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A061026 Smallest number m such that phi(m) is divisible by n, where phi = Euler totient function A000010. 9


%S 1,3,7,5,11,7,29,15,19,11,23,13,53,29,31,17,103,19,191,25,43,23,47,35,

%T 101,53,81,29,59,31,311,51,67,103,71,37,149,191,79,41,83,43,173,69,

%U 181,47,283,65,197,101,103,53,107,81,121,87,229,59,709,61,367,311,127,85

%N Smallest number m such that phi(m) is divisible by n, where phi = Euler totient function A000010.

%D M. J. Knight, Comment with Solution to 10837, American Mathematical Monthly, 2001.

%H T. D. Noe, <a href="/A061026/b061026.txt">Table of n, a(n) for n=1..1000</a>

%H Ho-joo Lee and Gerald Myerson, <a href="http://www.jstor.org/stable/3647787">Consecutive Integers Whose Totients Are Multiples of n: 10837</a>, The American Mathematical Monthly, Vol. 110, No. 2 (Feb., 2003), pp. 158-159.

%H P. Moree, <a href="http://www.fq.math.ca/Scanned/33-4/moree.pdf">On an arithmetical function related to Euler's totient and the discriminator</a> Fib. Quart. (1995).

%H József Sándor, <a href="http://nntdm.net/volume-15-2009/number-3/01-08/">On the Euler minimum and maximum functions</a>, Notes on Number Theory and Discrete Mathematics, Volume 15, 2009, Number 3, Pages 1—8.

%F Sequence is unbounded; a(n) <= n^2 since phi(n^2) is always divisible by n.

%F If n+1 is prime a(n)=n+1.

%F a(n) = min( k : phi(k) == 0 mod(n) )

%e a(48) = 65 because phi(65) = phi(5)phi(13) = (4)(12) = 48 and no smaller integer has phi(n) = 48.

%t a = ConstantArray[1, 64]; k = 1; While[Length[vac = Rest[Flatten[Position[a, 1]]]] > 0, k++; a[[Intersection[Divisors[EulerPhi[k]], vac]]] *= k]; a (* _Ivan Neretin_, May 15 2015 *)

%o (PARI) for(n=1,100,s=1; while(eulerphi(s)%n>0,s++); print1(s,","))

%Y Cf. A000010, A066674, A066675, A066676, A066678, A067005.

%Y Cf. A233516, A233517 (records).

%Y Cf. A005179 (analog for number of divisors), A070982 (analog for sum of divisors).

%K nonn

%O 1,2

%A Melvin J. Knight (knightmj(AT)juno.com), May 25 2001

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