

A023194


Numbers whose sum of divisors is prime.


77



2, 4, 9, 16, 25, 64, 289, 729, 1681, 2401, 3481, 4096, 5041, 7921, 10201, 15625, 17161, 27889, 28561, 29929, 65536, 83521, 85849, 146689, 262144, 279841, 458329, 491401, 531441, 552049, 579121, 597529, 683929, 703921, 707281, 734449, 829921, 1190281
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OFFSET

1,1


COMMENTS

All terms except the first are squares. Why?  Zak Seidov, Jun 10 2005
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.
"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.
"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."
All terms in this sequence also have a prime number of divisors.  Howard Berman (howard_berman(AT)hotmail.com), Oct 29 2008
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
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
Each term 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
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
Numbers of the form p^(q  1), where p is a prime, such that (p^q  1)/(p  1) is a prime. Then q must be a prime that does not divide p  1.  Thomas Ordowski, Nov 18 2017


LINKS



MAPLE

N:= 10^8: # to get all entries <= N
Primes:= select(isprime, [2, seq(2*i+1, i=1..floor((sqrt(N)1)/2))]):
P2:= select(t > (t > 2 and t < 1 + ilog2(N)), Primes):
cands:= {seq(seq([p, q], p=Primes), q=P2)} union {[2, 2]}:
f:= proc(pq) local t, j;
t:= pq[1]^(pq[2]1);
if t <= N and isprime((t*pq[1]1)/(pq[1]1)) then t else NULL fi
end proc:
map(f, cands);
# if using Maple 11 or earlier, uncomment the next line


MATHEMATICA

Select[ Range[ 100000 ], PrimeQ[ DivisorSigma[ 1, # ] ]& ] (* David W. Wilson *)
Prepend[Select[Range[1100]^2, PrimeQ[DivisorSigma[1, #]]&], 2] (* Harvey P. Dale, Dec 12 2010 *)


PROG

(PARI) for(x=1, 1000, if(isprime(sigma(x)), print(x))) /* Jorge Coveiro, Dec 23 2004 */
(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
(Magma) [n: n in [1..2*10^6]  IsPrime(SumOfDivisors(n))]; // Vincenzo Librandi, May 05 2015
(Python)
from sympy import isprime, divisor_sigma
A023194_list = [2]+[n**2 for n in range(1, 10**3) if isprime(divisor_sigma(n**2))] # Chai Wah Wu, Jul 14 2016


CROSSREFS

Cf. A055638 (the square roots of the squares in this sequence).
Cf. A023195 (the primes produced by these n).


KEYWORD

nonn,easy,nice


AUTHOR



STATUS

approved



