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A293432
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Sum of Jacobsthal numbers that divide n.
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4
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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
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OFFSET
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1,3
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COMMENTS
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a(n) is the sum of the divisors of n that are Jacobsthal numbers (A001045).
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LINKS
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FORMULA
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EXAMPLE
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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.
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MATHEMATICA
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With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Array[DivisorSum[#, # &, MemberQ[s, #] &] &, 105]] (* Michael De Vlieger, Oct 09 2017 *)
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PROG
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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));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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