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%I #9 Oct 09 2017 21:52:46
%S 0,1,1,1,1,4,1,1,4,6,1,4,1,1,9,1,1,4,1,6,4,12,1,4,6,1,4,1,1,9,1,1,15,
%T 1,6,4,1,1,4,6,1,25,1,12,9,1,1,4,1,6,4,1,1,4,17,1,4,1,1,9,1,1,25,1,6,
%U 15,1,1,4,6,1,4,1,1,9,1,12,4,1,6,4,1,1,25,6,44,4,12,1,9,1,1,4,1,6,4,1,1,15,6,1,4,1,1,30
%N a(n) is the sum of the proper divisors of n that are Jacobsthal numbers (A001045).
%H Antti Karttunen, <a href="/A293434/b293434.txt">Table of n, a(n) for n = 1..21845</a>
%H <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>
%F a(n) = Sum_{d|n, d<n} A147612(d)*d.
%F a(n) = A293432(n) - (A147612(n)*n).
%e For n = 15, whose proper divisors are [1, 3, 5], all of them are in A001045, thus a(15) = 1 + 3 + 5 = 9.
%e For n = 21, whose proper divisors are [1, 3, 7], both 1 and 3 are in A001045, thus a(21) = 1 + 3 = 4.
%e For n = 21845, whose proper divisors are [1, 5, 17, 85, 257, 1285, 4369], only 1, 5, 85 are in A001045, thus a(21845) = 1 + 5 + 85 = 91.
%t With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Table[DivisorSum[n, # &, And[MemberQ[s, #], # != n] &], {n, 105}]] (* _Michael De Vlieger_, Oct 09 2017 *)
%o (PARI)
%o A147612aux(n,i) = if(!(n%2),n,A147612aux((n+i)/2,-i));
%o A147612(n) = 0^(A147612aux(n,1)*A147612aux(n,-1));
%o A293434(n) = sumdiv(n,d,(d<n)*A147612(d)*d);
%Y Cf. A001045, A147612, A293432, A293433.
%Y Cf. also A293436.
%K nonn
%O 1,6
%A _Antti Karttunen_, Oct 09 2017