A097842 -id:A097842 - cp's OEIS frontend

A001946 a(n) = 11*a(n-1) + a(n-2).

2, 11, 123, 1364, 15127, 167761, 1860498, 20633239, 228826127, 2537720636, 28143753123, 312119004989, 3461452808002, 38388099893011, 425730551631123, 4721424167835364, 52361396397820127, 580696784543856761, 6440026026380244498, 71420983074726546239

Offset: 0

N. J. A. Sloane, Simon Plouffe

For odd n there is the Aurifeuillian factorization a(n) = Lucas[5n] = Lucas[n]*A[n]*B[n] = A000032[n]*A124296[n]*A124297[n], where A[n] = A124296[n] = 5*F(n)^2 - 5*F(n) + 1 and B[n] = A124297[n] = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci[n]. The largest prime divisors of a(n) for n>0 are listed in A121171[n] = {11, 41, 31, 2161, 151, 2521, 911, ...}. - Alexander Adamchuk, Oct 25 2006

For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 139.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
Tanya Khovanova, Recursive Sequences
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
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (11, 1).

Cf. A000032, A000045, A121171, A124296, A124297.

[ Lucas(5*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

A001946:=(-2+11*z)/(-1+11*z+z**2); # Conjectured by Simon Plouffe in his 1992 dissertation

Table[Fibonacci[5n-1]+Fibonacci[5n+1],{n,0,30}] (* Alexander Adamchuk, Oct 25 2006 *)
LinearRecurrence[{11,1},{2,11},20] (* Harvey P. Dale, Jan 25 2024 *)

a(n) = Lucas(5n) = Fibonacci(5n-1) + Fibonacci(5n+1). - Alexander Adamchuk, Oct 25 2006
a(n) = ((11 + 5*sqrt(5))/2)^n + ((11 - 5*sqrt(5))/2)^n. - Tanya Khovanova, Feb 06 2007
Contribution from Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 11*A097842(n), a(2n) = A065705(n).
a(3n+1) = A041226(5n), a(3n+2) = A041226(5n+3), a(3n+3) = 2* A041226(5n+4).
Limit(a(n+k)/a(k), k=infinity) = (A001946(n) + A049666(n)*sqrt(125))/2.
Limit(A001946(n)/A049666(n), n=infinity) = sqrt(125). (End)
From Peter Bala, Mar 22 2015: (Start)
a(n) = Fibonacci(10*n)/Fibonacci(5*n) for n >= 1.
a(n) = ( Fibonacci(5*n + 2*k) - F(5*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.
a(n) = ( Fibonacci(5*n + 2*k + 1) + F(5*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k.
a(n) = Sum_{k = 0..2*n} binomial(2*n,k)*Lucas(n + k). (End)
a(n) = [x^n] ( (1 + 11*x + sqrt(1 + 22*x + 125*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 26 2015
E.g.f.: 2*cosh(5*sqrt(5)*x/2)*(cosh(11*x/2) + sinh(11*x/2)). - Stefano Spezia, Jan 18 2025

A049670 a(n) = Fibonacci(10*n)/55.

Original entry on oeis.org

0, 1, 123, 15128, 1860621, 228841255, 28145613744, 3461681649257, 425758697244867, 52364858079469384, 6440451785077489365, 792123204706451722511, 97424713727108484379488, 11982447665229637126954513, 1473743638109518258131025611, 181258485039805516112989195640

Offset: 0

Clark Kimberling

Chebyshev polynomials S(n-1,123).
Used for all positive integer solutions of Pell equation x^2 - 5*(5*y)^2 = -4. See A097842 with A097843.
This is the k = 10 member of the k-family of sequences {F(k*n)/F(k)}, n >= 0 for k >= 1, with o.g.f. x/(1 - L(k)*x + (-1)^k*x^2). Proof: Binet-de Moivre formula for F and L. See also A028412. - Wolfdieter Lang, Aug 26 2012

Robert Israel, Table of n, a(n) for n = 0..383
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (123,-1).

A column of array A028412.

Cf. A000045.

[ Fibonacci(10*n)/55: n in [0..30]]; // G. C. Greubel, Dec 02 2017

seq(combinat:-fibonacci(10*n)/55, n=0..20); # Robert Israel, Apr 03 2015

Table[Fibonacci[10 n]/55, {n, 12}] (* Michael De Vlieger, Apr 03 2015 *)
LinearRecurrence[{123,-1},{0,1},20] (* Harvey P. Dale, Dec 03 2019 *)

numlib::fibonacci(10*n)/55 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(10*n)/55 \\ Charles R Greathouse IV, Oct 07 2016

G.f.: x/(1-123*x+x^2), 123=L(10)=A000032(10) (Lucas).
a(n+1) = S(n, 123) = U(n, 123/2) = S(2*n+1, 5*sqrt(5))/(5*sqrt(5)), n>=0, with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x).
a(n) = 123*a(n-1) - a(n-2), n >= 2; a(0)=0, a(1)=1.
a(n) = (ap^n - am^n)/(ap-am) with ap := (123+55*sqrt(5))/2 and am := (123-55*sqrt(5))/2 = 1/ap.
From Peter Bala, Nov 29 2013: (Start)
a(n) = 1/(11*55)*(F(10*n + 5) - F(10*n - 5)).
Sum_{n >= 1} 1/( 11*a(n) + 1/(11*a(n)) ) = 1/11. Compare with A001906 and A049660. (End)
From Peter Bala, Apr 03 2015: (Start)
For integer k, 1 + k*(22 - k)*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + k/5*Sum_{n >= 1} Fibonacci(5*n)*x^n )*( 1 + k/5*Sum_{n >= 1} Fibonacci(5*n)*(-x)^n ).
1 + 4*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n+5)*x^n )*( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n+5)*(-x)^n ) = ( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n-5)*x^n )*( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n-5)*(-x)^n ).
1 + 25*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Fibonacci(5*n+3)*x^n )*( 1 + Sum_{n >= 1} Fibonacci(5*n+3)*(-x)^n ) = ( 1 + Sum_{n >= 1} Fibonacci(5*n-3)*x^n )*( 1 + Sum_{n >= 1} Fibonacci(5*n-3)*(-x)^n ).
1 + 100*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + 2*Sum_{n >= 1} Fibonacci(5*n+1)*x^n )*( 1 + 2*Sum_{n >= 1} Fibonacci(5*n+1)*(-x)^n ) = ( 1 + 2*Sum_{n >= 1} Fibonacci(5*n-1)*x^n )*( 1 + 2*Sum_{n >= 1} Fibonacci(5*n-1)*(-x)^n ).
1 + 125*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Lucas(5*n)*x^n )*( 1 + Sum_{n >= 1} Lucas(5*n)*(-x)^n ). (End)

More terms from James Sellers, Jan 20 2000

Chebyshev and Pell comments from Wolfdieter Lang, Sep 10 2004

A097843 First differences of Chebyshev polynomials S(n,123) = A049670(n+1) with Diophantine property.

Original entry on oeis.org

1, 122, 15005, 1845493, 226980634, 27916772489, 3433536035513, 422297015595610, 51939099382224517, 6388086926998019981, 785682752921374233146, 96632590522402032656977, 11885022951502528642575025, 1461761190444288621004071098, 179784741401695997854858170029

Offset: 0

Wolfdieter Lang, Sep 10 2004

(11*b(n))^2 - 5*(5*a(n))^2 = -4 with b(n)=A097842(n) give all positive solutions of this Pell equation.

			All positive solutions of Pell equation x^2 - 125*y^2 = -4 are (11 = 11*1,1), (1364 = 11*124,122), (167761 = 11*15251,15005), (20633239 = 11*1875749,1845493), ...

Colin Barker, Table of n, a(n) for n = 0..478
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (123,-1).

a:=[1,122];; for n in [3..20] do a[n]:=123*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2019

m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-123*x+x^2) )); // G. C. Greubel, Jan 14 2019

LinearRecurrence[{123,-1}, {1,122}, 20] (* G. C. Greubel, Jan 14 2019 *)

Vec((1-x)/(1-123*x+x^2) + O(x^30)) \\ Colin Barker, Jun 15 2015

((1-x)/(1-123*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 14 2019

a(n) = ((-1)^n)*S(2*n, 11*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.
G.f.: (1-x)/(1-123*x+x^2).
a(n) = S(n, 123) - S(n-1, 123) = T(2*n+1, 5*sqrt(5)/2)/(5*sqrt(5)/2), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x) and T(n, x) Chebyshev's polynomials of the first kind, A053120.
a(n) = 123*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=122. - Philippe Deléham, Nov 18 2008
a(n) = (F(10*(n+1)) - F(10*n))/F(10), with F=A000045 (Fibonacci). F(10*n)/F(10) = A049670. - Wolfdieter Lang, Oct 11 2012
a(n) = (1/5)*F(10*n + 5). Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/11^2. Compare with A001519 and A007805. - Peter Bala, Nov 29 2013
From Peter Bala, Mar 23 2015: (Start)
a(n) = A049666(2*n + 1).
a(n) = ( Fibonacci(10*n + 10 - 2*k) - Fibonacci(10*n + 2*k) )/( Fibonacci(10 - 2*k) - Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(10*n + 10 - 2*k - 1) + Fibonacci(10*n + 2*k + 1) )/( Fibonacci(10 - 2*k - 1) + Fibonacci(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n))n>=1 = [1, 0, 122, 0, 15005, 0, 1845493, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -125, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

cp's OEIS Frontend

A001946 a(n) = 11*a(n-1) + a(n-2).

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A049670 a(n) = Fibonacci(10*n)/55.

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A097843 First differences of Chebyshev polynomials S(n,123) = A049670(n+1) with Diophantine property.

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A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

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