A304442 Number of partitions of n in which the sequence of the sum of the same summands is constant.
1, 1, 2, 2, 4, 2, 5, 2, 7, 3, 5, 2, 13, 2, 5, 4, 11, 2, 13, 2, 12, 4, 5, 2, 28, 3, 5, 5, 12, 2, 18, 2, 17, 4, 5, 4, 44, 2, 5, 4, 24, 2, 18, 2, 12, 10, 5, 2, 63, 3, 9, 4, 12, 2, 34, 4, 24, 4, 5, 2, 67, 2, 5, 10, 27, 4, 18, 2, 12, 4, 14, 2, 120, 2, 5, 7, 12, 4, 18, 2, 54
Offset: 0
Keywords
Examples
a(72) = binomial(d(72),1) + binomial(d(36),2) + binomial(d(24),3) + binomial(d(18),4) + binomial(d(12),6) = 12 + 36 + 56 + 15 + 1 = 120, where d(n) is the number of divisors of n. --+----------------------+----------------------------------------- n | | Sequence of the sum of the same summands --+----------------------+----------------------------------------- 1 | 1 | 1 2 | 2 | 2 | 1+1 | 2 3 | 3 | 3 | 1+1+1 | 3 4 | 4 | 4 | 2+2 | 4 | 2+1+1 | 2, 2 | 1+1+1+1 | 4 5 | 5 | 5 | 1+1+1+1+1 | 5 6 | 6 | 6 | 3+3 | 6 | 3+1+1+1 | 3, 3 | 2+2+2 | 6 | 1+1+1+1+1+1 | 6
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000
Crossrefs
These partitions are ranked by A353833.
Programs
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Mathematica
Table[Length[Select[IntegerPartitions[n],SameQ@@Total/@Split[#]&]],{n,0,15}] (* Gus Wiseman, Jun 25 2022 *)
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PARI
a(n) = if (n==0, 1, sumdiv(n, d, binomial(numdiv(n/d), d))); \\ Michel Marcus, May 13 2018
Formula
a(n) >= 2 for n > 1.
a(n) = Sum_{d|n} binomial(A000005(n/d), d) for n > 0.
Comments