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A365377
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Number of subsets of {1..n} without a subset summing to n.
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17
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0, 1, 2, 3, 6, 9, 17, 26, 49, 72, 134, 201, 366, 544, 984, 1436, 2614, 3838, 6770, 10019, 17767, 25808, 45597, 66671, 116461, 169747, 295922, 428090, 750343, 1086245, 1863608, 2721509, 4705456, 6759500, 11660244, 16877655, 28879255, 41778027, 71384579, 102527811, 176151979
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OFFSET
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0,3
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LINKS
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FORMULA
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EXAMPLE
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The a(1) = 1 through a(6) = 17 subsets:
{} {} {} {} {} {}
{1} {1} {1} {1} {1}
{2} {2} {2} {2}
{3} {3} {3}
{1,2} {4} {4}
{2,3} {1,2} {5}
{1,3} {1,2}
{2,4} {1,3}
{3,4} {1,4}
{2,3}
{2,5}
{3,4}
{3,5}
{4,5}
{1,3,4}
{2,3,5}
{3,4,5}
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MATHEMATICA
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Table[Length[Select[Subsets[Range[n]], FreeQ[Total/@Subsets[#], n]&]], {n, 0, 10}]
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PROG
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(PARI) isok(s, n) = forsubset(#s, ss, if (vecsum(vector(#ss, k, s[ss[k]])) == n, return(0))); return(1);
a(n) = my(nb=0); forsubset(n, s, if (isok(s, n), nb++)); nb; \\ Michel Marcus, Sep 09 2023
(Python)
from itertools import combinations, chain
from sympy.utilities.iterables import partitions
if n == 0: return 0
nset = set(range(1, n+1))
s, c = [set(p) for p in partitions(n, m=n, k=n) if max(p.values(), default=1) == 1], 1
for a in chain.from_iterable(combinations(nset, m) for m in range(2, n+1)):
if sum(a) >= n:
aset = set(a)
for p in s:
if p.issubset(aset):
c += 1
break
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CROSSREFS
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The complement w/ re-usable parts is A365073.
The complement is counted by A365376.
The version with re-usable parts is A365380.
A000124 counts distinct possible sums of subsets of {1..n}.
A124506 appears to count combination-free subsets, differences of A326083.
A364350 counts combination-free strict partitions, complement A364839.
A365381 counts subsets of {1..n} with a subset summing to k.
Cf. A007865, A085489, A088809, A093971, A103580, A151897, A236912, A237113, A237668, A326080, A364349, A364534.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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