A116665 Total number of parts that appear exactly once in the partitions of n into odd parts.
0, 1, 0, 1, 2, 2, 3, 4, 6, 7, 10, 12, 16, 20, 25, 31, 39, 47, 58, 71, 85, 103, 124, 148, 176, 210, 248, 293, 345, 405, 474, 555, 645, 751, 872, 1009, 1166, 1346, 1549, 1781, 2044, 2341, 2678, 3060, 3488, 3973, 4520, 5132, 5822, 6597, 7464, 8436, 9525, 10740
Offset: 0
Keywords
Examples
a(8) = 6 because in the partitions of 8 into odd parts, namely, [(7),(1)], [(5),(3)], [(5),1,1,1], [3,3,1,1], [(3),1,1,1,1,1] and [1,1,1,1,1,1,1,1], we have 6 parts that appear exactly once (shown between parentheses).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Crossrefs
Cf. A116664.
Programs
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Maple
f:=x*(1-x+x^2)/(1-x^4)/product(1-x^(2*j-1),j=1..40): fser:=series(f,x=0,61): seq(coeff(fser,x,n),n=0..57); # second Maple program: b:= proc(n, i) option remember; `if`(n=0, [1, 0], `if`(i<1, 0, add((p-> p+`if`(j=1, [0, p[1]], 0)) (b(n-i*j, i-2)), j=0..n/i))) end: a:= n-> b(n, n-irem(n+1, 2))[2]: seq(a(n), n=0..60); # Alois P. Heinz, Mar 16 2014 -
Mathematica
b[n_, i_] := b[n, i] = If[n == 0, {1, 0}, If[i<1, {0, 0}, Sum[Function[{p}, p + If[j == 1, {0, p[[1]]}, 0]][b[n-i*j, i-2]], {j, 0, n/i}]]]; a[n_] := b[n, n - Mod[n+1, 2]][[2]]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, May 13 2015, after Alois P. Heinz *) nmax = 60; CoefficientList[Series[x*(1-x+x^2)/(1-x^4) * Product[1/(1-x^(2*k-1)), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Mar 07 2016 *)
Formula
G.f.: x(1-x+x^2)/[(1-x^4)product(1-x^(2j-1), j=1..infinity)].
a(n) ~ 3^(1/4) * exp(sqrt(n/3)*Pi) / (8*Pi*n^(1/4)). - Vaclav Kotesovec, Mar 07 2016
Comments