OFFSET
0,6
COMMENTS
Number of up-steps (== number of parts - 1) in all partitions of n into distinct parts (represented as increasing lists), see example. - Joerg Arndt, Sep 03 2014
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..2000 (first 201 terms from Vincenzo Librandi)
FORMULA
EXAMPLE
a(8) = 7 because in the 6 partitions of 8 into distinct parts
1: [ 1 2 5 ]
2: [ 1 3 4 ]
3: [ 1 7 ]
4: [ 2 6 ]
5: [ 3 5 ]
6: [ 8 ]
there are 2+2+1+1+1+0 = 7 up-steps. - Joerg Arndt, Sep 03 2014
MAPLE
b:= proc(n, i) option remember; `if`(n=0, [1, 0], `if`(i<1, 0,
b(n, i-1)+`if`(i>n, 0, (p->p+[0, p[1]])(b(n-i, i-1)))))
end:
a:= n-> `if`(n=0, 0, (p-> p[2]-p[1])(b(n$2))):
seq(a(n), n=0..80); # Alois P. Heinz, Sep 03 2014
MATHEMATICA
max=80; s=Sum[(n-1)*q^(n*(n+1)/2)/QPochhammer[q, q, n], {n, Sqrt[max+1]}]+ O[q]^max; CoefficientList[s, q] (* Jean-François Alcover, Jan 17 2016 *)
PROG
(PARI)
N=66; q='q+O('q^N);
gf=sum(n=1, N, (n-1)*q^(n*(n+1)/2) / prod(k=1, n, 1-q^k ) );
v=Vec(gf+'a0); v[1]-='a0; v /* include initial zeros */
CROSSREFS
Cf. A015723, Sum_{n>=0} (n * q^(n*(n+1)/2) / Product_{k=1..n} (1 - q^k)).
Cf. A032020, Sum_{n>=0} (n! * q^(n*(n+1)/2) / Product_{k=1..n} (1 - q^k)).
Cf. A032153, Sum_{n>=1} ((n-1)! * q^(n*(n+1)/2) / Product_{k=1..n} (1 - q^k)).
Cf. A072576, Sum_{n>=0} ((n+1)! * q^(n*(n+1)/2) / Product_{k=1..n} (1 - q^k)).
Cf. A058884 (up-steps in all partitions).
KEYWORD
nonn
AUTHOR
Joerg Arndt, Oct 20 2012
STATUS
approved