A108492 Sum of NumberOfParts!/NumberOfDifferentParts! for all integer partitions of n.

1, 1, 3, 8, 31, 141, 819, 5562, 43773, 389203, 3858136, 42152116, 503098359, 6511429138, 90824834615, 1358130449902, 21672033893102, 367570633594883, 6602838664294634, 125228962373218571, 2500582942246200527, 52437202425839368049, 1152133477802718430790

Offset: 0

Author

Thomas Wieder, Jun 05 2005

Keywords

• nonn

Examples

			a(4) = 31 because n=4 has the following A000041(4) = 5 partitions:
i=1: (1111) with p(1,4)=4, d(1,4)=1; 4!/1! = 24;
i=2: (112)  with p(2,4)=3, d(2,4)=2; 3!/2! =  3;
i=3: (13)   with p(3,4)=2, d(3,4)=2; 2!/2! =  1;
i=4: (4)    with p(4,4)=1, d(4,4)=1; 1!/1! =  1;
i=5: (22)   with p(5,4)=2, d(5,4)=1; 2!/1! =  2;
Their contributions sum up to 24+3+1+1+2 = 31 = a(4).
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..450

Crossrefs

Cf. A000041, A007318, A008277, A339006.

Programs

Maple

A108492 := proc(n::integer) local i,prttnlst,prttn,ZahlTeile,liste,ZahlVerschiedenerTeile,A108492;
# Procedure A108492 calculates the sequence A108492 for the integer partitions of n. prttn = an integer partition of n. See also http://www.thomas-wieder.privat.t-online.de/default.html
prttnlst:=partition(n); A108492 := 0; for i from 1 to nops(prttnlst) do prttn := prttnlst[i]; ZahlTeile := nops(prttn); liste := convert(prttn,multiset); ZahlVerschiedenerTeile := nops(liste); A108492 := A108492 + (ZahlTeile!/ZahlVerschiedenerTeile!); od; print(n,A108492); end proc;
# Second Maple program:
b:= proc(n,i,l,p,d) option remember;
      if n<0 then 0
    elif n=0 then p!/d!
    elif i=0 then 0
    else b(n, i-1, l, p, d) +b(n-i, i, i, p+1, `if`(i=l, d, d+1))
      fi
    end:
a:= n-> b(n, n, 0, 0, 0):
seq(a(n), n=0..30);  # Alois P. Heinz, Apr 25 2011

Mathematica

b[n_, i_, l_, p_, d_] := b[n, i, l, p, d] = Which[n<0, 0, n==0, p!/d!, i==0, 0, True, b[n, i-1, l, p, d]+b[n-i, i, i, p+1, If[i==l, d, d+1]]];
a[n_] := b[n, n, 0, 0, 0];
a /@ Range[0, 30] (* Jean-François Alcover, Nov 18 2020, after Alois P. Heinz *)

Formula

a(n) = Sum_{i=1..A000041(n)} p(i,n)!/d(i,n)! with p(i,n) = number of parts of the i-th partition of n and d(i,n) = number of different parts of the i-th partition of n.

a(n) ~ n! * (1 + 1/(2*n) + 1/n^2 + 13/(6*n^3) + 19/(3*n^4) + 22/n^5 + 2057/(24*n^6)). - Vaclav Kotesovec, Sep 06 2015