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A293432 Sum of Jacobsthal numbers that divide n.

%I A293432 #20 Apr 18 2025 14:52:40
%S A293432 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,
%T A293432 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,
%U A293432 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
%N A293432 Sum of Jacobsthal numbers that divide n.
%C A293432 a(n) is the sum of the divisors of n that are Jacobsthal numbers (A001045).
%H A293432 Antti Karttunen, <a href="/A293432/b293432.txt">Table of n, a(n) for n = 1..21845</a>
%H A293432 <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>
%F A293432 a(n) = Sum_{d|n} A147612(d)*d.
%F A293432 a(n) = A293434(n) + (A147612(n)*n).
%e A293432 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.
%e A293432 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.
%e A293432 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.
%t A293432 With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Array[DivisorSum[#, # &, MemberQ[s, #] &] &, 105]] (* _Michael De Vlieger_, Oct 09 2017 *)
%o A293432 (PARI)
%o A293432 A147612aux(n,i) = if(!(n%2),n,A147612aux((n+i)/2,-i));
%o A293432 A147612(n) = 0^(A147612aux(n,1)*A147612aux(n,-1));
%o A293432 A293432(n) = sumdiv(n,d,A147612(d)*d);
%o A293432 (Python)
%o A293432 from sympy import divisors
%o A293432 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
%Y A293432 Cf. A000203, A001045, A147612, A293431, A293434.
%Y A293432 Cf. also A005092.
%K A293432 nonn
%O A293432 1,3
%A A293432 _Antti Karttunen_, Oct 09 2017