A167766 Minimum numbers whose phi of phi are multiples of the n-th prime: the n-th term is the minimum integer x such that: prime(n) | phi(phi(x)), prime(n) being the n-th prime.

5, 19, 23, 59, 47, 107, 479, 383, 283, 467, 1367, 1187, 167, 347, 1319, 643, 2837, 2203, 2153, 3413, 587, 5693, 1997, 359, 5827, 1619, 2063, 2999, 4799, 3167, 1019, 1579, 5483, 3343, 7159, 3023, 12569, 1307, 4679, 2083, 719, 3623, 4597, 3863, 18917, 4783

Offset

1,1

Comments

These minimal integers are always prime. To be clear, the phi function referred to here is Euler's totient function.

Links

Donovan Johnson, Table of n, a(n) for n = 1..10000

Eric Weisstein's World of Mathematics, Euler's Totient Function

Example

The first term is 5. phi(5) = 4 and phi(4)=2. 2 is a multiple of the first prime 2. 5 is the lowest such number x where 2 divides phi(phi(x)).

Maple

with(numtheory): P:=proc(n) local a, k; a:=ithprime(n);
for k from 1 to 10^3 do if frac(phi(phi(ithprime(k)))/a)=0
then RETURN(ithprime(k)); break; fi; od; end:
seq(P(i), i=1..46); # Paolo P. Lava, Oct 10 2018

Mathematica

a[n_] := (p=Prime[n]; k=1; While[k++; x=Prime[k]; Mod[ EulerPhi[ EulerPhi[x]], p] != 0]; x); Table[a[n], {n, 50}] (* Jean-François Alcover, Sep 14 2011 *)

Prog

(PARI) /* not the most efficient implementation */ ppp(a, b)= { forprime(p=a, b, v = 2*p + 1; v2 = 1; minv = 100000000; while (v2 <= minv || v <=minv, /* print ("Checking ", v, " for ", p); */ while(!isprime(v), v += 2*p /*; print ("Checking ", v, " for ", p)*/ ); if (v%(p*p)==1, /* don't do this step if: p^2 | v-1 */ v2 = v , v2 = 2*v + 1; while (!isprime(v2) && v2 < minv, v2 += 2*v ) ); if (v2 < minv, minv = v2; ); v += 2*p ); print (p, " => ", minv) ) }

Crossrefs

Cf. A010554.

Sequence in context: A191084 A146509 A062340 * A106957 A236167 A022143

Adjacent sequences: A167763 A167764 A167765 * A167767 A167768 A167769

Keyword

easy,nice,nonn

Author

Fred Schneider, Nov 11 2009

Status

approved