 12.
a(18) = 5 from the five partitions [18], [13,5], [12,6], [11,7], [10,8] of n = 18.
(End)
		

References

• G. H. Hardy, Ramanujan, AMS Chelsea Publ., Providence, RI, 2002, pp. 91-92.
• G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth ed., Clarendon Press, Oxford, 2003, pp. 290-291.

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
• George E. Andrews; R. J. Baxter, A motivated proof of the Rogers-Ramanujan identities, Amer. Math. Monthly 96 (1989), no. 5, 401-409.
• Shashank Kanade, Some results on the representation theory of vertex operator algebras and integer partition identities, PhD Handout, Math. Dept., Rutgers University, April 2015.
• Shashank Kanade, Some results on the representation theory of vertex operator algebras and integer partition identities, PhD Dissertation, Math. Dept., Rutgers University, April 2015.

Crossrefs

For the generalized Rogers-Ramanujan series G[0], G[1], G[2], G[3], G[4], G[5], G[6], G[7], G[8] see A003113, A003114, A003106, A006141, A264591, A264592, A264593, A264594, A264595.

Programs

Mathematica

nmax = 100; CoefficientList[Series[Sum[x^(k*(k+4))/Product[1-x^j, {j, 1, k}], {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 24 2016 *)

Formula

From Wolfdieter Lang, Nov 03 2016: (Start)

G.f.: G[5](q) = GI_2(q) = Sum_{m>=0} q^(m*(m+4))/Product_{j=1..m} (1 - q^j).

See [A-B], eq. (5.1) for i=5.

a(0) = 1 and a(n) gives the number of partitions of n without part 1 and 2, the parts differing by at least 2.

G.f.: Sum_{m=0} ((-1)^m*(1 - q^(m+1))*(1 - q^(m+2))*(1 - q^(m+3))*(1 - q^(2*(m+2))) * q^(5*(n+3)*n/2)) / Product_{j>=1} (1 - q^j). See [A-B], eq. (3.8) for i=5. (End)

a(n) ~ exp(2*Pi*sqrt(n/15)) / (2 * 3^(1/4) * sqrt(5) * phi^(7/2) * n^(3/4)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Dec 24 2016