cp's OEIS Frontend

Proof:

Proof is by induction. We assume as our induction hypothesis that the given multiplicative formula for A046644 (resp. additive formula for A046645) holds for all proper divisors d|n, dA046645(p) = 1. [Remark: for squares of primes, f(p^2) = (4*b(p^2) - 1)/8, thus a(p^2) = 8.]

First we note that A005187(x+y) <= A005187(x) + A005187(y), with equivalence attained only when A004198(x,y) = 0, that is, when x and y do not have any 1-bits in the shared positions. Let m = Sum_{e} A005187(e), with e ranging over the exponents in prime factorization of n.

For [case A] any n in A268388 it happens that only when d (and thus also n/d) are infinitary divisors of n will Sum_{e} A005187(e) [where e now ranges over the union of multisets of exponents in the prime factorizations of d and n/d] attain value m, which is the maximum possible for such sums computed for all divisor pairs d and n/d. For any n in A268388, A037445(n) = 2^k, k >= 2, thus A037445(n) - 2 = 2 mod 4 (excluding 1 and n from the count, thus -2). Thus, in the recursive formula above, the maximal denominator that occurs in the sum is 2^m which occurs k times, with k being an even number, but not a multiple of 4, thus the factor (1/2) in the front of the whole sum will ensure that the denominator of the whole expression is 2^m [which thus is equal to 2^A046645(n) = a(n)].

On the other hand [case B], for squares in A050376 (A082522, numbers of the form p^(2^k) with p prime and k>0), all the sums A005187(x)+A005187(y), where x+y = 2^k, 0 < x <= y < 2^k are less than A005187(2^k), thus it is the lonely "middle pair" f(p^(2^(k-1))) * f(p^(2^(k-1))) among all the pairs f(d)*f(n/d), 1 < d < n = p^(2^k) which yields the maximal denominator. Furthermore, as it occurs an odd number of times (only once), the common factor (1/2) for the whole sum will increase the exponent of 2 in denominator by one, which will be (2*A005187(2^(k-1))) + 1 = A005187(2^k) = A046645(p^(2^k)).

(End)

From Antti Karttunen, Aug 21 2018: (Start)

The following list gives a few such pairs num(n), b(n) for which b(n) is Dirichlet convolution of num(n)/a(n). Here ε stands for sequence A063524 (1, 0, 0, ...).

Numerators Dirichlet convolution of numerator(n)/a(n) yields

------- -----------

A046643 A000012

A257098 A008683

A317935 A003557

A318321 A003961

A317830 A175851

A317833 A078898

A317834 A078899

A317847 A303757

A317835 ε + A003415

A317845 ε + A001065

A317846 ε + A051953

A317936 ε + A100995

A317937 ε + A001221

A317938 ε + A001222

A317939 ε + A010051 (= A080339)

(End)

This sequence gives an upper bound for the denominators of any rational-valued sequence obtained as the "Dirichlet Square Root" of any integer-valued sequence. - Andrew Howroyd, Aug 23 2018