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%I #17 Sep 08 2022 08:45:57
%S 3,5,6,9,12,13,25,37,48,61,73,81,121,157,193,277,313,361,397,421,457,
%T 541,613,625,661,673,733,757,768,841,877,997,1093,1153,1201,1213,1237,
%U 1321,1381,1453,1621,1657,1753,1873,1933,1993,2017,2137,2341,2401,2473
%N Numbers n such that the arithmetic mean of the unitary divisors of n is a prime number.
%C Subsequence of A103826.
%C Similar to A187073, but considering unitary divisors, not prime divisors.
%C The odd terms of the sequence are: (1) the terms of A005383 (numbers n such that both n and (n+1)/2 are primes) and (2) the terms of A192618 (prime powers p^k with even exponents k>0 such that (1+p^k)/2 is prime).
%C [Note that A034448(n) and A034444(n) are multiplicative, so the arithmetic mean A034448(n)/A034444(n) is multiplicative with a(p^e) = (1+p^e)/2.]
%C The even terms of the sequence are 6, 12, 48, 768, 196608,... (no others < 10^10) with formula n = 3*2^(2^(k-1)) and averages 3, 5, 17, 257, 65537, ... (Fermat numbers, A000215).
%H Klaus Brockhaus, <a href="/A192577/b192577.txt">Table of n, a(n) for n = 1..10000</a>
%H A. Roldan Martinez, <a href="http://hojaynumeros.blogspot.com/">Numeros y hoja de calculo</a>
%e 48 has unitary divisors 1, 3, 16, 48 and (1+3+16+48)/4 = 17 is prime, therefore 48 is in the sequence.
%o (Magma) UnitaryDivisors:=func< n | [ d: d in Divisors(n) | Gcd(d, n div d) eq 1 ] >; [ n: n in [1..2500] | IsPrime(k) and s mod #U eq 0 where k is s div #U where s is &+U where U is UnitaryDivisors(n) ]; // _Klaus Brockhaus_, Jul 09 2011
%o (PARI) usigma(n)= {local(f, u=1); f=factor(n); for(i=1, matsize(f)[1], u*=(1+ f[i, 1]^f[i, 2])); return(u)}
%o ud(n)= {local (f, u); f=factor(n); u=2^(matsize(f)[1]); return(u) }
%o { for (n=2, 10^4, c=usigma(n)/ud(n); if (c==truncate(c),if(isprime(c), print1(n, ", ")))) }
%o // _Antonio Roldán_, Oct 08 2012
%Y Cf. A103826, A187073, A005383, A192618, A056798, A000215.
%K nonn
%O 1,1
%A _Antonio Roldán_, Jul 04 2011
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