A000129 -id:A000129 - cp's OEIS frontend

A238736 Balancing Wieferich primes: primes p that divide their Pell quotients, where the Pell quotient of p is A000129(p - (2/p))/p and (2/p) is a Jacobi symbol.

13, 31, 1546463

Offset: 1

John Blythe Dobson, Mar 04 2014

Williams 1982 (p. 86), notes that p = 13, 31 and 1546463 are the only primes less than 10^8 for which the Pell quotient vanishes mod p. Elsenhans and Jahnel, "The Fibonacci sequence modulo p^2," p. 5, report in effect that there are no more such primes p < 10^9.
Williams 1991 (p. 440), and Sun 1995 pt. 3, Theorem 3.3, together prove a set of formulas connecting the Pell quotient with the Fermat quotient (base 2) (A007663) and harmonic numbers like H(floor(p/8)) (see example in the Formula section below). As is well known, the vanishing of the Fermat quotient (base 2) is a necessary condition for the failure of the first case of Fermat's Last Theorem (see discussion under A001220); and in light of a corresponding result of Dilcher and Skula concerning this type of harmonic number, the vanishing of the Pell quotient mod p is also a necessary condition for the failure of the first case of Fermat's Last Theorem.
There are no more terms up to 10^10.
Using the PARI script by Charles R Greathouse IV, I have extended the search from 10^10 to 10^12 without finding a further solution. - John Blythe Dobson, Mar 30 2015
Also primes p such that p^2 divides A001109((p - (2/p))/2). - Jianing Song, Oct 08 2018
From Felix Fröhlich, May 18 2019: (Start)
The term "balancing Wieferich prime" comes from Rout, 2016.
Primes p that satisfy the congruence B_{p-(8/p)} == 0 (mod p^2), where B_i denotes the i-th balancing number A001109(i) and (a/b) denotes the Jacobi symbol (cf. Rout, 2016, (1.6)).
Primes p such that the period of the balancing sequence (A001109) modulo p is equal to the period of the balancing sequence modulo p^2 (cf. Panda, Rout, 2014, p. 275).
Under the abc conjecture for the number field Q(sqrt(2)) there exist at least (log(x)/log(log(x)))*(log(log(log(x))))^m balancing non-Wieferich primes <= x such that p == 1 (mod k) for any integers k > 2, m > 0 (cf. Dutta, Patel, Ray, 2019). This is an improvement of an earlier result stating there are at least log(x)/log(log(x)) balancing non-Wieferich primes p == 1 (mod k) less than x (cf. Theorem 3.2 in Rout 2016). (End)

			PellQuotient(13) = 6214 = 13*478; PellQuotient(31) = 3470274850 = 31*111944350.

Zakariae Bouazzaoui, On Periods of Fibonacci Sequences and Real Quadratic p-rational Fields, Fibonacci Quart. 58 (2020), no. 5, 103-110. See p. 7.
Karl Dilcher and Ladislav Skula, A new criterion for the first case of Fermat's Last Theorem, Mathematics of Computation, 64 (1995), 363-392.
Utkal Keshari Dutta, Bijan Kumar Patel and Prasanta Kumar Ray, A brief remark on balancing-Wieferich primes, Mathematica, Vol. 60 (83), No. 1 (2018), 48-53 [Subscription required].
Utkal Keshari Dutta, Bijan Kumar Patel and Prasanta Kumar Ray, Balancing non-Wieferich primes in arithmetic progressions, Proceedings - Mathematical Sciences, Vol. 129, No. 2 (2019), Article 21, DOI:10.1007/s12044-018-0459-3.
Andreas-Stephan Elsenhans and Jörg Jahnel, The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 10^14, arXiv 1006.0824 [math.NT], 2010.
Georges Gras, On the structure of the Galois group of the Abelian closure of a number field, arXiv 1212.3588 [math.NT], 2013.
Hao Pan, Lehmer's type congruences for lacunary harmonic sums, arXiv 0905.0941 [math.NT], 2009.
G. K. Panda and S. S. Rout, Periodicity of Balancing Numbers, Acta Mathematica Hungarica 143 (2014), 274-286. Also on ResearchGate.
Sudhansu Sekhar Rout, Balancing non-Wieferich primes in arithmetic progression and abc conjecture, Proc. Japan Acad. Ser. A Math. Sci., Volume 92, Number 9 (2016), 112-116.
Zhi-Hong Sun, Combinatorial sum ... and its applications in Number Theory, III (English version), originally published in Chinese in Journal of Nanjing University Mathematical Biquarterly, 12 (1995), 90-102.
Zhi-Hong Sun, Five congruences for primes, Fibonacci Quarterly, 40 (2002), 345-351.
H. C. Williams, The influence of computers in the development of number theory, Computers & Mathematics with Applications, 8 (1982), 75-93.
H. C. Williams, Some formulas concerning the fundamental unit of a real quadratic field, Discrete Mathematics, 92 (1991), 431--440.

Cf. A000129, A001109.

Select[Prime[Range[1000]], Mod[Fibonacci[# - JacobiSymbol[2, #], 2]/#, #] == 0 &]

is(n)=isprime(n) && (Mod([2,1;1,0],n^2)^(n-kronecker(2,n)))[2,1]==0 \\ Charles R Greathouse IV, Mar 04 2014

The condition for p to be a member of this sequence is A000129(p-e)/p == F(p-e, 2)/p == 0 (mod p), where F(p-e, 2) is the p-e'th Fibonacci polynomial evaluated at the argument 2, and e = (2/p) is a Jacobi Symbol.

Let PellQuotient(p) = A000129(p-e)/p, q_2 = (2^(p-1) - 1)/p = A007663(p) be the corresponding Fermat quotient of base 2, H(floor(p/8)) be a harmonic number, and e = (2/p) be a Jacobi Symbol. Then a result of Williams (1991), as refined by Sun (1995), shows that 2*PellQuotient(p) == -4*q_2 - H(floor(p/8)) (mod p).

Name amended by Felix Fröhlich, May 26 2019

A272040 a(n) = A000010(A000129(n)).

Original entry on oeis.org

1, 1, 4, 4, 28, 24, 156, 128, 784, 1120, 5740, 2880, 33460, 37128, 150080, 147456, 1128256, 931392, 6446016, 4677120, 28514304, 44450560, 224075664, 106168320, 1265644800, 1560708240, 5970392064, 5588803584, 44560482148, 33497856000, 255263424000, 196368924672, 1210784762880

Offset: 1

Altug Alkan, May 06 2016

nonn

			a(3) = 4 because a(3) = A000010(A000129(3)) = A000010(5) = 4.

Amiram Eldar, Table of n, a(n) for n = 1..630

Cf. A000010, A000129, A000203, A065449, A363829, A363831, A363833, A364818.

EulerPhi[LinearRecurrence[{2, 1}, {1, 2}, 33]] (* Amiram Eldar, Oct 21 2023 *)

a000129(n) = ([2,1;1,0]^n)[2,1];
a(n) = eulerphi(a000129(n));

A111954 a(n) = A000129(n) + (-1)^n.

Original entry on oeis.org

1, 0, 3, 4, 13, 28, 71, 168, 409, 984, 2379, 5740, 13861, 33460, 80783, 195024, 470833, 1136688, 2744211, 6625108, 15994429, 38613964, 93222359, 225058680, 543339721, 1311738120, 3166815963, 7645370044, 18457556053, 44560482148, 107578520351

Offset: 0

Creighton Dement, Aug 23 2005

a(n) + a(n+1) = A001333(n+1). Inverse binomial transform of A007070 (with prepended 1). Inverse invert transform of A077995.

Floretion Algebra Multiplication Program, FAMP Code: -4ibasejseq[J*D] with J = - .25'i + .25'j + .5'k - .25i' + .25j' + .5k' - .5'kk' - .25'ik' - .25'jk' - .25'ki' - .25'kj' - .5e and D = + .5'i - .25'j + .25'k + .5i' - .25j' + .25k' - .5'ii' - .25'ij' - .25'ik' - .25'ji' - .25'ki' - .5e

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,1).

Cf. A000129, A001333, A111955, A111956, A007070, A077995.

LinearRecurrence[{1,3,1},{1,0,3},40] (* Harvey P. Dale, Nov 24 2014 *)

a(n) = a(n-1) + 3*a(n-2) + a(n-3), n >= 3.
G.f.: (x-1)/((x+1)*(x^2+2*x-1)).
a(n) = (sqrt(2)/4)*((1 + sqrt(2))^n - (1 - sqrt(2))^n) + (-1)^n.
E.g.f.: cosh(x) - sinh(x) + exp(x)*sinh(sqrt(2)*x)/sqrt(2). - Stefano Spezia, May 26 2024

A270342 Positive integers n such that the sum of the Pell numbers A000129(0) + ... + A000129(n-1) is divisible by n.

Original entry on oeis.org

3, 4, 5, 7, 8, 11, 13, 16, 17, 19, 23, 24, 29, 31, 32, 37, 41, 43, 47, 48, 53, 59, 61, 64, 67, 71, 72, 73, 79, 83, 89, 96, 97, 101, 103, 107, 109, 113, 120, 127, 128, 131, 137, 139, 144, 149, 151, 157, 163, 167, 168, 169, 173, 179, 181, 191, 192, 193, 197, 199, 211, 216, 223, 227, 229

Offset: 1

Altug Alkan, Mar 15 2016

nonn

Sequence contains all odd primes because of the fact that ((1-sqrt(2))^p + (1+sqrt(2))^p - 2) is divisible by p where p is an odd prime.

			3 is a term because 0 + 1 + 2 = 3 is divisible by 3.
4 is a term because 0 + 1 + 2 + 5 = 8 is divisible by 4.
5 is a term because 0 + 1 + 2 + 5 + 12 = 20 is divisible by 5.
7 is a term because 0 + 1 + 2 + 5 + 12 + 20 + 79 = 119 is divisible by 7.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000040, A000129, A048739.

Module[{nn=250,pell},pell=LinearRecurrence[{2,1},{0,1},nn];Position[ Table[ Total[Take[pell,n]]/n,{n,nn}],?(IntegerQ[#]&)]]//Flatten (* _Harvey P. Dale, Nov 11 2021 *)

a048739(n) = local(w=quadgen(8)); -1/2+(3/4+1/2*w)*(1+w)^n+(3/4-1/2*w)*(1-w)^n;
for(n=1, 1e3, if(a048739(n-1) % (n+1) == 0, print1(n+1, ", ")));

A072280 Product representation of the Pell numbers A000129 and A002203.

Original entry on oeis.org

2, 1, 7, 6, 41, 5, 239, 34, 199, 29, 8119, 33, 47321, 169, 961, 1154, 1607521, 197, 9369319, 1121, 32641, 5741, 318281039, 1153, 45245801, 33461, 7761799, 38081, 63018038201, 1345, 367296043199, 1331714, 37667521, 1136689, 1273319041, 39201, 72722761475561

Offset: 1

Miklos Kristof, Jul 10 2002

nonn

Define the silver mean constants h=1+sqrt(2) = A014176, h^2=1+2h = A156035, and 1/h=h-2.
Let Phi(n,x) be the n-th cyclotomic polynomial A013595, so that x^n-1 = Product_{d | n} Phi(d, x). Let g(n) be the order of Phi(n, x), A000010. Then a(n)=(h-2)^g(n)*Phi(n, h^2) if n <> 2.
The Binet representations of the Pell numbers yields:
For even n, A000129(n) = Product_{d|n} a(d).
For odd n, A000129(n)=Product_{ d|n} a(2d).
For odd prime p, a(p)=A002203(p)/2, a(2p)=A000129(p).
a(2^(k+1))=A002203(2^k).
For odd n, A002203(n)=Product_{ d|n} a(d).
For k>0 and odd n, A002203(n*2^k)=Product_{ d | n} a(d*2^(k+1)).

			For even n=12, A000129(12) = a(1)*a(2)*a(3)*a(4)*a(6)*a(12) = 2*1*7*6*5*33 = 13860.
For odd n=9, A000129(9) = a(2)*a(6)*a(18)= 1*5*197 = 985.
For even n=8, A002203(12) = a(8)*a(24)=34*1153 = 39202.
For odd n=21, A002203(21) = a(1)*a(3)*a(7)*a(21) = 2*7*239*32641 = 109216786.

Dan Kalman and Robert Mena, The Fibonacci Numbers: Exposed, Math. Mag. 76 (3) (2003) 167-181.
Index to sequences related to cyclotomic polynomials.

Cf. A000129, A002203.

A072280 := proc(n) if n <= 2 then 3-n ; else g := numtheory[phi](n) ; h := 1+sqrt(2) ; (h-2)^g*numtheory[cyclotomic](n,h^2) ; simplify(expand(%)) ; end if; end proc:
seq(A072280(n),n=1..80) ; # R. J. Mathar, Nov 27 2009

a[n_] := If[n <= 2, 3-n, g = EulerPhi[n]; h = 1 + Sqrt[2]; (h - 2)^g*Cyclotomic[n, h^2] // Expand];
Table[a[n], {n, 1, 80}] (* Jean-François Alcover, May 08 2023, after R. J. Mathar *)

Edited and extended by R. J. Mathar, Nov 27 2009

A098586 a(n) = (1/2) * (5*P(n+1) + P(n) - 1), where P(k) are the Pell numbers A000129.

Original entry on oeis.org

2, 5, 13, 32, 78, 189, 457, 1104, 2666, 6437, 15541, 37520, 90582, 218685, 527953, 1274592, 3077138, 7428869, 17934877, 43298624, 104532126, 252362877, 609257881, 1470878640, 3551015162, 8572908965, 20696833093, 49966575152, 120629983398, 291226541949

Offset: 0

Creighton Dement, Oct 03 2004

Colin Barker, Table of n, a(n) for n = 0..1000
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (3,-1,-1).

Cf. A006451, A124124.

I:=[2,5,13]; [n le 3 select I[n] else 3*Self(n-1) - Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Feb 03 2018

A:= LREtools[REtoproc](a(n) = 3*a(n-1) - a(n-2) - a(n-3), a(n), {a(0)=2, a(1)=5, a(2)=13}):
seq(A(n),n=0..100); # Robert Israel, Aug 26 2014

LinearRecurrence[{3, -1, -1}, {2, 5, 13}, 28] (* Hermann Stamm-Wilbrandt, Aug 26 2014 *)
CoefficientList[Series[(2-x)/((1-x)*(1-2*x-x^2)), {x,0,50}], x] (* G. C. Greubel, Feb 03 2018 *)

Vec((2-x)/((1-x)*(1-2*x-x^2)) + O(x^50)) \\ Colin Barker, Mar 16 2016

a(n) = 3*a(n-1) - a(n-2) - a(n-3) with a(0)=2, a(1)=5, a(2)=13. - Hermann Stamm-Wilbrandt, Aug 26 2014
G.f.: (2-x)/((1-x)*(1-2*x-x^2)). - Robert Israel, Aug 26 2014
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 27 2014
a(2*n-1) = A006451(2*n), for n>0. - Hermann Stamm-Wilbrandt, Aug 27 2014
a(2*n) = A124124(2*n+2). - Hermann Stamm-Wilbrandt, Aug 27 2014
a(n) = (-2+(5-3*sqrt(2))*(1-sqrt(2))^n + (1+sqrt(2))^n*(5+3*sqrt(2)))/4. - Colin Barker, Mar 16 2016

Formula supplied by Thomas Baruchel, Oct 03 2004

More terms from Emeric Deutsch, Nov 17 2004

A110327 Triangle read by rows: T(n,k) = n!*Pell(n-k+1)/k!, where Pell(n)=A000129(n).

Original entry on oeis.org

1, 2, 1, 10, 4, 1, 72, 30, 6, 1, 696, 288, 60, 8, 1, 8400, 3480, 720, 100, 10, 1, 121680, 50400, 10440, 1440, 150, 12, 1, 2056320, 851760, 176400, 24360, 2520, 210, 14, 1, 39715200, 16450560, 3407040, 470400, 48720, 4032, 280, 16, 1, 862928640

Offset: 0

Paul Barry, Jul 20 2005

The row polynomials form an Appell sequence (see Wikipedia). - Tom Copeland, Dec 03 2013

			Rows begin:
  1;
  2,1;
  10,4,1;
  72,30,6,1;
  696,288,60,8,1;
  8400,3480,720,100,10,1;
  121680,50400,10440,1440,150,12,1;

Wikipedia, Appell sequence

Cf. A000129, A110328 (row sums), A110329 (diagonal sums), A110330 (matrix inverse).

Column k has e.g.f.: x^k/(k!*(1-2*x-x^2)).

E.g.f.: Sum_{n>=0, k>=0} T(n,k)*x^n*y^k/n! = e^(x*y)/(1-2*x-x^2). - Franklin T. Adams-Watters, Jan 12 2007

Edited by Franklin T. Adams-Watters, Jan 12 2007

A135153 Repeat Pell numbers A000129.

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 5, 5, 12, 12, 29, 29, 70, 70, 169, 169, 408, 408, 985, 985, 2378, 2378, 5741, 5741, 13860, 13860, 33461, 33461, 80782, 80782, 195025, 195025, 470832, 470832, 1136689, 1136689, 2744210, 2744210, 6625109, 6625109, 15994428, 15994428, 38613965

Offset: 0

Paul Curtz, Feb 14 2008

The binomial transform is 0, 0, 1, 4, 12, 32,... (n>=0), i.e. A135248 without one of the leading zeros. - R. J. Mathar, Jul 10 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1)

I:=[0,0,1,1]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Mar 03 2014

CoefficientList[Series[x^2 (1 + x)/(1 - 2 x^2 - x^4), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 03 2014 *)
LinearRecurrence[{0,2,0,1},{0,0,1,1},50] (* Harvey P. Dale, May 28 2023 *)

G.f.: x^2*(1+x)/(1-2*x^2-x^4). - Philippe Deléham, Feb 25 2014

a(n) = 2*a(n-2) + a(n-4), a(0) = a(1) = 0, a(2) = a(3) = 1. - Philippe Deléham, Feb 25 2014

Corrected and extended by Vincenzo Librandi, Mar 03 2014

A246556 a(n) = smallest prime which divides Pell(n) = A000129(n) but does not divide any Pell(k) for k

Original entry on oeis.org

2, 5, 3, 29, 7, 13, 17, 197, 41, 5741, 11, 33461, 239, 269, 577, 137, 199, 37, 19, 45697, 23, 229, 1153, 1549, 79, 53, 113, 44560482149, 31, 61, 665857, 52734529, 103, 1800193921, 73, 593, 9369319, 389, 241, 1746860020068409, 4663, 11437, 43, 6481, 47, 3761, 97, 293, 45245801, 101, 22307, 68480406462161287469, 7761799, 109, 1535466241

Offset: 2

Eric Chen, Nov 15 2014

nonn

First differs from A264137 (Largest prime factor of the n-th Pell number) at n=17; see Example section. - Jon E. Schoenfield, Dec 10 2016

			a(2) = 2 because Pell(2) = 2 and Pell(k) < 2 for k < 2.
a(4) = 3 because Pell(4) = 12 = 2^2 * 3, but 2 is not a primitive prime factor since Pell(2) = 2, so therefore 3 is the primitive prime factor.
a(5) = 29 because Pell(5) = 29, which is prime.
a(6) = 7 because Pell(6) = 70 = 2 * 5 * 7, but neither 2 nor 5 is a primitive prime factor, so therefore 7 is the primitive prime factor.
a(17) = 137 because Pell(17) = 1136689 = 137 * 8297, and both of them are primitive factors, we choose the smallest. (Pell(17) is the smallest Pell number with more than one primitive prime factor.)

Table of n, a(n) for n = 2..630
R. D. Carmichael, On the numerical factors of the arithmetic forms α^n ± β^n, Annals of Math., 15 (1/4) (1913), 30-70.

Cf. A001578 (for Fibonacci(n)), A000129 (Pell numbers), A008555, A086383, A096650, A120947, A175181, A214028, A264137.

prms={}; Table[f=First/@FactorInteger[Pell[n]]; p=Complement[f, prms]; prms=Join[prms, p]; If[p=={}, 1, First[p]], {n, 36}]

a(n) >= 2 for all n >= 2, by Carmichael's theorem. - Jonathan Sondow, Dec 08 2017

Edited by N. J. A. Sloane, Nov 29 2014
Terms up to a(612) in b-file added by Sean A. Irvine, Sep 23 2019
Terms a(613)-a(630) in b-file added by Max Alekseyev, Aug 26 2021

A264137 Largest prime factor of the n-th Pell number, A000129(n).

Original entry on oeis.org

2, 5, 3, 29, 7, 13, 17, 197, 41, 5741, 11, 33461, 239, 269, 577, 8297, 199, 179057, 59, 45697, 5741, 982789, 1153, 29201, 33461, 146449, 337, 44560482149, 269, 3272609, 665857, 52734529, 15607, 1800193921, 199, 1101341, 9369319, 4605197, 5521, 1746860020068409

Offset: 2

Jon E. Schoenfield, Dec 29 2015

nonn

First differs from A246556 at n = 17. Since Pell(17) = 1136689 = 137 * 8297, we find that 137 does not divide any earlier Pell number, and hence A246556(17) = 137, but 8297 is also prime, and so a(17) = 8297.

Daniel Suteu and Jon E. Schoenfield, Table of n, a(n) for n = 2..630 (terms a(2)..a(240) from Jon E. Schoenfield)

Cf. A000129, A006530, A060385, A246556.

Table[FactorInteger[Fibonacci[n, 2]][[-1, 1]], {n, 25}] (* Alonso del Arte, Dec 10 2016 *)
FactorInteger[#][[-1,1]]&/@LinearRecurrence[{2,1},{2,5},60] (* Harvey P. Dale, Jun 08 2019 *)

a(n) = vecmax(factor(([2, 1; 1, 0]^n)[2, 1])[,1]); \\ Daniel Suteu, May 26 2022

a(n) = A006530(A000129(n)).

cp's OEIS Frontend

A238736 Balancing Wieferich primes: primes p that divide their Pell quotients, where the Pell quotient of p is A000129(p - (2/p))/p and (2/p) is a Jacobi symbol.

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A272040 a(n) = A000010(A000129(n)).

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A111954 a(n) = A000129(n) + (-1)^n.

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A270342 Positive integers n such that the sum of the Pell numbers A000129(0) + ... + A000129(n-1) is divisible by n.

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A072280 Product representation of the Pell numbers A000129 and A002203.

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A098586 a(n) = (1/2) * (5*P(n+1) + P(n) - 1), where P(k) are the Pell numbers A000129.

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A110327 Triangle read by rows: T(n,k) = n!*Pell(n-k+1)/k!, where Pell(n)=A000129(n).

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