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Numbers n such that tau(n) - tau(n+1) = 1.
4

%I #26 Dec 27 2020 03:34:01

%S 4,8,81,441,625,1089,2024,2401,3025,3968,5624,5929,6561,6723,7569,

%T 8281,8463,12321,13225,13688,14161,14641,14884,15375,16641,20164,

%U 21608,24025,25921,26895,28561,34225,46225,55225,55695,61009,62001,67081

%N Numbers n such that tau(n) - tau(n+1) = 1.

%C a(n) cannot be a prime.

%C a(n) or a(n) + 1 is a perfect square as its number of divisors must be odd. - _David A. Corneth_, Dec 26 2020

%H David A. Corneth, <a href="/A068208/b068208.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Enrique PĂ©rez Herrero)

%F n such that A000005(n) - A000005(n+1) = 1.

%t Select[Range[10^5], DivisorSigma[0, #] - DivisorSigma[0, # + 1] == 1 &] (* _Michael De Vlieger_, Dec 02 2015 *)

%t Position[Differences[DivisorSigma[0,Range[70000]]],-1]//Flatten (* _Harvey P. Dale_, May 22 2020 *)

%o (PARI) isok(n) = numdiv(n) - numdiv(n+1) == 1; \\ _Michel Marcus_, Dec 02 2015

%Y Cf. A000005.

%K nonn,easy

%O 1,1

%A _Benoit Cloitre_, Mar 23 2002