cp's OEIS Frontend

A141799 Number of repeated integer partitions of n.

%I A141799 #26 Jun 01 2018 01:53:06
%S A141799 1,3,8,25,66,192,511,1418,3812,10383,27958,75758,204215,551821,
%T A141799 1488561,4018722,10842422,29262357,78955472,213063551,574905487,
%U A141799 1551325859,4185959285,11295211039,30478118079,82240300045,221911189754,598790247900,1615732588962
%N A141799 Number of repeated integer partitions of n.
%C A141799 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).
%H A141799 Alois P. Heinz, <a href="/A141799/b141799.txt">Table of n, a(n) for n = 1..1000</a>
%F A141799 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.
%F A141799 a(n) ~ c * d^n, where d = 2.69832910647421123126399866... (see A246828), c = 0.5088820425072641934222229579416714164592334575899644931509447692360546... . - _Vaclav Kotesovec_, Sep 04 2014
%e A141799 For the integers 1, 2, 3 and 4 we have
%e A141799 [1] -> 1,
%e A141799 thus a(1)=1.
%e A141799 [2] -> 1,
%e A141799 [1,1] => [1] ->, [1] -> 1.
%e A141799 thus a(2)=3.
%e A141799 [3] -> 1,
%e A141799 [1,2] => [1] -> 1, [2] -> 3,
%e A141799 [1,1,1] => [1] -> 1, [1] -> 1, [1] -> 1,
%e A141799 thus a(3)=8.
%e A141799 [4] -> 1,
%e A141799 [1,3] => [1] -> 1, [3] -> 8,
%e A141799 [2,2] => [2] -> 3, [2] -> 3,
%e A141799 [1,1,2] => [1] -> 1, [1] -> 1, [2] -> 3,
%e A141799 [1,1,1,1] => [1] -> 1, [1] -> 1, [1] -> 1, [1] -> 1,
%e A141799 thus a(4)=25.
%p A141799 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
%p A141799 # second Maple program
%p A141799 a:= proc(n) option remember;
%p A141799       1+ `if`(n>1, b(n, n-1)[2], 0)
%p A141799     end:
%p A141799 b:= proc(n, i) option remember; local f, g;
%p A141799       if n=0 or i=1 then [1, n]
%p A141799     else f:= b(n, i-1); g:= `if`(i>n, [0, 0], b(n-i, i));
%p A141799          [f[1]+g[1], f[2]+g[2] +g[1]*a(i)]
%p A141799       fi
%p A141799     end:
%p A141799 seq(a(n), n=1..40); # _Alois P. Heinz_, Apr 05 2012
%t A141799 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-François Alcover_, Oct 29 2015, after _Alois P. Heinz_ *)
%Y A141799 Cf. A000041, A131407, A131408, A137732, A246828.
%K A141799 nonn
%O A141799 1,2
%A A141799 _Thomas Wieder_, Jul 05 2008
%E A141799 Extended by _R. J. Mathar_, Aug 25 2008