A181796 a(n) = number of divisors of n whose canonical prime factorizations contain no repeated positive exponents (cf. A130091).
1, 2, 2, 3, 2, 3, 2, 4, 3, 3, 2, 5, 2, 3, 3, 5, 2, 5, 2, 5, 3, 3, 2, 7, 3, 3, 4, 5, 2, 4, 2, 6, 3, 3, 3, 7, 2, 3, 3, 7, 2, 4, 2, 5, 5, 3, 2, 9, 3, 5, 3, 5, 2, 7, 3, 7, 3, 3, 2, 7, 2, 3, 5, 7, 3, 4, 2, 5, 3, 4, 2, 10, 2, 3, 5, 5, 3, 4, 2, 9, 5, 3, 2, 7, 3, 3, 3, 7, 2, 7, 3, 5, 3, 3, 3, 11, 2, 5, 5, 7, 2, 4, 2, 7, 4
Offset: 1
Keywords
Examples
12 has a total of six divisors (1, 2, 3, 4, 6 and 12). Of those divisors, the number 1 has no prime factors, hence, no positive exponents at all (and no repeated positive exponents) in its canonical prime factorization. The lists of positive exponents for 2, 3, 4, 6 and 12 are (1), (1), (2), (1,1) and (2,1) respectively (cf. A124010). Of all six divisors, only the number 6 (2^1*3^1) has at least one positive exponent repeated (namely, 1). The other five do not; hence, a(12) = 5. For n = 90 = 2 * 3^2 * 5, the divisors that satisfy the condition are: 1, 2, 3, 3^2, 5, 2 * 3^2, 3^2 * 5, altogether 7, (but for example 90 itself is not included), thus a(90) = 7.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Prime Factorization.
- Gus Wiseman, Sequences counting and encoding certain classes of multisets
Crossrefs
Partitions with distinct multiplicities are A098859.
Sorted prime signature is A118914.
Unsorted prime signature is A124010.
a(n) is the number of divisors of n in A130091.
Factorizations with distinct multiplicities are A255231.
The largest of the counted divisors is A327498.
Factorizations using the counted divisors are A327523.
Programs
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Mathematica
Table[DivisorSum[n, 1 &, Length@ Union@ # == Length@ # &@ FactorInteger[#][[All, -1]] &], {n, 105}] (* Michael De Vlieger, May 28 2017 *)
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PARI
no_repeated_exponents(n) = { my(es = factor(n)[, 2]); if(length(Set(es)) == length(es),1,0); } A181796(n) = sumdiv(n,d,no_repeated_exponents(d)); \\ Antti Karttunen, May 27 2017
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Python
from sympy import factorint, divisors def ok(n): f=factorint(n) ex=[f[i] for i in f] for i in ex: if ex.count(i)>1: return 0 return 1 def a(n): return sum([1 for i in divisors(n) if ok(i)]) # Indranil Ghosh, May 27 2017
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