A000129 -id:A000129 - cp's OEIS frontend

A002965 Interleave denominators (A000129) and numerators (A001333) of convergents to sqrt(2).

0, 1, 1, 1, 2, 3, 5, 7, 12, 17, 29, 41, 70, 99, 169, 239, 408, 577, 985, 1393, 2378, 3363, 5741, 8119, 13860, 19601, 33461, 47321, 80782, 114243, 195025, 275807, 470832, 665857, 1136689, 1607521, 2744210, 3880899, 6625109, 9369319, 15994428, 22619537

Offset: 0

N. J. A. Sloane

Denominators of Farey fraction approximations to sqrt(2). The fractions are 1/0, 0/1, 1/1, 2/1, 3/2, 4/3, 7/5, 10/7, 17/12, .... See A082766(n+2) or A119016 for the numerators. "Add" (meaning here to add the numerators and add the denominators, not to add the fractions) 1/0 to 1/1 to make the fraction bigger: 2/1. Now 2/1 is too big, so add 1/1 to make the fraction smaller: 3/2, 4/3. Now 4/3 is too small, so add 3/2 to make the fraction bigger: 7/5, 10/7, ... Because the continued fraction for sqrt(2) is all 2's, it will always take exactly two terms here to switch from a number that's bigger than sqrt(2) to one that's less. A097545/A097546 gives the similar sequence for Pi. A119014/A119015 gives the similar sequence for e. - Joshua Zucker, May 09 2006
The principal and intermediate convergents to 2^(1/2) begin with 1/1, 3/2 4/3, 7/5, 10/7; essentially, numerators=A143607, denominators=A002965. - Clark Kimberling, Aug 27 2008
(a(2n)*a(2n+1))^2 is a triangular square. - Hugh Darwen, Feb 23 2012
a(2n) are the interleaved values of m such that 2*m^2+1 and 2*m^2-1 are squares, respectively; a(2n+1) are the interleaved values of their corresponding integer square roots. - Richard R. Forberg, Aug 19 2013
Coefficients of (sqrt(2)+1)^n are a(2n)*sqrt(2)+a(2n+1). - John Molokach, Nov 29 2015
Apart from the first two terms, this is the sequence of denominators of the convergents of the continued fraction expansion sqrt(2) = 1/(1 - 1/(2 + 1/(1 - 1/(2 + 1/(1 - ....))))). - Peter Bala, Feb 02 2017
Limit_{n->infinity} a(2n+1)/a(2n) = sqrt(2); lim_{n->infinity} a(2n)/a(2n-1) = (2+sqrt(2))/2. - Ctibor O. Zizka, Oct 28 2018

			The convergents are rational numbers given by the recurrence relation p/q -> (p + 2*q)/(p + q). Starting with 1/1, the next three convergents are (1 + 2*1)/(1 + 1) = 3/2, (3 + 2*2)/(3 + 2) = 7/5, and (7 + 2*5)/(7 + 5) = 17/12. The sequence puts the denominator first, so a(2) through a(9) are 1, 1, 2, 3, 5, 7, 12, 17. - _Michael B. Porter_, Jul 18 2016

C. Brezinski, History of Continued Fractions and Padé Approximants. Springer-Verlag, Berlin, 1991, p. 24.
Jay Kappraff, Musical Proportions at the Basis of Systems of Architectural Proportion both Ancient and Modern, in Volume I of K. Williams and M.J. Ostwald (eds.), Architecture and Mathematics from Antiquity to the Future, DOI 10.1007/978-3-319-00143-2_27, Springer International Publishing Switzerland 2015. See Eq. 32.7.
Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Guelena Strehler, Chess Fractal, April 2016, p. 24.

T. D. Noe, Table of n, a(n) for n = 0..500
Damanvir Singh Binner, Proofs of Chappelon and Alfonsín Conjectures On Square Frobenius Numbers and its Relationship to Simultaneous Pell's Equations, arXiv:2112.15474 [math.NT], 2021.
Jonathan Chappelon and Jorge Luis Ramírez Alfonsín, The Square Frobenius Number, arXiv:2006.14219 [math.NT], 2020.
H. S. M. Coxeter, The role of intermediate convergents in Tait's explanation for phyllotaxis, J. Algebra 20 (1972), 167-175.
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Pierre Lamothe, En marge du problème des cercles tangents
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Dave Rusin, Farey fractions on sci.math [Broken link]
Dave Rusin, Farey fractions on sci.math [Cached copy]
K. Williams, The sacred cult revisited: the pavement of the baptistery of San Giovanni, Florence, Math. Intellig., 16 (No. 2, 1994), 18-24.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

Cf. A000129(n) = a(2n), A001333(n) = a(2n+1).

Cf. A001109, A155046.

a:=[0,1];; for n in [3..45] do a[n]:=a[n-1]+a[n-2-((n-1) mod 2)]; od; a; # Muniru A Asiru, Oct 28 2018

import Data.List (transpose)
a002965 n = a002965_list !! n
a002965_list = concat $ transpose [a000129_list, a001333_list]
-- Reinhard Zumkeller, Jan 01 2014

a=new Array(); a[0]=0; a[1]=1;
for (i=2;i<50;i+=2) {a[i]=a[i-1]+a[i-2];a[i+1]=a[i]+a[i-2];}
document.write(a); // Jon Perry, Sep 12 2012

I:=[0,1,1,1]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Nov 30 2015

A002965 := proc(n) option remember; if n <= 0 then 0; elif n <= 3 then 1; else 2*A002965(n-2)+A002965(n-4); fi; end;
A002965:=-(1+2*z+z**2+z**3)/(-1+2*z**2+z**4); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence except for two leading terms

LinearRecurrence[{0, 2, 0, 1}, {0, 1, 1, 1}, 42] (* Vladimir Joseph Stephan Orlovsky, Feb 13 2012 *)
With[{c=Convergents[Sqrt[2],20]},Join[{0,1},Riffle[Denominator[c], Numerator[c]]]] (* Harvey P. Dale, Oct 03 2012 *)

PARI
```
a(n)=if(n<4,n>0,2*a(n-2)+a(n-4))
```

x='x+O('x^100); concat(0, Vec((x+x^2-x^3)/(1-2*x^2-x^4))) \\ Altug Alkan, Dec 04 2015

a(n) = 2*a(n-2) + a(n-4) if n>3; a(0)=0, a(1)=a(2)=a(3)=1.
a(2*n) = a(2*n-1) + a(2*n-2) and a(2*n+1) = 2*a(2*n) - a(2*n-1).
G.f.: (x+x^2-x^3)/(1-2*x^2-x^4).
a(0)=0, a(1)=1, a(n) = a(n-1) + a(2*[(n-2)/2]). - Franklin T. Adams-Watters, Jan 31 2006
For n > 0, a(2*n) = a(2*n-1) + a(2*n-2) and a(2*n+1) = a(2*n) + a(2*n-2). - Jon Perry, Sep 12 2012
a(n) = (((sqrt(2) - 2)*(-1)^n + 2 + sqrt(2))*(1 + sqrt(2))^(floor(n/2)) - ((2 + sqrt(2))*(-1)^n -2 + sqrt(2))*(1 - sqrt(2))^(floor(n/2)))/8. - Ilya Gutkovskiy, Jul 18 2016
a(n) = a(n-1) + a(n-2-(n mod 2)); a(0)=0, a(1)=1. - Ctibor O. Zizka, Oct 28 2018

Thanks to Michael Somos for several comments which improved this entry.

A175181 Pisano period of the 2-Fibonacci numbers A000129.

Original entry on oeis.org

1, 2, 8, 4, 12, 8, 6, 8, 24, 12, 24, 8, 28, 6, 24, 16, 16, 24, 40, 12, 24, 24, 22, 8, 60, 28, 72, 12, 20, 24, 30, 32, 24, 16, 12, 24, 76, 40, 56, 24, 10, 24, 88, 24, 24, 22, 46, 16, 42, 60, 16, 28, 108, 72, 24, 24, 40, 20, 40, 24, 124, 30, 24, 64, 84, 24, 136

Offset: 1

R. J. Mathar, Mar 01 2010

nonn

Period of the sequence defined by reading A000129 modulo n.

T. D. Noe, Table of n, a(n) for n = 1..1000
Sergio Falcon and Ángel Plaza, k-Fibonacci sequences modulo m, Chaos, Solit. Fractals 41 (2009), 497-504.
Eric Weisstein's World of Mathematics, Pisano period.
Wikipedia, Pisano period.

Cf. A000129, A001175, A175182, A175183, A175184, A175185.

F := proc(k,n) option remember; if n <= 1 then n; else k*procname(k,n-1)+procname(k,n-2) ; end if; end proc:
Pper := proc(k,m) local cha, zer,n,fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k,n) mod m ; cha := [op(cha),fmodm] ; if fmodm = 0 then zer := [op(zer),n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2],cha) ] = [ op(zer[2]+1..zer[3],cha) ] and [op(1..zer[2],cha)] = [ op(zer[3]+1..zer[4],cha) ] and [op(1..zer[2],cha)] = [ op(zer[4]+1..zer[5],cha) ] then return zer[2] ; elif [op(1..zer[3],cha) ] = [ op(zer[3]+1..zer[5],cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
k := 2 ; seq( Pper(k,m),m=1..80) ;

Table[s = t = Mod[{0, 1}, n]; cnt = 1; While[tmp = Mod[2*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s != t, cnt++]; cnt, {n, 100}] (* T. D. Noe, Jul 09 2012 *)

A214027 The number of zeros in the fundamental Pisano period of the sequence A000129 mod n.

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 1, 1, 2, 2, 2, 2, 4, 1, 2, 1, 2, 2, 2, 1, 2, 2, 1, 1, 4, 2, 2, 1, 4, 2, 1, 1, 2, 2, 2, 2, 4, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 4, 2, 2, 1, 2, 2, 2, 2, 4, 1, 2, 1, 4, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 2, 2, 2, 2, 2, 1

Offset: 1

Art DuPre, Jul 04 2012

nonn

This is intimately connected with A175181 and A214028, much as A001176 is intimately connected with A001175 and A001177. In fact, A175181(n)/a(n) = A214028(n). This is the same divisibility relation that holds between A001175, A001176 and A001177.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A175181, A214028.

Similar sequences: A001176, A322906.

Join[{1}, Table[s = t = Mod[{0, 1}, n]; zeros = 0; While[tmp = Mod[2*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s != t, If[tmp == 0, zeros++]]; zeros, {n, 2, 100}]] (* T. D. Noe, Jul 09 2012 *)

A000129(m) = ([2, 1; 1, 0]^m)[2, 1]
a(n) = my(i=1); while(A000129(i)%n!=0, i++); znorder(Mod(A000129(i+1), n)) \\ Jianing Song, Aug 10 2019

From Jianing Song, Sep 12 2018: (Start)
For odd primes p, a(p^e) = 4 if A214028(p) is odd; 1 if A214028(p) is even but not divisible by 4; 2 if A214028(p) is divisible by 4.
a(n) = 2 for n == 3 (mod 8). For primes p, a(p^e) = 1 if p == 7 (mod 8), 4 if p == 5 (mod 8). Conjecture: 1/6 of the primes congruent to 1 mod 8 satisfy a(p^e) = 1, 2/3 of them satisfy a(p^e) = 2 and 1/6 of them satisfy a(p^e) = 4.
(End)

A246692 Numbers k such that k | A000129(k).

Original entry on oeis.org

1, 2, 4, 8, 12, 16, 24, 32, 36, 48, 60, 64, 72, 84, 96, 108, 120, 128, 132, 136, 144, 168, 180, 192, 216, 240, 252, 256, 264, 272, 288, 300, 324, 336, 360, 384, 396, 408, 420, 432, 480, 504, 512, 528, 540, 544, 576, 588, 600, 648, 660, 672, 720, 756, 768

Offset: 1

Clark Kimberling, Sep 01 2014

These are the numbers k such that mean of the k-th row of the Delannoy triangle at A027926 is an integer. All such k except 0 and 1 are multiples of 4.

Is A181824 a subsequence? The first 31 terms appear in this sequence. - Jaycob Coleman, Mar 08 2015 [The first 323 terms of A181824 appear in this sequence. - Amiram Eldar, Mar 31 2021]

			Row 4 of the Delannoy triangle is (1,5,5,1), with sum 12 = A000129(4) divisible by 4.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A246693, A000129, A027926.

Cf. A181824.

z = 1000; t = LinearRecurrence[{2, 1}, {1, 2}, z]; (* A000129 *)
Select[Range[1, z], IntegerQ[t[[#]]/#] &]   (* A246692 *)
Table[t[[u[[n]]]]/u[[n]], {n, 1, 17}]  (* A246693 *)

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, ", ")))

cp's OEIS Frontend

A002965 Interleave denominators (A000129) and numerators (A001333) of convergents to sqrt(2).

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A175181 Pisano period of the 2-Fibonacci numbers A000129.

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A214027 The number of zeros in the fundamental Pisano period of the sequence A000129 mod n.

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A246692 Numbers k such that k | A000129(k).

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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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