%I
%S 2,4,9,16,25,64,289,729,1681,2401,3481,4096,5041,7921,10201,15625,
%T 17161,27889,28561,29929,65536,83521,85849,146689,262144,279841,
%U 458329,491401,531441,552049,579121,597529,683929,703921,707281,734449,829921,1190281
%N Numbers n such that sigma(n) (sum of divisors of n) is prime.
%C All numbers except the first are squares. Why?  _Zak Seidov_, Jun 10 2005
%C Answer from Gabe Cunningham (gcasey(AT)MIT.EDU): "From the fact that the sigma (the sumofdivisors function) is multiplicative, we can derive that the sigma(n) is even except when n is a square or twice a square.
%C "If n = 2*(2*k + 1)^2, that is, n is twice an odd square, then sigma(n) = 3*sigma((2*k + 1)^2). If n = 2*(2*k)^2, that is, n is twice an even square, then sigma(n) is only prime if n is a power of 2; otherwise we have sigma(n) = sigma(8*2^m) * sigma(k/2^m) for some positive integer m.
%C "So the only possible candidates for values of n other than squares such that sigma(n) is prime are odd powers of 2. But sigma(2^(2*m + 1)) = 2^(2*m + 2)  1 = (2^(m + 1) + 1) * (2^(m + 1)  1), which is only prime when m = 0, that is, when n = 2. So 2 is the only nonsquare n such that sigma(n) is prime."
%C All numbers on this list also have a prime number of divisors.  Howard Berman (howard_berman(AT)hotmail.com), Oct 29 2008
%C This is because 1 + p + ... + p^k is divisible by 1 + p + ... + p^j if k + 1 is divisible by j + 1.  _Robert Israel_, Jan 13 2015
%C From Gabe Cunningham's comment it follows that the alternate Mathematica program provided below is substantially more efficient as it only tests squares.  _Harvey P. Dale_, Dec 12 2010
%C Each number of this sequence is a prime power. This follows from the facts that sigma is multiplicative and sigma(n) > 1 for n > 1. Thus, for n > 1, a(n) is of the form a(n) = k^2 where k = p^m, with p prime, so the divisors of a(n) are {1, p, p^2, p^3, ..., (p^m)^2}, and this set is a multiplicative group (modulo q); if q is prime, q = sigma(k^2). Reciprocally, if q is a prime of the form 1 + p + p^2 + ... + p^(2*m), then q = sigma(p^(2*m)) (definition of sigma).  _Michel Lagneau_, Aug 17 2011, edited by _Franklin T. AdamsWatters_, Aug 17 2011
%C The sums of divisors of the even numbers in this sequence are the Mersenne primes, A000668. These even numbers are in A061652.  _Hartmut F. W. Hoft_, May 04 2015
%C Numbers n = p^(q  1), where p is a prime, such that (p^q  1)/(p  1) is prime. Then q must be a prime that does not divide p  1.  _Thomas Ordowski_, Nov 18 2017
%H T. D. Noe and David W. Wilson, <a href="/A023194/b023194.txt">Table of n, a(n) for n = 1..10000</a>
%p N:= 10^8: # to get all entries <= N
%p Primes:= select(isprime, [2,seq(2*i+1, i=1..floor((sqrt(N)1)/2))]):
%p P2:= select(t > (t > 2 and t < 1 + ilog2(N)), Primes):
%p cands:= {seq(seq([p,q],p=Primes), q=P2)} union {[2,2]}:
%p f:= proc(pq) local t,j;
%p t:= pq[1]^(pq[2]1);
%p if t <= N and isprime((t*pq[1]1)/(pq[1]1)) then t else NULL fi
%p end proc:
%p map(f,cands);
%p # if using Maple 11 or earlier, uncomment the next line
%p # sort(convert(%,list)); # _Robert Israel_, Jan 13 2015
%t Select[ Range[ 100000 ], PrimeQ[ DivisorSigma[ 1, # ] ]& ] (* _David W. Wilson_ *)
%t Prepend[Select[Range[1100]^2, PrimeQ[DivisorSigma[1,#]]&],2] (* _Harvey P. Dale_, Dec 12 2010 *)
%o (PARI) for(x=1,1000,if(isprime(sigma(x)),print(x))) /* Jorge Coveiro (jorgecoveiro(AT)yahoo.com), Dec 23 2004 */
%o (PARI) list(lim)=my(v=List([2])); forprime(p=2,sqrtint(lim\=1), if(isprime(p^2+p+1), listput(v,p^2))); forstep(e=4,logint(lim,2),2, forprime(p=2,sqrtnint(lim,e), if(isprime((p^(e+1)1)/(p1)), listput(v,p^e)))); Set(v) \\ _Charles R Greathouse IV_, Aug 17 2011; updated Jul 22 2016
%o (MAGMA) [n: n in [1..2*10^6]  IsPrime(SumOfDivisors(n))]; // _Vincenzo Librandi_, May 05 2015
%o (Python)
%o from sympy import isprime, divisor_sigma
%o A023194_list = [2]+[n**2 for n in range(1,10**3) if isprime(divisor_sigma(n**2))] # _Chai Wah Wu_, Jul 14 2016
%Y Cf. A055638 (the square roots of the squares in this sequence).
%Y Cf. A023195 (the primes produced by these n).
%K nonn,easy,nice
%O 1,1
%A _David W. Wilson_
