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

1, 4, 18, 50, 294, 1210, 1014, 578, 2166, 58190, 35322, 28830, 8214, 16810, 77658, 5588770, 219102, 4239858, 111630, 1481370, 1058610, 31974, 486798, 2824490, 871310, 56454, 102010, 1082118, 47330166, 71286, 536298, 677418, 6692790, 638146, 146646390, 4928622

Offset

1,2

Comments

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.

Links

Andrew Howroyd, Table of n, a(n) for n = 1..72

Mathematica

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 *)

Prog

(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

Crossrefs

Sequence in context: A180805 A120656 A256431 * A225263 A092349 A027659

Adjacent sequences: A256427 A256428 A256429 * A256431 A256432 A256433

Keyword

nonn

Author

David A. Corneth, Mar 28 2015

Extensions

a(16), a(18) and other terms corrected by Andrew Howroyd, Mar 01 2018

Status

approved