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A256430 The least positive integer in A055744 divisible by A008578(n).

%I A256430 #29 Mar 02 2018 11:53:50
%S A256430 1,4,18,50,294,1210,1014,578,2166,58190,35322,28830,8214,16810,77658,
%T A256430 5588770,219102,4239858,111630,1481370,1058610,31974,486798,2824490,
%U A256430 871310,56454,102010,1082118,47330166,71286,536298,677418,6692790,638146,146646390,4928622
%N A256430 The least positive integer in A055744 divisible by A008578(n).
%C A256430 Subset of A256431. Elements from this sequence can be used to find elements from A256431. For example, 18 and 50 are the least number in this sequence divisible by 3 and 5 respectively. These numbers can be used to find the least number in A055744 divisible by both 3 and 5 as follows: 18 = 2^1 * 3^2 and 50 = 2^1 * 5^2. 'Order' these factors together: 2^1|2^1|3^2|5^2. For two consecutive factors, if they have the same base, remove the one with the highest exponent. Leaves 2^1|3^2|5^2. Multiply these factors together. Gives 2 * 3^2 * 5^2 = 450. So 450 is in A256431. This method can be applied recursively to find the least n in A055744 divisible by 3, 5 and 7, for example; applying this to 294 and 450 gives 7350 which is the least element in A055744 divisible by primes 3, 5 and 7.
%H A256430 Andrew Howroyd, <a href="/A256430/b256430.txt">Table of n, a(n) for n = 1..72</a>
%t A256430 With[{s = Select[Range[10^6], SameQ @@ Map[FactorInteger[#][[All, 1]] &, {#, EulerPhi@ #}] &]}, TakeWhile[#, IntegerQ] &@ Table[SelectFirst[s, Divisible[#, p] &], {p, {1}~Join~Prime@ Range@ 30}]] (* _Michael De Vlieger_, Feb 22 2018 *)
%o A256430 (PARI) a(n)={my(m=0, p=if(n==1,1,prime(n-1))); until(my(f=factor(m)); f[, 1]==factor(eulerphi(f))[, 1], m+=p); m} \\ _Andrew Howroyd_, Mar 01 2018
%K A256430 nonn
%O A256430 1,2
%A A256430 _David A. Corneth_, Mar 28 2015
%E A256430 a(16), a(18) and other terms corrected by _Andrew Howroyd_, Mar 01 2018