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From square root of Riemann zeta function: form Dirichlet series Sum b_n/n^s whose square is zeta function; sequence gives numerator of b_n.
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%I #28 Jul 10 2017 03:45:34

%S 1,1,1,3,1,1,1,5,3,1,1,3,1,1,1,35,1,3,1,3,1,1,1,5,3,1,5,3,1,1,1,63,1,

%T 1,1,9,1,1,1,5,1,1,1,3,3,1,1,35,3,3,1,3,1,5,1,5,1,1,1,3,1,1,3,231,1,1,

%U 1,3,1,1,1,15,1,1,3,3,1,1,1,35,35,1,1,3,1,1,1,5,1,3

%N From square root of Riemann zeta function: form Dirichlet series Sum b_n/n^s whose square is zeta function; sequence gives numerator of b_n.

%C b(n) = A046643(n)/A046644(n) is multiplicative with b(p^n) = (2n-1)!!/2^n/n!. Dirichlet g.f. of A046643(n)/A046644(n) is sqrt(zeta(x)). - _Christian G. Bower_, May 16 2005

%C That is, b(p^n) = A001147(n) / (A000079(n)*A000142(n)) = A010050(n)/A000290(A000165(n)) = (2n)!/((2^n*n!)^2). - _Antti Karttunen_, Jul 08 2017

%H G. C. Greubel, <a href="/A046643/b046643.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>

%F Sum_{b|d} b(d)b(n/d) = 1. Also b_{2^j} = A001790[ j ]/2^A005187[ j ].

%F From _Antti Karttunen_, Jul 08 2017: (Start)

%F Multiplicative with a(p^n) = A001790(n).

%F a(1) = 1; for n > 1, a(n) = A001790(A067029(n)) * a(A028234(n)).

%F (End)

%e b_1, b_2, ... = 1, 1/2, 1/2, 3/8, 1/2, 1/4, 1/2, 5/16, 3/8, 1/4, 1/2, 3/16, ...

%p b := proc(n) option remember; local c,i,t1; if n = 1 then 1 else c := 1; t1 := divisors(n);

%p for i from 2 to nops(t1)-1 do c := c-b(t1[ i ])*b(n/t1[ i ]); od; c/2; fi; end;

%t b[1] = 1; b[n_] := b[n] = (dn = Divisors[n]; c = 1;

%t Do[c = c - b[dn[[i]]]*b[n/dn[[i]]], {i, 2, Length[dn] - 1}]; c/2); a[n_] := Numerator[b[n]]; a /@ Range[90] (* _Jean-François Alcover_, Apr 04 2011, after Maple version *)

%o (PARI)

%o A046643perA046644(n) = { my(c=1); if(1==n,c,fordiv(n,d, if((d>1)&&(d<n), c -= (A046643perA046644(d)*A046643perA046644(n/d)))); (c/2)); }

%o A046643(n) = numerator(A046643perA046644(n)); \\ After Maple-program, _Antti Karttunen_, Jul 08 2017

%o (Scheme)

%o (define (A046643 n) (if (= 1 n) n (* (A001790 (A067029 n)) (A046643 (A028234 n)))))

%o ;; Or, after _Christian G. Bower_'s May 16 2005 comment:

%o (definec (A046643perA046644 n) (if (= 1 n) n (* (/ (A010050 (A067029 n)) (A000290 (A000165 (A067029 n)))) (A046643perA046644 (A028234 n)))))

%o (define (A046643 n) (numerator (A046643perA046644 n)))

%o (define (A046644 n) (denominator (A046643perA046644 n)))

%o ;; _Antti Karttunen_, Jul 08 2017

%Y Cf. A000079, A000165, A001147, A001790, A005187, A010050, A028234, A067029.

%Y Cf. A046644, A046645.

%K nonn,easy,frac,nice,mult

%O 1,4

%A _N. J. A. Sloane_, Dec 11 1999