[final] Sum of divisors of n = prime ?

Wouter Meeussen eu000949 at pophost.eunet.be
Sat Aug 21 20:04:21 CEST 1999

hi all,

futility: learning after a day which bits are hard, and which are easy.

Easy: 1+x+x^2+...+x^(p*q-1) can of course always be split into
[x^(p*q)-1]/(x-1) == (x^p -1)/(x-1)  *  [( (x^p) ^q)-1]/((x^p) -1)

Hard: ( A023194 )
 = {2,4,9,16,25,64,289,729,1681,2401,3481,4096,5041,7921,10201,

these are {base,exponent}:

here I tried all integers, not just those that are a power of a (single) prime.
so I get some big bases with a small exponent like 167^2.

The reverse method checks the powers of the (low) primes
in a Mersenne-like search:

Select[Range[1,100],PrimeQ[(5^# -1)/4]&] gives {3,7,11,13,47}

{2,{2,3,5,7,13,17,19,31,61,89,107,127,521}},=A000043 (* =Mersenne *)
{3,{3,7,13,71,103}},  =A004060
{5,{3,7,11,13,47,127,149,181}}, =A004061

DivisorSigma[1,  19^(107 -1)]= ((19^107) -1)/(19 -1) is prime.

If there is no sane method to generate Mersenne primes,
then neither is there a chance for these babies.
Realm of the number-crunchers.

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