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Primes such that z(p) is odd where z(n) is A214028(n).
20

%I #54 Jun 26 2024 09:38:26

%S 5,13,29,37,53,61,101,109,137,149,157,173,181,197,229,269,277,293,317,

%T 349,373,389,397,421,461,509,521,541,557,569,593,613,653,661,677,701,

%U 709,733,757,773,797,821,829,853,857,877,941,953,997,1013,1021,1061,1069

%N Primes such that z(p) is odd where z(n) is A214028(n).

%C From _Jianing Song_, Aug 13 2019: (Start)

%C Primes p with 4 zeros in a fundamental period of A000129 mod p, that is, primes p such that A214027(p) = 4. For a proof of the equivalence between A214027(p) = 4 and A214028(p) being odd, see Section 2 of my link below.

%C For p > 2, p is in this sequence if and only if A175181(p) == 4 (mod 8).

%C This sequence contains all primes congruent to 5 modulo 8. This corresponds to case (1) for k = 6 in the Conclusion of Section 1 of my link below.

%C Conjecturely, this sequence has density 7/24 in the primes. (End) [Comment rewritten by _Jianing Song_, Jun 16 2024 and Jun 20 2024]

%H Jianing Song, <a href="/A261580/b261580.txt">Table of n, a(n) for n = 1..1280</a>

%H Bernadette Faye and Florian Luca, <a href="http://arxiv.org/abs/1508.05714">Pell Numbers whose Euler Function is a Pell Number</a>, arXiv:1508.05714 [math.NT], 2015.

%H Jianing Song, <a href="/A053027/a053027.pdf">Lucas sequences and entry point modulo p</a>

%e The smallest Pell number divisible by the prime 5 has index 3, which is odd, so 5 is in the sequence.

%t f[n_] := Block[{k = 1}, While[Mod[Simplify[((1 + Sqrt@ 2)^k - (1 - Sqrt@ 2)^k)/(2 Sqrt@ 2)], n] != 0, k++]; k]; Select[Prime@ Range@ 180, OddQ@ f@ # &] (* _Michael De Vlieger_, Aug 25 2015 *)

%o (PARI) pell(n) = polcoeff(Vec(x/(1-2*x-x^2) + O(x^(n+1))), n);

%o z(n) = {k=1; while (pell(k) % n, k++); k;}

%o lista(nn) = {forprime(p=2, nn, if (z(p) % 2, print1(p, ", ")););}

%o (PARI) forprime(p=2, 1100, if(A214027(p)==4, print1(p, ", "))) \\ _Jianing Song_, Aug 13 2019

%Y Cf. A214028, A261581.

%Y Cf. also A175181.

%Y Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.

%Y | m=1 | m=2 | m=3

%Y -----------------------------+----------+----------+---------

%Y The sequence {x(n)} | A000045 | A000129 | A006190

%Y The sequence {w(k)} | A001176 | A214027 | A322906

%Y Primes p such that w(p) = 1 | A112860* | A309580 | A309586

%Y Primes p such that w(p) = 2 | A053027 | A309581 | A309587

%Y Primes p such that w(p) = 4 | A053028 | this seq | A309588

%Y Numbers k such that w(k) = 1 | A053031 | A309583 | A309591

%Y Numbers k such that w(k) = 2 | A053030 | A309584 | A309592

%Y Numbers k such that w(k) = 4 | A053029 | A309585 | A309593

%Y * and also A053032 U {2}

%K nonn

%O 1,1

%A _Michel Marcus_, Aug 25 2015