OFFSET
1,1
COMMENTS
a(n) cannot be a prime.
a(n) or a(n) + 1 is a perfect square as its number of divisors must be odd. - David A. Corneth, Dec 26 2020
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from Enrique Pérez Herrero)
MATHEMATICA
Select[Range[10^5], DivisorSigma[0, #] - DivisorSigma[0, # + 1] == 1 &] (* Michael De Vlieger, Dec 02 2015 *)
Position[Differences[DivisorSigma[0, Range[70000]]], -1]//Flatten (* Harvey P. Dale, May 22 2020 *)
PROG
(PARI) isok(n) = numdiv(n) - numdiv(n+1) == 1; \\ Michel Marcus, Dec 02 2015
(PARI) list(lim) = {my(d, e, k2); for(k = 2, lim, e = factor(k)[, 2]; d = vecprod(apply(x -> 2*x + 1, e)); k2 = k^2; if(numdiv(k2-1) == d+1, print1(k2-1, ", ")); if(numdiv(k2+1) == d-1, print1(k2, ", "))); } \\ Amiram Eldar, Apr 20 2025
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Benoit Cloitre, Mar 23 2002
STATUS
approved
