A195012 Sum of positive cranks minus the sum of positive ranks of all partitions of n.

1, 1, 1, 2, 2, 4, 5, 7, 10, 13, 17, 24, 31, 40, 53, 69, 88, 113, 144, 183, 231, 290, 362, 453, 563, 696, 859, 1058, 1296, 1587, 1935, 2354, 2856, 3458, 4175, 5033, 6051, 7259, 8692, 10390, 12391, 14756, 17537, 20808, 24648, 29151, 34417, 40581, 47773, 56158

Offset: 1

Author

Omar E. Pol, Jan 10 2012

Keywords

• nonn

Comments

It appears this is also the column 0 of triangle A195011 without the first one (see the Andrews-Garvan-Liang paper, page 16).

Is this also the ospt(n) function mentioned in the Andrews-Chan-Kim paper? Is A115995(n) the first crank momment? Is A209616(n) the first rank moment? - Omar E. Pol, Apr 07 2012

From Jeremy Lovejoy, Oct 14 2022: (Start)

a(n) is also the number of rank 0 strongly unimodal sequences of size n. A strongly unimodal sequence is a sequence of positive integers which are strictly increasing up to a point (the peak) and then strictly decreasing thereafter. The size is the sum of all of the parts and the rank is the number of parts to the left of the peak minus the number of parts to the right of the peak.

For example, there are 10 strongly unimodal sequences of size 6: (6), (1,5), (5,1), (2,4), (4,2), (1,4,1), (3,2,1), (1,2,3), (1,3,2), and (2,3,1). The sequences (6), (1,4,1), (1,3,2), and (2,3,1) have rank 0, and so a(6) = 4. (End)

Examples

			For n = 6 we have:
------------------------------------------------
Partitions
of 6                  Crank             Rank
------------------------------------------------
6                           6        6 - 1 =  5
3+3                         3        3 - 2 =  1
4+2                         4        4 - 2 =  2
2+2+2                       2        2 - 3 = -1
5+1                1 - 1 =  0        5 - 2 =  3
3+2+1              2 - 1 =  1        3 - 3 =  0
4+1+1              1 - 2 = -1        4 - 3 =  1
2+2+1+1            0 - 2 = -2        2 - 4 = -2
3+1+1+1            0 - 3 = -3        3 - 4 = -1
2+1+1+1+1          0 - 4 = -4        2 - 5 = -3
1+1+1+1+1+1        0 - 6 = -6        1 - 6 = -5
------------------------------------------------
The sum of positive cranks is 6+3+4+2+1 = 16 and the sum of positive ranks is 5+1+2+3+1 = 12 therefore a(6) = 16 - 12 = 4.
		

Links

• G. E. Andrews, S. H. G. Chan, and B. Kim, The odd moments of ranks and cranks (This is the function C_1 - R_1 of that paper), Journal of Combinatorial Theory, Series A, Volume 120, Issue 1, January 2013, Pages 77-91.
• G. E. Andrews, F. G. Garvan, and J. Liang, Combinatorial interpretation of congruences for the spt-function, The Ramanujan Journal, December 2012, Volume 29, Issue 1-3, pp 321-338.
• A. O. L. Atkin and F. G. Garvan, Relations between the ranks and cranks of partitions, arXiv:math/0208050 [math.NT], 2002.
• K. Bringmann, C. Jennings-Shaffer, K. Mahlburg, and R. Rhoades, Peak positions of strongly unimodal sequences, Trans. Amer. Math. Soc. 372 (2019), 7087-7109.
• Frank Garvan, Dyson's rank function and Andrews's SPT-function [Broken link?]
• K. Hikami and J. Lovejoy, Torus knots and quantum modular forms, Res. Math. Sci. 2, Article 2 (2015).

Crossrefs

Cf. A092269, A115995, A195011, A208482, A209616.

Cf. A059618.

Programs

Maple

# Based on Theorem 1 of Andrews-Chan-Kim:
M:=101;
qinf:=mul(1-q^i,i=1..M);
qinf:=series(qinf,q,M);
C1:=add((-1)^(n+1)*q^(n*(n+1)/2)/(1-q^n),n=1..M);
C1:=series(C1/qinf,q,M);
R1:=add((-1)^(n+1)*q^(n*(3*n+1)/2)/(1-q^n),n=1..M);
R1:=series(R1/qinf,q,M);
series(C1-R1,q,M);
seriestolist(%); # N. J. A. Sloane, Sep 04 2012

Mathematica

M = 101;
qinf = Product[1-q^i, {i, 1, M}];
qinf = Series[qinf, {q, 0, M}];
C1 = Sum[(-1)^(n+1) q^(n(n+1)/2)/(1-q^n), {n, 1, M}];
C1 = Series[C1/qinf, {q, 0, M}];
R1 = Sum[(-1)^(n+1) q^(n(3n+1)/2)/(1-q^n), {n, 1, M}];
R1 = Series[R1/qinf, {q, 0, M}];
CoefficientList[Series[C1-R1, {q, 0, M}], q] // Rest (* Jean-François Alcover, Aug 18 2018, translated from Maple *)

Formula

a(n) = A115995(n) - A209616(n).

From Jeremy Lovejoy, Oct 14 2022: (Start)

G.f.: (1/Product_{n>=1}(1-x^n))*Sum_{n>=1} x^(n*(n+1)/2)*(-1)^(n-1)*(1-x^(n^2))/(1-x^n).

G.f.: (1/Product_{n>=1}(1-x^n))*Sum_{n,r>=0} (-1)^(n+r)*x^(n*(3*n+5)/2+2*n*r+r*(r+3)/2). (End)

a(n) ~ exp(Pi*sqrt(2*n/3)) / (16*sqrt(3)*n). - Vaclav Kotesovec, Jul 06 2025

Extensions

New name, example and more terms from Omar E. Pol, Apr 06 2012

More terms a(44)-a(50) from Alois P. Heinz, Apr 08 2012