A355746 Number of different multisets that can be obtained by choosing a prime index (or a prime factor) of each integer from 2 to n.
1, 1, 1, 1, 1, 2, 2, 2, 2, 4, 4, 6, 6, 12, 20, 20, 20, 26, 26, 36, 58, 116, 116, 140, 140, 280, 280, 384, 384, 536, 536, 536, 844, 1688, 2380, 2716, 2716, 5432, 8484, 10152, 10152, 13308, 13308, 18064, 21616, 43232, 43232, 47648, 47648, 54656, 84480, 114304, 114304
Offset: 1
Keywords
Examples
The a(n) multisets for n = 2, 6, 10, 12: {1} {1,1,1,2,3} {1,1,1,1,1,2,2,3,4} {1,1,1,1,1,1,2,2,3,4,5} {1,1,2,2,3} {1,1,1,1,2,2,2,3,4} {1,1,1,1,1,2,2,2,3,4,5} {1,1,1,1,2,2,3,3,4} {1,1,1,1,1,2,2,3,3,4,5} {1,1,1,2,2,2,3,3,4} {1,1,1,1,2,2,2,2,3,4,5} {1,1,1,1,2,2,2,3,3,4,5} {1,1,1,2,2,2,2,3,3,4,5}
Links
- Michael S. Branicky, Table of n, a(n) for n = 1..75
Crossrefs
Programs
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Mathematica
primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; Table[Length[Union[Sort/@Tuples[primeMS/@Range[2,n]]]],{n,15}]
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Python
from sympy import factorint from itertools import count, islice def agen(): s = {(1,)} for n in count(2): yield len(s) s = set(tuple(sorted(t+(d,))) for t in s for d in factorint(n)) print(list(islice(agen(), 53))) # Michael S. Branicky, Aug 03 2022
Formula
a(p) = a(p-1) for p prime. - Michael S. Branicky, Aug 03 2022
Extensions
a(28) and beyond from Michael S. Branicky, Aug 03 2022