|
%I #15 Jan 09 2018 00:25:16
%S 8,27,32,125,243,512,1331,2048,32768,50653,79507,103823,131072,161051,
%T 177147,357911,1419857,2097152,2248091,3869893,11089567,15813251,
%U 16974593,20511149,28934443,69343957,115501303,147008443,263374721,536870912,844596301,1284365503,1305751357
%N Prime powers p^e with odd exponent e such that rho(p^(e+1)) is prime, where rho is A206369.
%C Along with A065508, these are the integers mentioned at the bottom of page 4 of the Iannucci link. Let x = p^e, and q = rho(p^(e+1)), then x/rho(x) = (x*p*q)/rho(x*p*q). An example with A065508 is 3, for which rho(3) is 7, so 3 and 3*3*7 have the same x/rho(x) ratio, 3/2.
%C Note that there are other "rho-friendly pairs" that have a different, yet simple, form like for instance 7^5 and 7^8*117307.
%C Number of terms < 10^k: 1, 3, 6, 8, 11, 16, 20, 26, 31, 46, 73, 110, 198, 327, 611, 1157, 2135, 4107, 7724, 14771, 28610, etc. - _Robert G. Wilson v_, Jan 07 2018
%H Robert G. Wilson v, <a href="/A297868/b297868.txt">Table of n, a(n) for n = 1..28610</a>
%H Douglas E. Iannucci, <a href="http://www.integers-ejcnt.org/g41/g41.Abstract.html">On a variation of perfect numbers</a>, INTEGERS: Electronic Journal of Combinatorial Number Theory, 6 (2006), #A41.
%e 8=2^3 is a term because rho(2*8)=11 is prime, so 8 and 8*2*11 have the same x/rho(x) ratio, 8/5.
%t rho[n_] := n*DivisorSum[n, LiouvilleLambda[#]/# &]; fQ[n_] := Block[{p = FactorInteger[n][[1, 1]]}, PrimeQ[ rho[p n]]]; mx = 10^9; lst = Sort@ Flatten@ Table[ Prime[n]^e, {n, PrimePi[mx^(1/3)]}, {e, 3, Floor@ Log[ Prime@ n, mx], 2}]; Select[lst, fQ] (* _Robert G. Wilson v_, Jan 07 2018 *)
%o (PARI) rhope(p, e) = my(s=1); for(i=1, e, s=s*p + (-1)^i); s;
%o lista(nn) = {for (n=1, nn, if ((e = isprimepower(n,&p)) && (e > 1) && (e % 2) && isprime(rhope(p,e+1)), print1(n, ", ");););}
%Y Cf. A065508, A074268, A206369.
%K nonn
%O 1,1
%A _Michel Marcus_, Jan 07 2018
|
