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A141799 Number of repeated integer partitions of n. 3
1, 3, 8, 25, 66, 192, 511, 1418, 3812, 10383, 27958, 75758, 204215, 551821, 1488561, 4018722, 10842422, 29262357, 78955472, 213063551, 574905487, 1551325859, 4185959285, 11295211039, 30478118079, 82240300045, 221911189754, 598790247900, 1615732588962 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

An integer n can be partitioned into P(n) partitions P([n],i) where i=1,...,P(n) counts the partitions. The partition P([n],i) consists of T(n,i) integer parts t(i,j) with j=1,...,T(n,i). Now we perform on each t(i,j) an integer partition again and arrive at new partitions. Their parts can be partitioned again and so forth. We count such repeated partitions of n. One convention is necessary to avoid an infinite loop: The trivial partition P([n],1)=[n] will not be partitioned again but just counted once (and therefore we also have a(1)=1).

LINKS

Alois P. Heinz, Table of n, a(n) for n = 1..1000

FORMULA

Let sum_{i=1}^P(n) denote the sum over all integer partitions P([n],i) of n. Let sum_{j=1}^T(i,j) denote the sum over all parts of the i-th integer partition. Then we have the recursive formula 1 if t(i,j)=n a(n) = sum_{i=1}^P(n) sum_{j=1}^T(i,j) { a(t(i,j)) else. E.g. a(4)=25 because [4] contributes 1, [1,3] contributes a(1)+a(3)=1+8=9, [2,2] contributes a(2)+a(2)=3+3=6, [1,1,2] contributes a(1)+a(1)+a(2)=1+1+3=5, [1,1,1,1] contributes a(1)+a(1)+a(1)+a(1)=1+1+1+1=4 which gives in total 25.

a(n) ~ c * d^n, where d = 2.69832910647421123126399866... (see A246828), c = 0.5088820425072641934222229579416714164592334575899644931509447692360546... . - Vaclav Kotesovec, Sep 04 2014

EXAMPLE

For the integers 1, 2, 3 and 4 we have

[1] -> 1,

thus a(1)=1.

[2] -> 1,

[1,1] => [1] ->, [1] -> 1.

thus a(2)=3.

[3] -> 1,

[1,2] => [1] -> 1, [2] -> 3,

[1,1,1] => [1] -> 1, [1] -> 1, [1] -> 1,

thus a(3)=8.

[4] -> 1,

[1,3] => [1] -> 1, [3] -> 8,

[2,2] => [2] -> 3, [2] -> 3,

[1,1,2] => [1] -> 1, [1] -> 1, [2] -> 3,

[1,1,1,1] => [1] -> 1, [1] -> 1, [1] -> 1, [1] -> 1,

thus a(4)=25.

MAPLE

A141799 := proc(n) option remember ; local a, P, i, p ; if n =1 then 1; else a := 0 ; for P in combinat[partition](n) do if nops(P) > 1 then for i in P do a := a+procname(i) ; od: else a := a+1 ; fi; od: RETURN(a) ; fi ; end: for n from 1 to 40 do printf("%d, ", A141799(n)) ; od: # R. J. Mathar, Aug 25 2008

# second Maple program

a:= proc(n) option remember;

      1+ `if`(n>1, b(n, n-1)[2], 0)

    end:

b:= proc(n, i) option remember; local f, g;

      if n=0 or i=1 then [1, n]

    else f:= b(n, i-1); g:= `if`(i>n, [0, 0], b(n-i, i));

         [f[1]+g[1], f[2]+g[2] +g[1]*a(i)]

      fi

    end:

seq(a(n), n=1..40); # Alois P. Heinz, Apr 05 2012

MATHEMATICA

a[n_] := a[n] = 1 + If[n>1, b[n, n-1][[2]], 0]; b[n_, i_] := b[n, i] = Module[{f, g}, If[n == 0 || i == 1, {1, n}, f = b[n, i-1]; g = If[i>n, {0, 0}, b[n-i, i]]; {f[[1]] + g[[1]], f[[2]] + g[[2]] + g[[1]]*a[i]}]]; Table[a[n], {n, 1, 40}] (* Jean-Fran├žois Alcover, Oct 29 2015, after Alois P. Heinz *)

CROSSREFS

Cf. A000041, A131407, A131408, A137732, A246828.

Sequence in context: A018789 A203413 A301604 * A093969 A259699 A259700

Adjacent sequences:  A141796 A141797 A141798 * A141800 A141801 A141802

KEYWORD

nonn

AUTHOR

Thomas Wieder, Jul 05 2008

EXTENSIONS

Extended by R. J. Mathar, Aug 25 2008

STATUS

approved

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Last modified October 22 14:44 EDT 2019. Contains 328318 sequences. (Running on oeis4.)