A293432 Sum of Jacobsthal numbers that divide n.
1, 1, 4, 1, 6, 4, 1, 1, 4, 6, 12, 4, 1, 1, 9, 1, 1, 4, 1, 6, 25, 12, 1, 4, 6, 1, 4, 1, 1, 9, 1, 1, 15, 1, 6, 4, 1, 1, 4, 6, 1, 25, 44, 12, 9, 1, 1, 4, 1, 6, 4, 1, 1, 4, 17, 1, 4, 1, 1, 9, 1, 1, 25, 1, 6, 15, 1, 1, 4, 6, 1, 4, 1, 1, 9, 1, 12, 4, 1, 6, 4, 1, 1, 25, 91, 44, 4, 12, 1, 9, 1, 1, 4, 1, 6, 4, 1, 1, 15, 6, 1, 4, 1, 1, 30
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) = 1 + 3 + 5 = 9. 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) = 1 + 3 + 5 + 21 = 30. For n = 21845, whose divisors are [1, 5, 17, 85, 257, 1285, 4369, 21845], the divisors 1, 5, 85 and 21845 are in A001045, thus a(21845) = 1 + 5 + 85 + 21845 = 21936.
Links
Programs
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Mathematica
With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Array[DivisorSum[#, # &, 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)); A293432(n) = sumdiv(n,d,A147612(d)*d);
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Python
from sympy import divisors def A293432(n): return sum(d for d in divisors(n,generator=True) if (m:=3*d+1).bit_length()>(m-3).bit_length()) # Chai Wah Wu, Apr 18 2025
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