A088314 Cardinality of set of sets of parts of all partitions of n.
1, 1, 2, 3, 5, 6, 10, 12, 18, 22, 30, 37, 51, 61, 79, 96, 124, 148, 186, 222, 275, 326, 400, 473, 575, 673, 811, 946, 1132, 1317, 1558, 1813, 2138, 2463, 2893, 3323, 3882, 4461, 5177, 5917, 6847, 7818, 8994, 10251, 11766, 13334, 15281, 17309, 19732, 22307
Offset: 0
Examples
The 7 partitions of 5 and their sets of parts are [ #] partition set of parts [ 1] [ 1 1 1 1 1 ] {1} [ 2] [ 2 1 1 1 ] {1, 2} [ 3] [ 2 2 1 ] {1, 2} (same as before) [ 4] [ 3 1 1 ] {1, 3} [ 5] [ 3 2 ] {2, 3} [ 6] [ 4 1 ] {1, 4} [ 7] [ 5 ] {5} so we have a(5) = |{{1}, {1, 2}, {1, 3}, {2, 3}, {1, 4}, {5}}| = 6.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..100
Crossrefs
Cf. A182410.
Programs
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Haskell
a066186 = sum . concat . ps 1 where ps _ 0 = [[]] ps i j = [t:ts | t <- [i..j], ts <- ps t (j - t)] -- Reinhard Zumkeller, Jul 13 2013
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Maple
list2set := L -> {op(L)}; a:= N -> list2set(map( list2set, combinat[partition](N) )); seq(nops(a(n)), n=0..30); # Yogy Namara (yogy.namara(AT)gmail.com), Jan 13 2010 b:= proc(n, i) option remember; `if`(n=0, {{}}, `if`(i<1, {}, {b(n, i-1)[], seq(map(x->{x[],i}, b(n-i*j, i-1))[], j=1..n/i)})) end: a:= n-> nops(b(n, n)): seq(a(n), n=0..40); # Alois P. Heinz, Aug 09 2012
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Mathematica
Table[Length[Union[Map[Union,IntegerPartitions[n]]]],{n,1,30}] (* Geoffrey Critzer, Feb 19 2013 *) (* Second program: *) b[n_, i_] := b[n, i] = If[n == 0, {{}}, If[i < 1, {}, Union@Flatten@{b[n, i - 1], Table[If[Head[#] == List, Append[#, i]]& /@ b[n - i*j, i - 1], {j, 1, n/i}]}]]; a[n_] := Length[b[n, n]]; a /@ Range[0, 40] (* Jean-François Alcover, Jun 04 2021, after Alois P. Heinz *) combp[n_,y_]:=With[{s=Table[{k,i},{k,y}, {i,1,Floor[n/k]}]}, Select[Tuples[s], Total[Times@@@#]==n&]]; Table[Length[Select[Join@@Array[IntegerPartitions,n], UnsameQ@@#&&combp[n,#]!={}&]], {n,0,15}] (* Gus Wiseman, Sep 11 2023 *)
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Python
from sympy.utilities.iterables import partitions def A088314(n): return len({tuple(sorted(set(p))) for p in partitions(n)}) # Chai Wah Wu, Sep 10 2023
Formula
a(n) = 2^(n-1) - A070880(n). - Alois P. Heinz, Feb 08 2019
a(n) = A365042(n) + 1. - Gus Wiseman, Sep 13 2023
Extensions
More terms and clearer definition from Vladeta Jovovic, Apr 21 2005
Comments