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A325565
a(n) is the number of such divisors d of n that A048720(A065621(d),n/d) is equal to n.
8
1, 2, 1, 3, 1, 2, 1, 4, 2, 2, 1, 3, 1, 2, 1, 5, 1, 4, 1, 3, 2, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 2, 2, 2, 6, 1, 2, 1, 4, 1, 4, 1, 3, 2, 2, 1, 5, 2, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 2, 4, 1, 3, 1, 4, 1, 8, 1, 2, 2, 3, 1, 2, 1, 5, 1, 2, 1, 6, 1, 2, 1, 4, 1, 4, 1, 3, 2, 2, 1, 6, 1, 4, 1, 3, 1, 2, 1, 4, 2
OFFSET
1,2
COMMENTS
Equally, a(n) is number of such pairs of natural numbers t, u that A048720(t,u) = n and A065620(t)*u = n.
FORMULA
a(n) = Sum_{d|n} [A048720(A065621(d),n/d) == n], where [ ] is the Iverson bracket.
a(n) / a(A000265(n)) = A001511(n).
a(n) <= A000005(n) for all n.
a(n) <= A091220(n) for all n.
PROG
(PARI)
A048720(b, c) = fromdigits(Vec(Pol(binary(b))*Pol(binary(c)))%2, 2);
A065621(n) = bitxor(n-1, n+n-1);
A325565(n) = sumdiv(n, d, A048720(A065621(d), n/d)==n);
(PARI)
A065620(n, c=1) = sum(i=0, logint(n+!n, 2), if(bittest(n, i), (-1)^c++<<i)); \\ From A065620
A325565(n) = { my(p = Pol(binary(n))*Mod(1, 2)); sum(d=1, n, my(q = Pol(binary(d))*Mod(1, 2)); (0==(p%q) && (n==(A065620(d)*fromdigits(Vec(lift(p/q)), 2))))); };
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 09 2019
STATUS
approved