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A116680
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Number of even parts in all partitions of n into distinct parts.
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8
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0, 0, 1, 1, 1, 2, 4, 5, 5, 8, 11, 14, 18, 23, 29, 37, 44, 55, 69, 83, 102, 124, 148, 178, 213, 253, 300, 356, 421, 494, 582, 680, 793, 926, 1074, 1246, 1446, 1668, 1922, 2215, 2545, 2918, 3345, 3823, 4366, 4982, 5668, 6445, 7321, 8300, 9401, 10639, 12021, 13566
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OFFSET
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0,6
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LINKS
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Vaclav Kotesovec, Table of n, a(n) for n = 0..10000
D. Herden, M. R. Sepanski, J. Stanfill, C. C. Hammon, J. Henningsen, H. Ickes, J. M. Menendez, T. Poe, I. Ruiz, and E. L. Smith, Counting the parts divisible by k in all the partitions of n whose parts have multiplicity less than k, arXiv:2010.02788 [math.CO], 2020. See also Integers (2022) Vol. 22, #A49.
Runqiao Li, Andrew Y. Z. Wang, On the combinatorics of the number of even parts in all partitions with distinct parts, The Raman. J. 56 (2021) 712-727
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FORMULA
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a(n) = Sum_{k >= 0} k*A116679(n,k).
G.f.: Product_{j >= 1} (1+x^j) * Sum_{k >= 1} (x^(2*k)/(1+x^(2*k)).
For n > 0, a(n) = A015723(n) - A116676(n). - Vaclav Kotesovec, May 26 2018
a(n) ~ 3^(1/4) * log(2) * exp(Pi*sqrt(n/3)) / (4*Pi*n^(1/4)). - Vaclav Kotesovec, May 26 2018
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EXAMPLE
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a(9)=8 because in the partitions of 9 into distinct parts, namely, [9], [8,1], [7,2], [6,3], [6,2,1], [5,4], [5,3,1], and [4,3,2], we have a total of 8 even parts. [edited by Rishi Advani, Jun 07 2019]
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MAPLE
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f:=product(1+x^j, j=1..70)*sum(x^(2*j)/(1+x^(2*j)), j=1..40): fser:=series(f, x=0, 65): seq(coeff(fser, x, n), n=0..60);
# second Maple program:
b:= proc(n, i) option remember; `if`(i*(i+1)/2<n, 0, `if`(n=0, [1, 0],
b(n, i-1)+(p-> p+`if`(i::odd, 0, [0, p[1]]))(b(n-i, min(n-i, i-1)))))
end:
a:= n-> b(n$2)[2]:
seq(a(n), n=0..60); # Alois P. Heinz, May 24 2022
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MATHEMATICA
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With[{m = 25}, CoefficientList[Series[Product[1+x^j, {j, 1, 4*m}]* Sum[x^(2*k)/(1+x^(2*k)), {k, 1, 2*m}], {x, 0, 3*m}], x]] (* G. C. Greubel, Jun 07 2019 *)
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PROG
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(PARI) my(m=25); my(x='x+O('x^(3*m))); concat([0, 0], Vec( prod(j=1, 4*m, 1+x^j)*sum(k=1, 2*m, x^(2*k)/(1+x^(2*k))) )) \\ G. C. Greubel, Jun 07 2019
(Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), 3*m); [0, 0] cat Coefficients(R!( (&*[1+x^j: j in [1..4*m]])*(&+[x^(2*k)/(1+x^(2*k)): k in [1..2*m]]) )); // G. C. Greubel, Jun 07 2019
(Sage)
m = 25
R = PowerSeriesRing(ZZ, 'x')
x = R.gen().O(3*m)
s = product(1+x^j for j in (1..4*m))*sum(x^(2*k)/(1+x^(2*k)) for k in (1..2*m))
[0, 0] + s.coefficients() # G. C. Greubel, Jun 07 2019
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CROSSREFS
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Cf. A116679, A305121, A305122.
Cf. A305082, A015723, A090867, A067588, A116676.
Sequence in context: A026404 A327326 A170882 * A138083 A181524 A240568
Adjacent sequences: A116677 A116678 A116679 * A116681 A116682 A116683
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KEYWORD
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nonn
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AUTHOR
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Emeric Deutsch, Feb 22 2006
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STATUS
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approved
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