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A256430
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The least positive integer in A055744 divisible by A008578(n).
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2
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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