A293431 a(n) is the number of Jacobsthal numbers dividing n.
1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 3, 1, 1, 2, 1, 2, 3, 2, 1, 2, 2, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 2, 1, 1, 2, 2, 1, 3, 2, 2, 3, 1, 1, 2, 1, 2, 2, 1, 1, 2, 3, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 3, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 2, 1, 2, 2, 1, 1, 3, 3, 2, 2, 2, 1, 3, 1, 1, 2, 1, 2, 2, 1, 1, 3, 2, 1, 2, 1, 1, 4
Offset: 1
Keywords
Examples
For n = 15, whose divisors are [1, 3, 5, 15], the first three, 1, 3 and 5 are all in A001045, thus a(15) = 3. For n = 21, whose divisors are [1, 3, 7, 21], 1, 3 and 21 are in A001045, thus a(21) = 3. For n = 105, whose divisors are [1, 3, 5, 7, 15, 21, 35, 105], only the divisors 1, 3, 5 and 21 are in A001045, thus a(105) = 4.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..21845
Programs
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Mathematica
With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Array[DivisorSum[#, 1 &, MemberQ[s, #] &] &, 105]] (* Michael De Vlieger, Oct 09 2017 *)
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PARI
A147612aux(n,i) = if(!(n%2),n,A147612aux((n+i)/2,-i)); A147612(n) = 0^(A147612aux(n,1)*A147612aux(n,-1)); A293431(n) = sumdiv(n,d,A147612(d));
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Python
from sympy import divisors def A293431(n): return sum(1 for d in divisors(n,generator=True) if (m:=3*d+1).bit_length()>(m-3).bit_length()) # Chai Wah Wu, Apr 18 2025
Formula
a(n) = Sum_{d|n} A147612(d).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{n>=2} 1/A001045(n) = 1.718591611927... . - Amiram Eldar, Jan 01 2024