A182340 - OEIS

%I #15 Apr 11 2020 06:11:02

%S 5,10,15,20,25,30,32,35,40,45,50,55,60,65,70,75,80,85,90,95,96,100,

%T 105,110,115,120,125,130,135,140,145,150,155,160,165,170,175,180,185,

%U 190,195,200,205,210,215,220,224,225,230,235,240,243,245,250,255,260

%N List of positive integers whose prime tower factorization, as defined in comments, contains the prime 5.

%C The prime tower factorization of a number can be recursively defined as follows:

%C (0) The prime tower factorization of 1 is itself

%C (1) To find the prime tower factorization of an integer n>1, let n = p1^e1 * p2^e2 * ... * pk^ek be the usual prime factorization of n. Then the prime tower factorization is given by p1^(f1) * p2^(f2) * ... * pk^(fk), where fi is the prime tower factorization of ei.

%H Amiram Eldar, <a href="/A182340/b182340.txt">Table of n, a(n) for n = 1..10000</a>

%H Patrick Devlin and Edinah Gnang, <a href="http://arxiv.org/abs/1204.5251">Primes Appearing in Prime Tower Factorization</a>, arXiv:1204.5251 [math.NT], 2012-2014.

%p # The integer n is in this sequence if and only if

%p # conatinsPrimeInTower(5, n) returns true

%p conatinsPrimeInTower:=proc(q, n) local i, L, currentExponent; option remember;

%p if n <= 1 then return false: end if;

%p if type(n/q, integer) then return true: end if;

%p L := ifactors(n)[2];

%p for i to nops(L) do currentExponent := L[i][2];

%p if containsPrimeInTower(q, currentExponent) then return true: end if

%p end do;

%p return false:

%p end proc:

%t containsPrimeInTower[q_, n_] := containsPrimeInTower[q, n] = Module[{i, L, currentExponent}, If[n <= 1, Return[False]]; If[IntegerQ[n/q], Return[True]]; L = FactorInteger[n]; For[i = 1, i <= Length[L], i++, currentExponent = L[[i, 2]]; If[containsPrimeInTower[q, currentExponent], Return[True]]]; Return[False]];

%t Select[Range[300], containsPrimeInTower[5, #]&] (* _Jean-François Alcover_, Jan 22 2019, from Maple *)

%Y Cf. A182318.

%K nonn

%O 1,1

%A _Patrick Devlin_, Apr 25 2012