|
%I #57 Jun 08 2015 02:27:57
%S 2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,18,21,24,29,30,32,33,35,38,41,
%T 42,50,53,54,56,58,59,66,69,89,90,94,95,97,99,101,104,117,118,120,135,
%U 138,160,167,181,191,210,221,237,242,247
%N Indices of Pell numbers having exactly one primitive prime factor.
%C Conjecture: The n-th Pell number A000129(n) has a primitive prime factor for all n > 1. (The n-th Fibonacci number A000045(n) has a primitive prime factor for all n except n = 0, 1, 2, 6, and 12.)
%C For prime p, all prime factors of Pell(p) are primitive. Hence the only primes in this sequence are the prime numbers in A096650, which gives the indices of prime Pell numbers.
%e Pell(1) = 1, which has no prime factors, so 1 is not in this sequence.
%e Pell(4) = 12 = 2^2 * 3, but 2 is not a primitive prime factor, and 3 is the only primitive prime factor of Pell(4), so 4 is in this sequence.
%e Pell(5) = 29, which is a prime and the only primitive prime factor of itself, so 5 is in this sequence.
%e Pell(12) = 13860 = 2^2 * 3^2 * 5 * 7 * 11, but none of 2, 3, 5, 7 is a primitive prime factor, and 11 is the only primitive prime factor of Pell(12), so 12 is in this sequence.
%e Pell(14) = 80782 = 2 * 13^2 * 239, but neither 2 nor 13 is a primitive prime factor, and 239 is the only primitive prime factor of Pell(14), so 14 is in this sequence.
%e Pell(19) = 6625109 = 37 * 179057, both of which are primitive prime factors of Pell(19), so 19 is not in this sequence.
%t Select[Range[1000], PrimePowerQ[(1-Sqrt[2])^EulerPhi[#]*Cyclotomic[#, (1+Sqrt[2])/(1-Sqrt[2])]/GCD[Cyclotomic[#, (1+Sqrt[2])/(1-Sqrt[2])], # ]]&] - _Eric Chen_, Dec 12 2014
%t pell[n_] := pell[n] = ((1+Sqrt[2])^n-(1-Sqrt[2])^n )/(2*Sqrt[2]) // Round; primitivePrimeFactors[n_] := Cases[FactorInteger[pell[n]][[All, 1]], p_ /; And @@ (GCD[p, #] == 1 & /@ Array[pell, n-1])]; Reap[For[n=2, n <= 200, n++, If[Length[primitivePrimeFactors[n]] == 1, Print[n]; Sow[n]]]][[2, 1]] (* _Jean-François Alcover_, Dec 12 2014 *)
%o (PARI) pell(n) = imag((1 + quadgen(8))^n);
%o isok(pf, vp) = sum(i=1, #pf, vecsearch(vp, pf[i]) == 0) == 1;
%o lista(nn) = {vp = []; for (n=2, nn, pf = factor(pell(n))[,1]; if (isok(pf, vp), print1(n, ", ")); vp = vecsort(concat(vp, pf),, 8););} \\ _Michel Marcus_, Nov 29 2014
%Y Cf. A152012 (for Fibonacci numbers).
%Y Cf. A000129, A246556, A096650, A086383, A008555, A175181.
%K nonn
%O 1,1
%A _Eric Chen_, Nov 29 2014
%E Two incorrect terms (72 and 110) deleted by _Colin Barker_, Nov 29 2014
%E More terms from _Colin Barker_, Nov 30 2014
|
