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A195012
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Sum of positive cranks minus the sum of positive ranks of all partitions of n.
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10
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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
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OFFSET
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1,4
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COMMENTS
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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
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)
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LINKS
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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.
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FORMULA
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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)
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EXAMPLE
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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.
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MAPLE
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# 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);
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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New name, example and more terms from Omar E. Pol, Apr 06 2012
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STATUS
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approved
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