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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.
11
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