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List of positive integers whose prime tower factorization, as defined in comments, does not contain the prime 3.
2

%I #23 Apr 11 2020 06:10:23

%S 1,2,4,5,7,10,11,13,14,16,17,19,20,22,23,25,26,28,29,31,32,34,35,37,

%T 38,41,43,44,46,47,49,50,52,53,55,58,59,61,62,65,67,68,70,71,73,74,76,

%U 77,79,80,82,83,85,86,89,91,92,94,95,97,98,100,101,103,106

%N List of positive integers whose prime tower factorization, as defined in comments, does not contain the prime 3.

%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.

%C As an alternative definition, let I(n) be the indicator function for the set of positive integers whose prime tower factorization does not contain a 3. Then I(n) is the multiplicative function satisfying I(p^k) = I(k) for p prime not equal to 3, and I(3^k) = 0.

%H Amiram Eldar, <a href="/A182337/b182337.txt">Table of n, a(n) for n = 1..10000</a>

%H Patrick Devlin and Edinah Gnang, <a href="https://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 # containsPrimeInTower(3, n) returns false

%p containsPrimeInTower:=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:

%p select(x-> not containsPrimeInTower(3,x), [$1..120])[];

%t indic[1] = 1; indic[n_] := indic[n] = Switch[f = FactorInteger[n], {{3, _}}, 0, {{_, _}}, indic[f[[1, 2]] ], _, Times @@ (indic /@ (Power @@@ f))]; Select[Range[120], indic[#] == 1&] (* _Jean-François Alcover_, Feb 25 2018 *)

%Y Cf. A182318.

%K nonn

%O 1,2

%A _Patrick Devlin_, Apr 25 2012