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A167766
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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.
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3
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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
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1,1
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COMMENTS
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These minimal integers are always prime. To be clear, the phi function referred to here is Euler's totient function.
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LINKS
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EXAMPLE
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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)).
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MAPLE
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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:
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MATHEMATICA
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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 *)
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PROG
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(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) ) }
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CROSSREFS
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KEYWORD
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easy,nice,nonn
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AUTHOR
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STATUS
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approved
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