

A167766


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


3



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
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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}] (* JeanFrançois 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  v1 */ 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



