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%I #59 Jan 11 2024 08:57:03
%S 1,2,2,2,2,4,2,4,2,4,2,4,2,4,4,4,2,4,2,4,4,4,2,8,2,4,4,4,2,8,2,6,4,4,
%T 4,4,2,4,4,8,2,8,2,4,4,4,2,8,2,4,4,4,2,8,4,8,4,4,2,8,2,4,4,6,4,8,2,4,
%U 4,8,2,8,2,4,4,4,4,8,2,8,4,4,2,8,4,4,4,8,2,8
%N a(n) is the number of bi-unitary divisors of n.
%C a(n) is the number of terms of the n-th row of A222266.
%H Antti Karttunen, <a href="/A286324/b286324.txt">Table of n, a(n) for n = 1..10000</a>
%H Vaclav Kotesovec, <a href="/A286324/a286324.jpg">Graph - the asymptotic ratio (100000 terms)</a>
%H D. Suryanarayana, <a href="https://doi.org/10.1007/BFb0058797">The number of bi-unitary divisors of an integer</a>, in The Theory of Arithmetic Functions pp 273-282, Lecture Notes in Mathematics book series (LNM, volume 251).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BiunitaryDivisor.html">Biunitary Divisor</a>.
%F Multiplicative with a(p^e) = e + (e mod 2). - _Andrew Howroyd_, Aug 05 2018
%F a(A340232(n)) = 2*n. - _Bernard Schott_, Mar 12 2023
%F a(n) = A000005(A350390(n)) (the number of divisors of the largest exponentially odd number dividing n). - _Amiram Eldar_, Sep 01 2023
%F From _Vaclav Kotesovec_, Jan 11 2024: (Start)
%F Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - (p^s - 1)/((p^s + 1)*p^(2*s))).
%F Let f(s) = Product_{p prime} (1 - (p^s - 1)/((p^s + 1)*p^(2*s))).
%F Sum_{k=1..n} a(k) ~ f(1) * n * (log(n) + 2*gamma - 1 + f'(1)/f(1)), where
%F f(1) = Product_{p prime} (1 - (p-1)/((p+1)*p^2)) = A306071 = 0.80733082163620503914865427993003113402584582508155664401800520770441381...,
%F f'(1) = f(1) * Sum_{p prime} 2*(p^2 - p - 1) * log(p) /(p^4 + 2*p^3 + 1) = f(1) * 0.40523703144422392508596509911218523410441417240419849262346362977537989... = f(1) * A306072
%F and gamma is the Euler-Mascheroni constant A001620. (End)
%e From _Michael De Vlieger_, May 07 2017: (Start)
%e a(1) = 1 since 1 is the empty product; all divisors of 1 (i.e., 1) have a greatest common unitary divisor that is 1. 1 is a unitary divisor of all numbers n.
%e a(p) = 2 since 1 and p have greatest common unitary divisor 1.
%e a(6) = 4 since the divisor pairs {1, 6} and {2, 3} have greatest common unitary divisor 1.
%e a(24) = 8 since {1, 24}, {2, 12}, {3, 8}, {4, 6} have greatest unitary divisors {1, {1, 3, 8, 24}}, {{1, 2}, {1, 3, 4, 12}}, {{1, 3}, {1, 8}}, {1, 4}, {1, 2, 3, 6}}: 1 is the greatest common unitary divisor among all 4 pairs.
%e (End)
%t f[n_] := Select[Divisors[n], Function[d, CoprimeQ[d, n/d]]]; Table[DivisorSum[n, 1 &, Last@ Intersection[f@ #, f[n/#]] == 1 &], {n, 90}] (* _Michael De Vlieger_, May 07 2017 *)
%t f[p_, e_] := If[OddQ[e], e + 1, e]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 120] (* _Amiram Eldar_, Dec 19 2018 *)
%o (PARI) udivs(n) = {my(d = divisors(n)); select(x->(gcd(x, n/x)==1), d); }
%o gcud(n, m) = vecmax(setintersect(udivs(n), udivs(m)));
%o biudivs(n) = select(x->(gcud(x, n/x)==1), divisors(n));
%o a(n) = #biudivs(n);
%o (PARI) a(n)={my(f=factor(n)[,2]); prod(i=1, #f, my(e=f[i]); e + e % 2)} \\ _Andrew Howroyd_, Aug 05 2018
%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (X^3 - X^2 + X + 1) / ((X-1)^2 * (X+1)))[n], ", ")) \\ _Vaclav Kotesovec_, Jan 11 2024
%Y Cf. A222266, A188999, A293185 (indices of records), A340232, A350390.
%Y Cf. A000005, A034444 (unitary), A037445 (infinitary).
%K nonn,easy,mult
%O 1,2
%A _Michel Marcus_, May 07 2017
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