login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Sum of Jacobsthal numbers that divide n.
4

%I #14 Oct 09 2017 21:52:38

%S 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 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 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 Sum of Jacobsthal numbers that divide n.

%C a(n) is the sum of the divisors of n that are Jacobsthal numbers (A001045).

%H Antti Karttunen, <a href="/A293432/b293432.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} A147612(d)*d.

%F a(n) = A293434(n) + (A147612(n)*n).

%e 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 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 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 With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Array[DivisorSum[#, # &, MemberQ[s, #] &] &, 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 A293432(n) = sumdiv(n,d,A147612(d)*d);

%Y Cf. A000203, A001045, A147612, A293431, A293434.

%Y Cf. also A005092.

%K nonn

%O 1,3

%A _Antti Karttunen_, Oct 09 2017