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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 (list; graph; refs; listen; history; text; internal format)
 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. LINKS Antti Karttunen, Table of n, a(n) for n = 1..16384 Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537 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++<

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Last modified July 27 23:16 EDT 2021. Contains 346316 sequences. (Running on oeis4.)