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Number of distinct ways to express n as a partial sums sum of 1 + [1,3] + [1,5] + [1,7] + [1,9] + ... that equal n. E.g., a(6)=2 because we have 6 = 1+1+1+1+1+1 = 1+1+3+1+1. - Jon Perry, Jan 01 2004
Number of different distinct partial sums of 1 + [1,3] + [1,5] + [1,7] + [1,9] + ... that equal n. E.g., a(6)=2 because we have 6 = 1+1+1+1+1+1 = 1+1+3+1. - Jon Perry, Jan 01 2004
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Also number of partitions of n such that the largest part occurs exactly once and all the other parts occur exactly twice. Example: a(9)=4 because we have [9], [7,1,1], [5,2,2] and [3,2,2,1,1]. - Emeric Deutsch, Mar 08 2006
Coefficients of the '3rd -order' mock theta function psi(q)
Number of different partial sums of 1+[1,3]+[1,5]+[1,7]+[1,9]+... E.g. , a(6)=2 because we have 6=1+1+1+1+1+1=1+1+3+1. - Jon Perry, Jan 01 2004
Also number of partitions of n such that the largest part occurs exactly once and all the other parts occur exactly twice. Example: a(9)=4 because we have [9],[7,1,1],[5,2,2] and [3,2,2,1,1]. - Emeric Deutsch, Mar 08 2006
For _Emeric Deutsch_'s comment above, (1) this appears to be an alternately equal case of A122130, (2) the ordered version (compositions) is A239327, (3) allowing any length gives A351006, (4) the even-length version is A351007. - Gus Wiseman, Feb 25 2022
Srinivasa Ramanujan, Collected Papers, Chelsea, New York, 1962, pp. 354-355.
Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, p. 31.
G.f.: psi(q) = sum(Sum_{n>=1, } q^(n^2) / ( (1-q)*(1-q^3)*...*(1-q^(2*n-1)) ) ).
G.f.: sum(Sum_{k>=1, } q^k*prod(Product_{j=1..k-1, } (1+q^(2*j) ) ), (see the Fine reference, p. 58, Eq. (26,53)). - Emeric Deutsch, Mar 08 2006
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