A157456 -id:A157456 - cp's OEIS frontend

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449

Offset: 0

Bruno Berselli, Feb 25 2014

First bisection of A041611.

Bruno Berselli, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (36,-1).

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

[n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020

CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]

a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016

m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
print(f.coefficients())

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

Original entry on oeis.org

1, 14, 209, 3121, 46606, 695969, 10392929, 155197966, 2317576561, 34608450449, 516809180174, 7717529252161, 115246129602241, 1720974414781454, 25699370092119569, 383769576967012081, 5730844284413061646, 85578894689228912609, 1277952576054020627489

Offset: 1

Paul Weisenhorn, May 23 2009

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.
The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,
with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.
Generic recurrences are:
A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.
B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.
k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).
x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;
Binet-type of solutions of these 2nd order recurrences are:
R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);
A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);
B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);
x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;
k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]
.C -A----- -B----- -k-----
01 A001519 A002878 A058038
02 A001653 A002315 A045899/2
03 A004253 A030221 A160695
04 A001653 A002315 A078522/4
05 A049685 A033890 A161582
06 A070997 A057080 A159683/2
07 A070998 A057081 A161585
08 A072256 A054320 A045502/4
09 A078922 A097783 A161586
10 A077417 A077416 A159681/2
11 A085260 A126816 A161588
12 A001570 A028230 A059989/4
13 A160682 A161591 A161584
14 A157456 A159678 A159679/2
15 A161595 A161599 A161583
16 A007805 A049629 A157459/4
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
J.-C. Novelli, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Index entries for the Pell equation
Index entries for linear recurrences with constant coefficients, signature (15,-1).

Cf. similar sequences listed in A238379.

I:=[1,14]; [n le 2 select I[n] else 15*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 12 2014

LinearRecurrence[{15,-1},{1,14},20] (* Harvey P. Dale, Oct 08 2012 *)
CoefficientList[Series[(1 - x)/(1 - 15 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)

a(n) = round((2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221)) \\ Colin Barker, Jul 25 2016

a(n) = 15*a(n-1)-a(n-2).
G.f.: (1-x)*x/(1-15*x+x^2).
a(n) = (2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221). - Colin Barker, Jul 25 2016

Edited, extended by R. J. Mathar, Sep 02 2009

First formula corrected by Harvey P. Dale, Oct 08 2012

A159678 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2-equation problem 7n(j) + 1 = a(j)a(j) and 9n(j) + 1 = b(j)b(j) with positive integer numbers.

Original entry on oeis.org

1, 17, 271, 4319, 68833, 1097009, 17483311, 278635967, 4440692161, 70772438609, 1127918325583, 17975920770719, 286486814005921, 4565813103324017, 72766522839178351, 1159698552323529599, 18482410314337295233, 294558866477073194129, 4694459453318833810831

Offset: 1

Paul Weisenhorn, Apr 19 2009

The sequence a(j) is A157456, the sequence n(j) is A159679, the sequence b(j) the sequence given here.

Numbers k such that 7*k^2 + 2 is a square. - Colin Barker, Mar 17 2014

Colin Barker, Table of n, a(n) for n = 1..800
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Index entries for linear recurrences with constant coefficients, signature (16,-1).

Cf. A077412, A157456, A159679, A266698.

[n le 2 select 17^(n-1) else 16*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Jun 03 2018

for a from 1 by 2 to 100000 do b:=sqrt((9*a*a-2)/7): if (trunc(b)=b) then
n:=(a*a-1)/7: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:
# Second program
seq(simplify(ChebyshevU(n-1,8) + ChebyshevU(n-2,8)), n=1..30); # G. C. Greubel, Sep 27 2022

Rest[CoefficientList[Series[x (1+x)/(1-16x+x^2),{x,0,30}],x]] (* or *) LinearRecurrence[{16,-1},{1,17},30] (* Harvey P. Dale, Dec 25 2011 *)

Vec(x*(1+x)/(1-16*x+x^2) + O(x^30)) \\ Michel Marcus, Jan 03 2016

a(n) = round((-(8-3*sqrt(7))^n*(3+sqrt(7))-(-3+sqrt(7))*(8+3*sqrt(7))^n)/(2*sqrt(7))) \\ Colin Barker, Jul 25 2016

[(lucas_number2(n,16,1)-lucas_number2(n-1,16,1))/14 for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

The b(j) recurrence (this sequence) is b(1)=1, b(2)=17, b(t+2) = 16*b(t+1) - b(t).
From R. J. Mathar, Oct 31 2011: (Start)
G.f.: x*(1+x) / ( 1-16*x+x^2 ).
a(n) = A077412(n-1) + A077412(n-2). (End)
a(n) = 16*a(n-1) - a(n-2), with a(1)=1, a(2)=17. - Harvey P. Dale, Dec 25 2011
a(n) = ( (3-sqrt(7))*(8+3*sqrt(7))^n - (3+sqrt(7))*(8-3*sqrt(7))^n )/(2*sqrt(7)). - Colin Barker, Jul 25 2016

More terms from Zerinvary Lajos, Nov 10 2009

A257711 Triangular numbers (A000217) that are the sum of seven consecutive triangular numbers.

Original entry on oeis.org

210, 3486, 51681, 883785, 13125126, 224476266, 3333728685, 57016086141, 846753959226, 14481861401910, 215072171913081, 3678335779997361, 54627484911961710, 934282806257926146, 13875166095466359621, 237304154453733242085, 3524237560763543380386

Offset: 1

Colin Barker, May 05 2015

			210 is in the sequence because T(20) = 210 = 10+15+21+28+36+45+55 = T(4)+ ... +T(10).

Colin Barker, Table of n, a(n) for n = 1..830
Index entries for linear recurrences with constant coefficients, signature (1,254,-254,-1,1).

Cf. A000217, A001110, A129803, A131557, A257712, A257713, A259413, A259414, A259415.

LinearRecurrence[{1, 254, -254, -1, 1}, {210, 3486, 51681, 883785, 13125126}, 30] (* Vincenzo Librandi, Jun 27 2015 *)

Vec(-21*x*(x^4-245*x^2+156*x+10) / ((x-1)*(x^2-16*x+1)*(x^2+16*x+1)) + O(x^100))

G.f.: -21*x*(x^4-245*x^2+156*x+10) / ((x-1)*(x^2-16*x+1)*(x^2+16*x+1)).

16*a(n) = 104 +225*A157456(n+1) +7*(-1)^n*A159678(n+1). - R. J. Mathar, Apr 28 2020

A243469 Denominators of the rational convergents to the periodic continued fraction 1/(2 + 1/(7 + 1/(2 + 1/(7 + ...)))).

Original entry on oeis.org

1, 2, 15, 32, 239, 510, 3809, 8128, 60705, 129538, 967471, 2064480, 15418831, 32902142, 245733825, 524369792, 3916322369, 8357014530, 62415424079, 133187862688, 994730462895, 2122648788478, 15853271982241, 33829192752960, 252657621252961, 539144435258882

Offset: 1

Peter Bala, Jun 06 2014

The sequence of convergents to the simple periodic continued fraction 1/(2 + 1/(7 + 1/(2 + 1/(7 + ...)))) begins [0/1, 1/2, 7/15, 15/32, 112/239, 239/510, ...]. Euler considers these convergents, in section 378 of the first volume of his textbook Introductio in Analysin Infinitorum, as a way of finding rational approximations to sqrt(7). The present sequence is the sequence of denominators of the convergents. It is a strong divisibility sequence, that is gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. The sequence is closely related to A041111, the Lehmer numbers U_n(sqrt(R),Q) with parameters R = 14 and Q = -1.

See A243470 for the sequence of numerators to the convergents.

Peter Bala, Notes on 2-periodic continued fractions and Lehmer sequences
Leonhard Euler, Introductio in analysin infinitorum, Vol.1, Chapter 18, section 378. French and German translations.
Eric W. Weisstein, MathWorld: Lehmer Number
Index to divisibility sequences

Cf. A041111, A243470.

Vec(x*(1+2*x-x^2)/(1-16*x^2+x^4)+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2015

Let alpha = ( sqrt(14) + sqrt(18) )/2 and beta = ( sqrt(14) - sqrt(18) )/2 be the roots of the equation x^2 - sqrt(14)*x - 1 = 0. Then a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, while a(n) = 2*(alpha^n - beta^n)/(alpha^2 - beta^2) for n even.
a(2*n + 1) = Product_{k = 1..n} (14 + 4*cos^2(k*Pi/(2*n+1)));
a(2*n) = 2*Product_{k = 1..n-1} (14 + 4*cos^2(k*Pi/(2*n))).
Recurrence equations: a(0) = 0, a(1) = 1 and for n >= 2, a(2*n) = 2*a(2*n - 1) + a(2*n - 2) and a(2*n + 1) = 7*a(2*n) + a(2*n - 1).
Fourth-order recurrence: a(n) = 16*a(n - 2) - a(n - 4) for n >= 5.
O.g.f.: x*(1 + 2*x - x^2)/(1 - 16*x^2 + x^4).
a(2n-1) = A157456, a(2n) = 2*A077412(n-1). - Ralf Stephan, Jun 13 2014

A159668 Expansion of (1 - x)/(1 - 28*x + x^2).

Original entry on oeis.org

1, 27, 755, 21113, 590409, 16510339, 461699083, 12911063985, 361048092497, 10096435525931, 282339146633571, 7895399670214057, 220788851619360025, 6174192445671866643, 172656599627192905979, 4828210597115729500769, 135017240119613233115553

Offset: 0

Paul Weisenhorn, Apr 19 2009

Previous name was: The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 13*n(j) + 1 = a(j)*a(j) and 15*n(j) + 1 = b(j)*b(j) with positive integer numbers.

Positive values of x (or y) satisfying x^2 - 28*x*y + y^2 + 26 = 0. - Colin Barker, Feb 23 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..199
Index entries for linear recurrences with constant coefficients, signature (28,-1).

Cf. A097311, A157456, A159669, A159673.

Cf. similar sequences listed in A238379.

[n le 2 select 27^(n-1) else 28*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 25 2014

for a from 1 by 2 to 100000 do b:=sqrt((15*a*a-2)/13): if (trunc(b)=b) then
n:=(a*a-1)/13: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: endif: enddo:
# Second program
seq(simplify(ChebyshevU(n,14) -ChebyshevU(n-1,14)), n=0..40); # G. C. Greubel, Sep 26 2022

CoefficientList[Series[(1-x)/(1-28x+x^2), {x,0,40}], x] (* Vincenzo Librandi, Feb 25 2014 *)
LinearRecurrence[{28,-1},{1,27},40] (* Harvey P. Dale, Apr 09 2014 *)

Vec((-x+1)/(x^2-28*x+1) + O(x^100)) \\ Colin Barker, Feb 23 2014

def A159668(n): return chebyshev_U(n,14) - chebyshev_U(n-1,14)
[A159668(n) for n in range(40)] # G. C. Greubel, Sep 26 2022

G.f.: (1 - x)/(1 - 28*x + x^2).
The a(j) recurrence is a(0)=1, a(1)=27, a(t+2) = 28*a(t+1) - a(t) resulting in terms 1, 27, 755, 21113, ... (this sequence).
The b(j) recurrence is b(0)=1, b(1)=29, b(t+2) = 28*b(t+1) - b(t) resulting in terms 1, 29, 811, 22679, ... (A159669).
The n(j) recurrence is n(0) = n(1) = 0, n(2) = 56, n(t+3) = 783*(n(t+2) -n(t+1)) + n(t) resulting in terms 0, 0, 56, 43848, 34289136, ... (A159673).
a(n) = (1/30)*(15-sqrt(195))*(1+(14+sqrt(195))^(2*n+1))/(14+sqrt(195))^n. - Bruno Berselli, Feb 25 2014
a(n) = 28*a(n-1) - a(n-2), a(0)=1, a(1)=27. - Harvey P. Dale, Apr 09 2014
a(n) = A097311(n) - A097311(n-1). - G. C. Greubel, Sep 26 2022

More terms from Colin Barker, Feb 23 2014

New name and offset changed to 0 from Joerg Arndt, Feb 23 2014

A159664 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 11n(j) + 1 = a(j)a(j) and 13n(j) + 1 = b(j)b(j), with positive integer numbers.

Original entry on oeis.org

1, 23, 551, 13201, 316273, 7577351, 181540151, 4349386273, 104203730401, 2496540143351, 59812759710023, 1433009692897201, 34332419869822801, 822545067182850023, 19706749192518577751, 472139435553263016001, 11311639704085793806273, 271007213462505788334551

Offset: 1

Paul Weisenhorn, Apr 19 2009

Positive values of x (or y) satisfying x^2 - 24*x*y + y^2 + 22 = 0. - Colin Barker, Feb 19 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (24,-1).

Cf. A077423, A157456, A159661, A159665.

Cf. similar sequences listed in A238379.

[n le 2 select 23^(n-1) else 24*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 21 2014

for a from 1 by 2 to 100000 do b:=sqrt((13*a*a-2)/11): if (trunc(b)=b) then
n:=(a*a-1)/11: La:=[La),a]:Lb:=[op(Lb),b]: Ln:=[op(Ln),n]: end if: end do:
# Second program
seq(simplify(ChebyshevU(n-1,12) - ChebyshevU(n-2,12)), n=1..30); # G. C. Greubel, Sep 27 2022

CoefficientList[Series[(1-x)/(1-24x+x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 21 2014 *)
LinearRecurrence[{24,-1}, {1,23}, 30] (* G. C. Greubel, Sep 27 2022 *)

Vec(x*(1-x)/(1-24*x+x^2) + O(x^100)) \\ Colin Barker, Feb 19 2014

a(n) = round((12+sqrt(143))^(-n)*(13+sqrt(143)-(-13+sqrt(143))*(12+sqrt(143))^(2*n))/26) \\ Colin Barker, Jul 25 2016

def A159664(n): return chebyshev_U(n-1,12) - chebyshev_U(n-2,12)
[A159664(n) for n in range(1,30)] # G. C. Greubel, Sep 27 2022

The a(j) recurrence is a(1)=1, a(2)=23, a(t+2) = 24*a(t+1) - a(t) resulting in terms 1, 23, 551, 13201, ... (this sequence).
The b(j) recurrence is b(1)=1, b(2)=25, b(t+2) = 24*b(t+1) - b(t) resulting in terms 1, 25, 599, 14351, ... (A159661).
The n(j) recurrence is n(0)=n(1)=1, n(2)=48, n(t+3) = 575*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 48, 27600, 15842400, ... (A159665).
G.f.: x*(1-x)/(1 - 24*x + x^2). - Colin Barker, Feb 19 2014
a(n) = (12+sqrt(143))^(-n)*(13+sqrt(143)-(-13+sqrt(143))*(12+sqrt(143))^(2*n))/26. - Colin Barker, Jul 25 2016
a(n) = A077423(n-1) - A077423(n-2). - G. C. Greubel, Sep 27 2022

More terms from Colin Barker, Feb 19 2014

A159674 Expansion of (1 - x)/(1 - 32*x + x^2).

Original entry on oeis.org

1, 31, 991, 31681, 1012801, 32377951, 1035081631, 33090234241, 1057852414081, 33818187016351, 1081124132109151, 34562154040476481, 1104907805163138241, 35322487611179947231, 1129214695752595173151, 36099547776471865593601, 1154056314151347103822081

Offset: 0

Paul Weisenhorn, Apr 19 2009

Previous name was: The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 15*n(j) + 1 = a(j)*a(j) and 17*n(j) + 1 = b(j)*b(j) with positive integers.

Positive values of x (or y) satisfying x^2 - 32*x*y + y^2 + 30 = 0. - Colin Barker, Feb 24 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (32,-1).

Cf. A029548, A157456, A159675, A159677.

Cf. similar sequences listed in A238379.

A029548:= func< n | Evaluate(ChebyshevSecond(n),16) >;
[A029548(n+1) -A029548(n): n in [0..30]]; // G. C. Greubel, Sep 25 2022

for a from 1 by 2 to 100000 do b:=sqrt((17*a*a-2)/15): if (trunc(b)=b) then
n:=(a*a-1)/15: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: endif: enddo:
# Second program
seq(simplify(ChebyshevU(n, 16) -ChebyshevU(n-1, 16)), n=0..30); # G. C. Greubel, Sep 25 2022

CoefficientList[Series[(1-x)/(1-32*x+x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 26 2014 *)
LinearRecurrence[{32,-1},{1,31},30] (* Harvey P. Dale, Mar 21 2017 *)

concat([0], Vec((-x+1)/(x^2-32*x+1) + O(x^100))) \\ Colin Barker, Feb 24 2014

def A159674(n): return chebyshev_U(n, 16) - chebyshev_U(n-1, 16)
[A159674(n) for n in range(31)] # G. C. Greubel, Sep 25 2022

The a(j) recurrence is: a(0)=1, a(1)=31, a(t+2) = 32*a(t+1) - a(t) resulting in terms 1, 31, 991, 31681, ... (this sequence).
The b(j) recurrence is: b(0)=1, b(1)=33, b(t+2) = 32*b(t+1) - b(t) resulting in terms 1, 33, 1055, 33727, ... (A159675).
The n(j) recurrence is: n(-1) = n(0) = 0, n(1) = 64, n(t+3) = 1023*(n(t+2) -n(t+1)) + n(t) resulting in terms 0, 0, 64, 65472, 66912384, ... (A159677).
a(n) = (1/34)*(17-sqrt(255))*(1+(16+sqrt(255))^(2*n+1))/(16+sqrt(255))^n. - Bruno Berselli, Feb 25 2014
a(n) = ChebyshevU(n, 16) - ChebyshevU(n-1, 16) = A029548(n) - A029548(n-1). - G. C. Greubel, Sep 25 2022

More terms and new name from Colin Barker, Feb 24 2014

Set offset to 0 by Joerg Arndt, Feb 25 2014

A159661 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 11n(j) + 1 = a(j)a(j) and 13n(j) + 1 = b(j)b(j) with positive integer elements. the solutions of the 2 equations problem: 11n(j) + 1 = a(j)a(j); 13n(j) + 1 = b(j)b(j); with integer numbers.

Original entry on oeis.org

1, 25, 599, 14351, 343825, 8237449, 197354951, 4728281375, 113281398049, 2714025271801, 65023325125175, 1557845777732399, 37323275340452401, 894200762393125225, 21423495022094552999, 513269679767876146751, 12297048819406932969025, 294615901985998515109849

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..725
Index entries for linear recurrences with constant coefficients, signature (24,-1).

Cf. A077423, A157456.

[n le 2 select 24*n-23 else 24*Self(n-1) -Self(n-2): n in [1..31]]; // G. C. Greubel, Jun 25 2022

for a from 1 by 2 to 100000 do b:=sqrt((13*a*a-2)/11): if (trunc(b)=b) then
n:=(a^2-1)/C: La:=[op(La),a]: Lb:=[op(Lb),b]: Ln:=[op(Ln),n]: endif: enddo:

LinearRecurrence[{24,-1}, {1,25}, 31] (* G. C. Greubel, Jun 25 2022 *)

Vec(x*(x+1)/(x^2-24*x+1) + O(x^20)) \\ Colin Barker, Sep 25 2015

a(n) = round((12+sqrt(143))^(-n)*(-11-sqrt(143)+(-11+sqrt(143))*(12+sqrt(143))^(2*n))/22) \\ Colin Barker, Jul 26 2016

[chebyshev_U(n-1, 12) + chebyshev_U(n-2, 12) for n in (1..30)] # G. C. Greubel, Jun 25 2022

The a(j) recurrence is a(1)=1; a(2)=23; a(t+2) = 24*a(t+1) - a(t); resulting in a(j) terms 1, 23, 551, 13201, 316273, 7577351, 181540151, 4349386273.
The b(j) recurrence is b(1)=1; b(2)=23; b(t+2) = 24*b(t+1) - b(t); resulting in b(j) terms 1, 25, 599, 14351, 343825, 8237449 as listed above.
The n(j) recurrence is n(0)=n(1)=0; n(2)=48; n(t+3) = 575*(n(t+2) - n(t+1)) + n(t) resulting in n(j) terms 0, 0, 48, 27600, 15842400, 9093510048, 5219658925200.
From Colin Barker, Sep 25 2015: (Start)
a(n) = 24*a(n-1)-a(n-2) for n>2.
G.f.: x*(1+x) / (1 - 24*x + x^2). (End)
a(n) = (12+sqrt(143))^(-n)*(-11 - sqrt(143) + (-11+sqrt(143))*(12+sqrt(143))^(2*n))/22. - Colin Barker, Jul 26 2016
From G. C. Greubel, Jun 25 2022: (Start)
a(n) = ChebyshevU(n-1, 12) + Chebyshev(n-2, 12).
E.g.f.: exp(12*x)*(cosh(sqrt(143)*x) + sqrt(13/11)*sinh(sqrt(143)*x)). (End)

A159669 Expansion of x(1 + x)/(1 - 28x + x^2).

Original entry on oeis.org

1, 29, 811, 22679, 634201, 17734949, 495944371, 13868707439, 387827863921, 10845311482349, 303280893641851, 8481019710489479, 237165271000063561, 6632146568291290229, 185462938641156062851, 5186330135384078469599, 145031780852113041085921

Offset: 1

Paul Weisenhorn, Apr 19 2009

Previous name was: The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 13*n(j) + 1 = a(j)*a(j) and 15*n(j) + 1 = b(j)*b(j) with positive integer numbers.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (28,-1).

Cf. A097311, A157456, A159668, A159673.

A097311:= func< n | Evaluate(ChebyshevSecond(n-1), 14) >;
[A097311(n+1) + A097311(n): n in [1..30]]; // G. C. Greubel, Sep 25 2022

for a from 1 by 2 to 100000 do b:=sqrt((15*a*a-2)/13): if (trunc(b)=b) then
n:=(a*a-1)/13: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: endif: enddo:
# Second program
seq(simplify(ChebyshevU(n, 14) +ChebyshevU(n-1, 14)), n=1..30); # G. C. Greubel, Sep 25 2022

CoefficientList[Series[(1+x)/(1-28x+x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 26 2014 *)
LinearRecurrence[{28,-1},{1,29},20] (* Harvey P. Dale, Jul 01 2019 *)

Vec(x*(x+1)/(x^2-28*x+1) + O(x^100)) \\ Colin Barker, Feb 24 2014

a(n) = round((14+sqrt(195))^(-n)*(-13-sqrt(195)+(-13+sqrt(195))*(14+sqrt(195))^(2*n))/26) \\ Colin Barker, Jul 25 2016

def A159669(n): return chebyshev_U(n-1, 14) + chebyshev_U(n-2, 14)
[A159669(n) for n in range(1,30)] # G. C. Greubel, Sep 25 2022

The a(j) recurrence is a(1)=1, a(2)=27, a(t+2) = 28*a(t+1) - a(t) resulting in terms 1, 27, 755, 21113, ... (A159668).
The b(j) recurrence is b(1)=1, b(2)=29, b(t+2) = 28*b(t+1) - b(t) resulting in terms 1, 29, 811, 22679, ... (this sequence).
The n(j) recurrence is n(0) = n(1) = 0, n(2) = 56, n(t+3) = 783*(n(t+2) -n(t+1)) + n(t) resulting in terms 0, 0, 56, 43848, 34289136, ... (A159673).
G.f.: x*(1+x)/(1-28*x+x^2). - Vincenzo Librandi, Feb 26 2014
a(n) = (14+sqrt(195))^(-n)*(-13-sqrt(195)+(-13+sqrt(195))*(14+sqrt(195))^(2*n))/26. - Colin Barker, Jul 25 2016
a(n) = chebyshev_U(n-1, 14) + chebyshev_U(n-2, 14) = A097311(n) + A097311(n-1). - G. C. Greubel, Sep 25 2022

More terms and new name from Colin Barker, Feb 24 2014

cp's OEIS Frontend

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

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A160682 The list of the A values in the common solutions to 13*k+1 = A^2 and 17*k+1 = B^2.

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A159678 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2-equation problem 7*n(j) + 1 = a(j)*a(j) and 9*n(j) + 1 = b(j)*b(j) with positive integer numbers.

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A257711 Triangular numbers (A000217) that are the sum of seven consecutive triangular numbers.

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A243469 Denominators of the rational convergents to the periodic continued fraction 1/(2 + 1/(7 + 1/(2 + 1/(7 + ...)))).

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A159668 Expansion of (1 - x)/(1 - 28*x + x^2).

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A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

A159678 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2-equation problem 7n(j) + 1 = a(j)a(j) and 9n(j) + 1 = b(j)b(j) with positive integer numbers.

A159664 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 11n(j) + 1 = a(j)a(j) and 13n(j) + 1 = b(j)b(j), with positive integer numbers.

A159669 Expansion of x(1 + x)/(1 - 28x + x^2).