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The g.f. for "number of k's" is (1/2)*(x^k/(1+x^k))*prod(Product_{n>=1,} 1 + x^n) - (1/2)*(x^k/(1-x^k))*prod(Product_{n>=1,} 1 - x^n).

0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 2, 2, 2, 3, 4, 5, 6, 7, 9, 10, 12, 15, 17, 20, 24, 27, 32, 38, 43, 50, 59, 67, 77, 90, 102, 117, 135, 153, 175, 200, 226, 257, 292, 330, 373, 422, 475, 535, 603, 677, 760, 853, 955, 1069, 1196, 1336, 1491, 1663, 1853, 2063, 2295

Andrew Howroyd, <a href="/A238218/b238218.txt">Table of n, a(n) for n = 0..1000</a>

a(n) =sum_ Sum_{j=1..round(n/6)} A067659(n-(2*j-1)*3) -sum_ Sum_{j=1..floor(n/6))} A067661(n-6*j).

G.f.: (1/2)*(x^3/(1+x^3))*prod(Product_{n>=1,} 1 + x^n) - (1/2)*(x^3/(1-x^3))*prod(Product_{n>=1,} 1 - x^n).

a(13) = 3 because the partitions in question are: 10+3, 7+3+2+1, 5+4+3+1.

(PARI) seq(n)={my(A=O(x^(n-2))); Vec(x*(eta(x^2 + A)/(eta(x + A)*(1+x^3)) - eta(x + A)/(1-x^3))/2, -(n+1))} \\ Andrew Howroyd, May 01 2020

Column k=3 of A238451.

Terms a(51) and beyond from Andrew Howroyd, May 01 2020

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allocated for Mircea MercaThe total number of 3's in all partitions of n into an even number of distinct parts.

0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 2, 2, 2, 3, 4, 5, 6, 7, 9, 10, 12, 15, 17, 20, 24, 27, 32, 38, 43, 50, 59, 67, 77, 90, 102, 117, 135, 153, 175, 200, 226, 257, 292, 330, 373, 422, 475, 535, 603, 677, 760

0,11

The g.f. for "number of k's" is (1/2)*x^k/(1+x^k)*prod(n>=1,1+x^n)-(1/2)*x^k/(1-x^k)*prod(n>=1,1-x^n).

a(n)=sum_{j=1..round(n/6)}A067659(n-(2*j-1)*3)-sum_{j=1..floor(n/6))}A067661(n-6*j).

G.f.: (1/2)*x^3/(1+x^3)*prod(n>=1,1+x^n)-(1/2)*x^3/(1-x^3)*prod(n>=1,1-x^n).

a(13)=3 because the partitions in question are: 10+3, 7+3+2+1, 5+4+3+1.

Cf. A067659, A067661.

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Mircea Merca, Feb 20 2014

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