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A116680
Number of even parts in all partitions of n into distinct parts.
8
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
OFFSET
0,6
LINKS
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
FORMULA
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
EXAMPLE
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]
MAPLE
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
MATHEMATICA
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 *)
PROG
(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
KEYWORD
nonn
AUTHOR
Emeric Deutsch, Feb 22 2006
STATUS
approved