|
|
A293431
|
|
a(n) is the number of Jacobsthal numbers dividing n.
|
|
5
|
|
|
1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 3, 1, 1, 2, 1, 2, 3, 2, 1, 2, 2, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 2, 1, 1, 2, 2, 1, 3, 2, 2, 3, 1, 1, 2, 1, 2, 2, 1, 1, 2, 3, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 3, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 2, 1, 2, 2, 1, 1, 3, 3, 2, 2, 2, 1, 3, 1, 1, 2, 1, 2, 2, 1, 1, 3, 2, 1, 2, 1, 1, 4
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
LINKS
|
|
|
FORMULA
|
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{n>=2} 1/A001045(n) = 1.718591611927... . - Amiram Eldar, Jan 01 2024
|
|
EXAMPLE
|
For n = 15, whose divisors are [1, 3, 5, 15], the first three, 1, 3 and 5 are all in A001045, thus a(15) = 3.
For n = 21, whose divisors are [1, 3, 7, 21], 1, 3 and 21 are in A001045, thus a(21) = 3.
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) = 4.
|
|
MATHEMATICA
|
With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Array[DivisorSum[#, 1 &, MemberQ[s, #] &] &, 105]] (* Michael De Vlieger, Oct 09 2017 *)
|
|
PROG
|
(PARI)
A147612aux(n, i) = if(!(n%2), n, A147612aux((n+i)/2, -i));
A147612(n) = 0^(A147612aux(n, 1)*A147612aux(n, -1));
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|