A117148 Number of parts in all partitions of n in which no part occurs more than 3 times.

1, 3, 6, 8, 15, 24, 36, 50, 75, 102, 143, 197, 264, 349, 467, 606, 789, 1016, 1299, 1656, 2100, 2634, 3302, 4117, 5106, 6306, 7772, 9523, 11639, 14185, 17216, 20839, 25166, 30280, 36361, 43551, 52022, 62004, 73753, 87510, 103638, 122507, 144496, 170133

Offset

1,2

Comments

a(n) = sum(A117147(n,k), k>=1).

Links

Vaclav Kotesovec, Table of n, a(n) for n = 1..15000 (terms 1..1000 from Alois P. Heinz)

Formula

G.f.: product(1+x^j+x^(2j)+x^(3j), j=1..infinity) * sum((x^j+2x^(2j)+3x^(3j)) / (1+x^j+x^(2j)+x^(3j)), j=1..infinity).

a(n) ~ log(2) * exp(Pi*sqrt(n/2)) / (Pi * 2^(1/4) * n^(1/4)). - Vaclav Kotesovec, May 27 2018

Example

a(4) = 8 because the partitions of 4 in which no part occurs more than 3 times are [4], [3,1], [2,2] and [2,1,1] ([1,1,1,1] does not qualify) with a total of 1+2+2+3=8 parts.

Maple

g:=product(1+x^j+x^(2*j)+x^(3*j), j=1..55) *sum((x^j+2*x^(2*j)+3*x^(3*j))/ (1+x^j+x^(2*j)+x^(3*j)), j=1..55): gser:=series(g, x=0, 53): seq(coeff(gser, x^n), n=1..50);
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, [1, 0], `if`(i<1, [0, 0],
      add((l->[l[1], l[2]+l[1]*j])(b(n-i*j, i-1)), j=0..min(n/i, 3))))
    end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..50);  # Alois P. Heinz, Jan 08 2013

Mathematica

b[n_, i_] := b[n, i] = If[n == 0, {1, 0}, If[i<1, {0, 0}, Sum[Function[{l}, {l[[1]], l[[2]] + l[[1]]*j}][b[n-i*j, i-1]], {j, 0, Min[n/i, 3]}]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 50}] (* Jean-François Alcover, May 26 2015, after Alois P. Heinz *)

Crossrefs

Cf. A001935, A117147.

Column k=3 of A210485. - Alois P. Heinz, Jan 23 2013

Sequence in context: A143869 A165298 A352507 * A343272 A305595 A320687

Adjacent sequences: A117145 A117146 A117147 * A117149 A117150 A117151

Keyword

nonn

Author

Emeric Deutsch, Mar 07 2006

Status

approved