A214027 -id:A214027 - cp's OEIS frontend

A261580 Primes p such that A214028(p) is odd.

5, 13, 29, 37, 53, 61, 101, 109, 137, 149, 157, 173, 181, 197, 229, 269, 277, 293, 317, 349, 373, 389, 397, 421, 461, 509, 521, 541, 557, 569, 593, 613, 653, 661, 677, 701, 709, 733, 757, 773, 797, 821, 829, 853, 857, 877, 941, 953, 997, 1013, 1021, 1061, 1069

Offset: 1

Michel Marcus, Aug 25 2015

nonn

From Jianing Song, Aug 13 2019: (Start)
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.
For p > 2, p is in this sequence if and only if A175181(p) == 4 (mod 8).
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.
Conjecturely, since (k+2)/2 = 4 is a square, this sequence has density 7/24 in the primes; see the end of Section 1 of my link. (End) [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 20 2024]
The conjecture above is an analog of Hasse's result that the set {p prime : ord(2,p) is odd} has density 7/24 in the primes, where ord(a,m) is the multiplicative order of a modulo m; see A014663. - Jianing Song, Jun 26 2025

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

Jianing Song, Table of n, a(n) for n = 1..1280
Bernadette Faye and Florian Luca, Pell Numbers whose Euler Function is a Pell Number, arXiv:1508.05714 [math.NT], 2015.
Jianing Song, Lucas sequences and entry point modulo p

Cf. A214028, A261581.
Cf. also A175181.
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.
                             |   m=1    |   m=2    |   m=3
-----------------------------+----------+----------+---------
The sequence {x(n)}          | A000045  | A000129  | A006190
The sequence {w(k)}          | A001176  | A214027  | A322906
Primes p such that w(p) = 1  | A112860* | A309580  | A309586
Primes p such that w(p) = 2  | A053027  | A309581  | A309587
Primes p such that w(p) = 4  | A053028  | this seq | A309588
Numbers k such that w(k) = 1 | A053031  | A309583  | A309591
Numbers k such that w(k) = 2 | A053030  | A309584  | A309592
Numbers k such that w(k) = 4 | A053029  | A309585  | A309593
* and also A053032 U {2}

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 *)

pell(n) = polcoeff(Vec(x/(1-2*x-x^2) + O(x^(n+1))), n);
z(n) = {k=1; while (pell(k) % n, k++); k;}
lista(nn) = {forprime(p=2, nn, if (z(p) % 2, print1(p, ", ")););}

forprime(p=2, 1100, if(A214027(p)==4, print1(p, ", "))) \\ Jianing Song, Aug 13 2019

A214028 Entry points for the Pell sequence: smallest k such that n divides A000129(k).

Original entry on oeis.org

1, 2, 4, 4, 3, 4, 6, 8, 12, 6, 12, 4, 7, 6, 12, 16, 8, 12, 20, 12, 12, 12, 22, 8, 15, 14, 36, 12, 5, 12, 30, 32, 12, 8, 6, 12, 19, 20, 28, 24, 10, 12, 44, 12, 12, 22, 46, 16, 42, 30, 8, 28, 27, 36, 12, 24, 20, 10, 20, 12, 31, 30, 12, 64, 21, 12, 68, 8, 44, 6, 70, 24, 36, 38

Offset: 1

Art DuPre, Jul 04 2012

nonn

Conjecture: A175181(n)/A214027(n) = a(n). This says that the zeros appear somewhat uniformly in a period. The second zero in a period is exactly where n divides the first Lucas number, so this relationship is not really surprising.
From Jianing Song, Aug 29 2018: (Start)
The comment above is correct, since n divides A000129(k*a(n)) for all integers k and clearly a(n) divides A175181(n), so the zeros appear uniformly.
a(n) <= 4*n/3 for all n, where the equality holds iff n is a power of 3.
(End)

			11 first divides the term A000129(12) = 13860 = 2*3*5*7*11.

G. C. Greubel, Table of n, a(n) for n = 1..10000
Bernadette Faye and Florian Luca, Pell Numbers whose Euler Function is a Pell Number, arXiv:1508.05714 [math.NT], 2015 (called z(n)).
N. Robbins, On Pell numbers of the form p*x^2, where p is prime, Fib. Quart. 22 (4) (1984) 340-348, definition 1.

Cf. A001175, A001176, A001177, A214027, A175181.

A214028 := proc(n)
    local a000129,k ;
    a000129 := [1,2,5] ;
    for k do
        if modp(a000129[1],n) = 0 then
            return k;
        end if;
        a000129[1] := a000129[2] ;
        a000129[2] := a000129[3] ;
        a000129[3] := 2*a000129[2]+a000129[1] ;
    end do:
end proc:
seq(A214028(n),n=1..40); # R. J. Mathar, May 26 2016

a[n_] := With[{s = Sqrt@ 2}, ((1 + s)^n - (1 - s)^n)/(2 s)] // Simplify; Table[k = 1; While[Mod[a[k], n] != 0, k++]; k, {n, 80}] (* Michael De Vlieger, Aug 25 2015, after Michael Somos at A000129 *)
Table[k = 1; While[Mod[Fibonacci[k, 2], n] != 0, k++]; k, {n, 100}] (* G. C. Greubel, Aug 10 2018 *)

pell(n) = polcoeff(Vec(x/(1-2*x-x^2) + O(x^(n+1))), n);
a(n) = {k=1; while (pell(k) % n, k++); k;} \\ Michel Marcus, Aug 25 2015

If p^2 does not divide A000129(a(p)) (that is, p is not in A238736) then a(p^e) = a(p)*p^(e - 1). If gcd(m, n) = 1 then a(mn) = lcm(a(m), a(n)). - Jianing Song, Aug 29 2018

A309580 Primes p with 1 zero in a fundamental period of A000129 mod p.

Original entry on oeis.org

2, 7, 23, 31, 41, 47, 71, 79, 103, 127, 151, 167, 191, 199, 223, 239, 263, 271, 311, 313, 353, 359, 367, 383, 409, 431, 439, 457, 463, 479, 487, 503, 599, 607, 631, 647, 719, 727, 743, 751, 761, 809, 823, 839, 863, 887, 911, 919, 967, 983, 991, 1031, 1039, 1063, 1087, 1103, 1129, 1151, 1201, 1223, 1231, 1279

Offset: 1

Jianing Song, Aug 10 2019

nonn

Primes p such that A214027(p) = 1.
For p > 2, p is in this sequence if and only if A175181(p) == 2 (mod 4), and if and only if A214028(p) == 2 (mod 4). For a proof of the equivalence between A214027(p) = 1 and A214028(p) == 2 (mod 4), see Section 2 of my link below.
This sequence contains all primes congruent to 7 modulo 8. This corresponds to case (3) for k = 6 in the Conclusion of Section 1 of my link below.
Conjecturely, since (k+2)/2 = 4 is a square, this sequence has density 7/24 in the primes; see the end of Section 1 of my link. [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]
The conjecture above is an analog of Hasse's result that the set {p prime : ord(2,p) is odd} has density 7/24 in the primes, where ord(a,m) is the multiplicative order of a modulo m; see A014663. - Jianing Song, Jun 26 2025

Jianing Song, Table of n, a(n) for n = 1..1280
Jianing Song, Lucas sequences and entry point modulo p

Cf. A175181, A214028.
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.
                             |   m=1    |   m=2    |   m=3
-----------------------------+----------+----------+---------
The sequence {x(n)}          | A000045  | A000129  | A006190
The sequence {w(k)}          | A001176  | A214027  | A322906
Primes p such that w(p) = 1  | A112860* | this seq | A309586
Primes p such that w(p) = 2  | A053027  | A309581  | A309587
Primes p such that w(p) = 4  | A053028  | A261580  | A309588
Numbers k such that w(k) = 1 | A053031  | A309583  | A309591
Numbers k such that w(k) = 2 | A053030  | A309584  | A309592
Numbers k such that w(k) = 4 | A053029  | A309585  | A309593
* and also A053032 U {2}

forprime(p=2, 1300, if(A214027(p)==1, print1(p, ", ")))

A309581 Primes p with 2 zeros in a fundamental period of A000129 mod p.

Original entry on oeis.org

3, 11, 17, 19, 43, 59, 67, 73, 83, 89, 97, 107, 113, 131, 139, 163, 179, 193, 211, 227, 233, 241, 251, 257, 281, 283, 307, 331, 337, 347, 379, 401, 419, 433, 443, 449, 467, 491, 499, 523, 547, 563, 571, 577, 587, 601, 617, 619, 641, 643, 659, 673, 683, 691

Offset: 1

Jianing Song, Aug 10 2019

nonn

Primes p such that A214027(p) = 2.
For p > 2, p is in this sequence if and only if 8 divides A175181(p), and if and only if 4 divides A214028(p). For a proof of the equivalence between A214027(p) = 2 and 4 dividing A214028(p), see Section 2 of my link below.
This sequence contains all primes congruent to 3 modulo 8. This corresponds to case (2) for k = 6 in the Conclusion of Section 1 of my link below.
Conjecturely, since (k+2)/2 = 4 is a square, this sequence has density 5/12 in the primes; see the end of Section 1 of my link. [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]
The conjecture above is an analog of Hasse's result that the set {p prime : ord(2,p) is odd} has density 7/24 in the primes, where ord(a,m) is the multiplicative order of a modulo m; see A014663. - Jianing Song, Jun 26 2025

Jianing Song, Table of n, a(n) for n = 1..1600
Jianing Song, Lucas sequences and entry point modulo p

Cf. A175181, A214028.
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.
                             |   m=1    |   m=2    |   m=3
-----------------------------+----------+----------+---------
The sequence {x(n)}          | A000045  | A000129  | A006190
The sequence {w(k)}          | A001176  | A214027  | A322906
Primes p such that w(p) = 1  | A112860* | A309580  | A309586
Primes p such that w(p) = 2  | A053027  | this seq | A309587
Primes p such that w(p) = 4  | A053028  | A261580  | A309588
Numbers k such that w(k) = 1 | A053031  | A309583  | A309591
Numbers k such that w(k) = 2 | A053030  | A309584  | A309592
Numbers k such that w(k) = 4 | A053029  | A309585  | A309593
* and also A053032 U {2}

forprime(p=2, 700, if(A214027(p)==2, print1(p, ", ")))

A309583 Numbers k with 1 zero in a fundamental period of A000129 mod k.

Original entry on oeis.org

1, 2, 4, 7, 8, 14, 16, 20, 23, 24, 28, 31, 32, 40, 41, 46, 47, 48, 49, 52, 56, 62, 64, 71, 72, 79, 80, 82, 88, 92, 94, 96, 98, 100, 103, 104, 112, 116, 120, 124, 127, 128, 140, 142, 144, 148, 151, 152, 158, 160, 161, 164, 167, 168, 176, 184, 188, 191, 192

Offset: 1

Jianing Song, Aug 10 2019

nonn

Numbers k such that A214027(k) = 1.

The odd numbers k satisfy A175181(k) == 2 (mod 4).

Jianing Song, Table of n, a(n) for n = 1..5000

Cf. A175181.
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.
                             |   m=1    |   m=2    |   m=3
-----------------------------+----------+----------+---------
The sequence {x(n)}          | A000045  | A000129  | A006190
The sequence {w(k)}          | A001176  | A214027  | A322906
Primes p such that w(p) = 1  | A112860* | A309580  | A309586
Primes p such that w(p) = 2  | A053027  | A309581  | A309587
Primes p such that w(p) = 4  | A053028  | A261580  | A309588
Numbers k such that w(k) = 1 | A053031  | this seq | A309591
Numbers k such that w(k) = 2 | A053030  | A309584  | A309592
Numbers k such that w(k) = 4 | A053029  | A309585  | A309593
* and also A053032 U {2}

for(k=1, 200, if(A214027(k)==1, print1(k, ", ")))

A309584 Numbers k with 2 zeros in a fundamental period of A000129 mod k.

Original entry on oeis.org

3, 6, 9, 10, 11, 12, 15, 17, 18, 19, 21, 22, 26, 27, 30, 33, 34, 35, 36, 38, 39, 42, 43, 44, 45, 50, 51, 54, 55, 57, 58, 59, 60, 63, 66, 67, 68, 69, 70, 73, 74, 75, 76, 77, 78, 81, 83, 84, 85, 86, 87, 89, 90, 91, 93, 95, 97, 99, 102, 105, 106, 107, 108, 110

Offset: 1

Jianing Song, Aug 10 2019

nonn

Numbers k such that A214027(k) = 2.
This sequence contains all numbers k such that 4 divides A214028(k). As a consequence, this sequence contains all numbers congruent to 3 modulo 8.
This sequence contains all odd numbers k such that 8 divides A175181(k).

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A175181, A214028.
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.
                             |   m=1    |   m=2    |   m=3
-----------------------------+----------+----------+---------
The sequence {x(n)}          | A000045  | A000129  | A006190
The sequence {w(k)}          | A001176  | A214027  | A322906
Primes p such that w(p) = 1  | A112860* | A309580  | A309586
Primes p such that w(p) = 2  | A053027  | A309581  | A309587
Primes p such that w(p) = 4  | A053028  | A261580  | A309588
Numbers k such that w(k) = 1 | A053031  | A309583  | A309591
Numbers k such that w(k) = 2 | A053030  | this seq | A309592
Numbers k such that w(k) = 4 | A053029  | A309585  | A309593
* and also A053032 U {2}

for(k=1, 100, if(A214027(k)==2, print1(k, ", ")))

A309585 Numbers k with 4 zeros in a fundamental period of A000129 mod k.

Original entry on oeis.org

5, 13, 25, 29, 37, 53, 61, 65, 101, 109, 125, 137, 145, 149, 157, 169, 173, 181, 185, 197, 229, 265, 269, 277, 293, 305, 317, 325, 349, 373, 377, 389, 397, 421, 461, 481, 505, 509, 521, 541, 545, 557, 569, 593, 613, 625, 653, 661, 677, 685, 689, 701, 709

Offset: 1

Jianing Song, Aug 10 2019

nonn

Numbers k such that A214027(k) = 4.

Also numbers k such that A214028(k) is odd.

Jianing Song, Table of n, a(n) for n = 1..1000

Cf. A214028.
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.
                             |   m=1    |   m=2    |   m=3
-----------------------------+----------+----------+---------
The sequence {x(n)}          | A000045  | A000129  | A006190
The sequence {w(k)}          | A001176  | A214027  | A322906
Primes p such that w(p) = 1  | A112860* | A309580  | A309586
Primes p such that w(p) = 2  | A053027  | A309581  | A309587
Primes p such that w(p) = 4  | A053028  | A261580  | A309588
Numbers k such that w(k) = 1 | A053031  | A309583  | A309591
Numbers k such that w(k) = 2 | A053030  | A309584  | A309592
Numbers k such that w(k) = 4 | A053029  | this seq | A309593
* and also A053032 U {2}

for(k=1, 700, if(A214027(k)==4, print1(k, ", ")))

A309586 Primes p with 1 zero in a fundamental period of A006190 mod p.

Original entry on oeis.org

2, 3, 23, 43, 53, 61, 79, 101, 103, 107, 127, 131, 139, 173, 179, 191, 199, 211, 251, 263, 277, 283, 311, 347, 367, 419, 433, 439, 443, 467, 491, 503, 523, 547, 563, 569, 571, 599, 607, 647, 659, 677, 719, 727, 751, 757, 823, 829, 859, 881, 883, 887, 907

Offset: 1

Jianing Song, Aug 10 2019

nonn

Primes p such that A322906(p) = 1.
For p > 2, p is in this sequence if and only if A175182(p) == 2 (mod 4), and if and only if A322907(p) == 2 (mod 4). For a proof of the equivalence between A322906(p) = 1 and A322907(p) == 2 (mod 4), see Section 2 of my link below.
This sequence contains all primes congruent to 3, 23, 27, 35, 43, 51 modulo 52. This corresponds to case (3) for k = 11 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

Jianing Song, Table of n, a(n) for n = 1..1200
Jianing Song, Lucas sequences and entry point modulo p

Cf. A175182, A322907.
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.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+----------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | this seq
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

forprime(p=2, 900, if(A322906(p)==1, print1(p, ", ")))

A309587 Primes p with 2 zeros in a fundamental period of A006190 mod p.

Original entry on oeis.org

7, 11, 17, 19, 31, 47, 59, 67, 71, 83, 113, 151, 163, 167, 223, 227, 239, 257, 271, 307, 313, 331, 337, 359, 379, 383, 431, 463, 479, 487, 499, 521, 587, 601, 619, 631, 641, 643, 673, 683, 691, 739, 743, 787, 809, 811, 827, 839, 863, 947, 967, 983

Offset: 1

Jianing Song, Aug 10 2019

nonn

Primes p such that A322906(p) = 2.
For p > 2, p is in this sequence if and only if 8 divides A175182(p), and if and only if 4 divides A322907(p). For a proof of the equivalence between A322906(p) = 2 and 4 dividing A322907(p), see Section 2 of my link below.
This sequence contains all primes congruent to 7, 11, 15, 19, 31, 47 modulo 52. This corresponds to case (2) for k = 11 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

Jianing Song, Table of n, a(n) for n = 1..1200
Jianing Song, Lucas sequences and entry point modulo p

Cf. A006190, A175182, A322907.
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.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+----------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | this seq
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

forprime(p=2, 1000, if(A322906(p)==2, print1(p, ", ")))

A309588 Primes p with 4 zeros in a fundamental period of A006190 mod p.

Original entry on oeis.org

5, 13, 29, 37, 41, 73, 89, 97, 109, 137, 149, 157, 181, 193, 197, 229, 233, 241, 269, 281, 293, 317, 349, 353, 373, 389, 397, 401, 409, 421, 449, 457, 461, 509, 541, 557, 577, 593, 613, 617, 653, 661, 701, 709, 733, 761, 769, 773, 797, 821, 853, 857, 877

Offset: 1

Jianing Song, Aug 10 2019

nonn

Primes p such that A322906(p) = 4.
For p > 2, p is in this sequence if and only if A175182(p) == 4 (mod 8), and if and only if A322907(p) is odd. For a proof of the equivalence between A322906(p) = 4 and A322907(p) being odd, see Section 2 of my link below.
This sequence contains all primes congruent to 5, 21, 33, 37, 41, 45 modulo 52. This corresponds to case (1) for k = 11 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

Jianing Song, Table of n, a(n) for n = 1..1200
Jianing Song, Lucas sequences and entry point modulo p

Cf. A006190, A175182, A322907.
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.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+----------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | this seq
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

forprime(p=2, 900, if(A322906(p)==4, print1(p, ", ")))

cp's OEIS Frontend

A261580 Primes p such that A214028(p) is odd.

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A214028 Entry points for the Pell sequence: smallest k such that n divides A000129(k).

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A309580 Primes p with 1 zero in a fundamental period of A000129 mod p.

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A309581 Primes p with 2 zeros in a fundamental period of A000129 mod p.

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A309583 Numbers k with 1 zero in a fundamental period of A000129 mod k.

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A309584 Numbers k with 2 zeros in a fundamental period of A000129 mod k.

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A309585 Numbers k with 4 zeros in a fundamental period of A000129 mod k.

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A309586 Primes p with 1 zero in a fundamental period of A006190 mod p.

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