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Numbers k such that sigma(sigma(k)) = 2*phi(k).
3

%I #16 Feb 27 2024 03:02:35

%S 13,43,109,151,883,2143,116581,388537,1711663,2498227,4004107,5550331,

%T 12641137,13617361,18591967,20755393,22998397,26838523,29308291,

%U 34564351,36300841,44829073,82368469,149460841,184988197,238225003,252757891,340428853

%N Numbers k such that sigma(sigma(k)) = 2*phi(k).

%C If p = 2^k + 3 and both numbers p & q = (1/2)*(p^2 - 3p - 2) are primes then q is in the sequence, because sigma(sigma(q)) = sigma(q+1) = sigma((1/2)*(p-3)*p) = sigma(2^(n-1)*p) = (2^n-1)*(p+1) = (p-4)*(p+1) = p^2 - 3p - 4 = 2q - 2 = 2*phi(q). 13, 43, 151 & 2143 are such terms corresponding to n = 2, 3, 4 & 6.

%H Donovan Johnson, <a href="/A135241/b135241.txt">Table of n, a(n) for n = 1..54</a> (terms < 10^10)

%e sigma(sigma(36300841)) = sigma(36313684) = 72576000 = 2*36288000 = 2*phi(36300841) so 36300841 is in the sequence.

%t lst = {}; fQ[n_] := DivisorSigma[1, DivisorSigma[1, n]] == 2 EulerPhi@n; Do[ If[ fQ@n, AppendTo[lst, n]; Print@n], {n, 252000000}] (* _Robert G. Wilson v_, Jan 01 2008 *)

%o (PARI) is(n) = sigma(sigma(n))==2*eulerphi(n) \\ _Felix Fröhlich_, May 18 2019

%K nonn

%O 1,1

%A _Farideh Firoozbakht_, Dec 30 2007

%E More terms from _Robert G. Wilson v_, Jan 01 2008