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A023194
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Numbers whose sum of divisors is prime.
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83
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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
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1,1
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COMMENTS
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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 sum-of-divisors 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. Adams-Watters, 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
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LINKS
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MAPLE
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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
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MATHEMATICA
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Select[ Range[ 100000 ], PrimeQ[ DivisorSigma[ 1, # ] ]& ] (* David W. Wilson *)
Prepend[Select[Range[1100]^2, PrimeQ[DivisorSigma[1, #]]&], 2] (* Harvey P. Dale, Dec 12 2010 *)
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PROG
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(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)/(p-1)), 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
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CROSSREFS
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Cf. A055638 (the square roots of the squares in this sequence).
Cf. A023195 (the primes produced by these n).
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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