A001541 -id:A001541 - cp's OEIS frontend

A090390 Repeatedly multiply (1,0,0) by ([1,2,2],[2,1,2],[2,2,3]); sequence gives leading entry.

1, 1, 9, 49, 289, 1681, 9801, 57121, 332929, 1940449, 11309769, 65918161, 384199201, 2239277041, 13051463049, 76069501249, 443365544449, 2584123765441, 15061377048201, 87784138523761, 511643454094369, 2982076586042449, 17380816062160329, 101302819786919521, 590436102659356801

Offset: 0

Vim Wenders, Jan 30 2004

The values of a and b in (a,b,c)*A give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1; the values of c are A000129(2n)
Binomial transform of A086348. - Johannes W. Meijer, Aug 01 2010
All values of a(n) are squares.  sqrt(a(n+1)) = A001333(n). The ratio a(n+1)/a(n) converges to 3 + 2*sqrt(2). - Richard R. Forberg, Aug 14 2013

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 60 at p. 123.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A000129, A001333, A001541, A079291, A086348, A095344, A123270, A302946.

a090390 n = a090390_list !! n
a090390_list = 1 : 1 : 9 : zipWith (-) (map (* 5) $
   tail $ zipWith (+) (tail a090390_list) a090390_list) a090390_list
-- Reinhard Zumkeller, Aug 17 2013

[Evaluate(DicksonFirst(n,-1),2)^2/4: n in [0..40]]; // G. C. Greubel, Aug 21 2022

a:= n-> (<<1|0|0>>. <<1|2|2>, <2|1|2>, <2|2|3>>^n)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Aug 17 2013

CoefficientList[Series[(1-4x-x^2)/((1+x)(1-6x+x^2)),{x, 0, 30}], x] (* Harvey P. Dale, May 20 2012 *)
LinearRecurrence[{5,5,-1}, {1,1,9}, 30] (* Harvey P. Dale, May 20 2012 *)
Table[(ChebyshevT[n,3]+(-1)^n)/2, {n,0,30}] (* Eric W. Weisstein, Apr 17 2018 *)
(LucasL[Range[0, 40], 2]/2)^2 (* G. C. Greubel, Aug 21 2022 *)

a(n)=polcoeff((1-4*x-x^2)/((1+x)*(1-6*x+x^2))+x*O(x^n),n)

a(n)=if(n<0,0,([1,2,2;2,1,2;2,2,3]^n)[1,1])

Vec( (1-4*x-x^2)/((1+x)*(1-6*x+x^2)) + O(x^66) ) \\ Joerg Arndt, Aug 16 2013

use Math::Matrix; use Math::BigInt; $a = new Math::Matrix ([ 1, 2, 2], [ 2, 1, 2], [ 2, 2, 3]); $p = new Math::Matrix ([1, 0, 0]); $p->print(); for ($i=1; $i<20;$i++) { $p = $p->multiply($a); $p->print(); }

[lucas_number2(n,2,-1)^2/4 for n in (0..40)] # G. C. Greubel, Aug 21 2022

G.f.: (1-4*x-x^2)/((1+x)*(1-6*x+x^2)).
a(n) = A001333(n)^2
(a, b, c) = (1, 0, 0). Recursively multiply (a, b, c)*( [1, 2, 2], [2, 1, 2], [2, 2, 3] ).
M^n * [ 1 1 1] = [a(n+1) q a(n)], where M = the 3 X 3 matrix [4 4 1 / 2 1 0 / 1 0 0]. E.g. M^5 * [1 1 1] = [9801 4059 1681] where 9801 = a(6), 1681 = a(5). Similarly, M^n * [1 0 0] generates A079291 (Pell number squares). - Gary W. Adamson, Oct 31 2004
a(n) = (((1+sqrt(2))^(2*n) + (1-sqrt(2))^(2*n)) + 2*(-1)^n)/4 - Lambert Klasen (lambert.klasen(AT)gmx.net), Oct 09 2005
a(n) = (A001541(n) + (-1)^n)/2. - R. J. Mathar, Nov 20 2009
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3), with a(0)=1, a(1)=1, a(2)=9. - Harvey P. Dale, May 20 2012
(a(n)) = tesseq(- .5'j + .5'k - .5j' + .5k' - 2'ii' + 'jj' - 'kk' + .5'ij' + .5'ik' + .5'ji' + 'jk' + .5'ki' + 'kj' + e), apart from initial term. - Creighton Dement, Nov 16 2004
a(n) = A302946(n)/4. - Eric W. Weisstein, Apr 17 2018
E.g.f.: exp(-x)*(1 + exp(4*x)*cosh(2*sqrt(2)*x))/2. - Stefano Spezia, Aug 03 2024

A143608 A005319 and A002315 interleaved.

Original entry on oeis.org

0, 1, 4, 7, 24, 41, 140, 239, 816, 1393, 4756, 8119, 27720, 47321, 161564, 275807, 941664, 1607521, 5488420, 9369319, 31988856, 54608393, 186444716, 318281039, 1086679440, 1855077841, 6333631924, 10812186007, 36915112104, 63018038201, 215157040700

Offset: 0

Originally submitted by Clark Kimberling, Aug 27 2008. Merged with an essentially identical sequence submitted by Kenneth J Ramsey, Jun 01 2012, by N. J. A. Sloane, Aug 02 2012

Also, numerators of the lower principal and intermediate convergents to 2^(1/2). The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators=A143608 and denominators=A079496.
Sequence a(n) such that a(2*n) = sqrt(2*A001108(2*n)) and a(2*n+1) = sqrt(A001108(2*n+1)).
For n > 0, a(n) divides A******(k+1,n+1)-A******(k,n+1) where A****** is any one of A182431, A182439, A182440, A182441 and k is any nonnegative integer.
If p is a prime of the form 8*r +/- 3 then a(p+1) == 0 (mod p); if p is a prime of the form 8*r +/- 1 then a(p-1) == 0 (mod p).
Numbers n such that sqrt(floor(n^2/2 + 1)) is an integer. The integer square roots are given by A079496. - Richard R. Forberg, Aug 01 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
a(2*n) = sqrt(2)*(1 o 1 o ... o 1) (2*n terms). Cf. A084068.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = sqrt(2)*U(2*n) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ). (End)

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Colin Barker, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Creighton Kenneth Dement, Comments on A143608 and A143609
Vincent Granville, Successive records in mathematical sequences: surprising result, Mathematics Stack Exchange, 2019.
Clark Kimberling, Best lower and upper approximations to irrational numbers, Elem. Math. vol. 52 iss. 3 (1997) 122-126.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A002315, A005319, A182431, A001108, A049629, A084068.

I:=[0,1,4,7]; [n le 4 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..30]]; // G. C. Greubel, Mar 27 2018

A143608 := proc(n)
    option remember;
    if n <= 3 then
        op(n+1,[0,1,4,7]) ;
    else
        6*procname(n-2)-procname(n-4) ;
    end if;
end proc: # R. J. Mathar, Jul 22 2012

a = -4; b = -1; Reap[While[b<2000000000, t = 4*b-a; Sow[t]; a=b; b=t; t = 2*b-a; Sow[t]; a=b; b=t]][[2,1]]
CoefficientList[Series[x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 24 2014 *)
LinearRecurrence[{0, 6, 0, -1}, {0, 1, 4, 7}, 31] (* Jean-François Alcover, Sep 21 2017 *)

a(n)=([0,1,0,0;0,0,1,0;0,0,0,1;-1,0,6,0]^n*[0;1;4;7])[1,1] \\ Charles R Greathouse IV, Jun 11 2015

concat(0, Vec(x*(1+4*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)) + O(x^50))) \\ Colin Barker, Mar 27 2016

a(2*n) = (a(2*n - 1) + a(2*n + 1))/2.
a(2*n + 1) = (a(2*n) + a(2*n + 2))/4.
a(2*n) = 4*A001109(n).
a(2*n + 1) = 4*A001109(n) + A001541(n).
From Colin Barker, Jun 29 2012: (Start)
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1 + 4*x + x^2)/((1 + 2*x - x^2)*(1 - 2*x - x^2)) = x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4). (End)
2*a(n) = A078057(n) - A123335(n-1). - R. J. Mathar, Jul 04 2012
a(2n) = A005319(n); a(2n+1) = A002315(n). - R. J. Mathar, Jul 17 2009
a(n)*a(n+1) + 1 = A001653(n+1). - Charlie Marion, Dec 11 2012
a(n) = (((-2 - sqrt(2) + (-1)^n * (-2+sqrt(2))) * ((-1+sqrt(2))^n - (1+sqrt(2))^n)))/(4*sqrt(2)). - Colin Barker, Mar 27 2016
a(n) = A084068(n) - A079496(n). - César Aguilera, Feb 14 2023

A077420 Bisection of Chebyshev sequence T(n,3) (odd part) with Diophantine property.

Original entry on oeis.org

1, 33, 1121, 38081, 1293633, 43945441, 1492851361, 50713000833, 1722749176961, 58522759015841, 1988051057361633, 67535213191279681, 2294209197446147521, 77935577499977736033, 2647515425801796877601

Offset: 0

Wolfdieter Lang, Nov 29 2002

(3*a(n))^2 - 2*(2*b(n))^2 = 1 with companion sequence b(n)= A046176(n+1), n>=0 (special solutions of Pell equation).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Z. Cerin and G. M. Gianella, On sums of squares of Pell-Lucas Numbers, INTEGERS 6 (2006) #A15
Tanya Khovanova, Recursive Sequences
S. Vidhyalakshmi, V. Krithika, and K. Agalya, On The Negative Pell Equation y^2 = 72x^2 - 8, International Journal of Emerging Technologies in Engineering Research (IJETER) Volume 4, Issue 2, February (2016).
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A056771 (even part).
Row 34 of array A094954.
Row 3 of array A188646.
Cf. similar sequences listed in A238379.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k are listed in A302329. This is the case k=3.

I:=[1,33]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

LinearRecurrence[{34,-1},{1,33},20] (* Vincenzo Librandi, Nov 22 2011 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041027 *)
a[18, 20] (* Gerry Martens, Jun 07 2015 *)

makelist(expand(((1+sqrt(2))^(4*n+2)+(1-sqrt(2))^(4*n+2))/6),n,0,14);  /* _Bruno Berselli, Nov 22 2011 */

Vec((1-x)/(1-34*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Nov 22 2011

a(n) = 34*a(n-1) - a(n-2), a(-1)=1, a(0)=1.
a(n) = T(2*n+1, 3)/3 = S(n, 34) - S(n-1, 34), with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 34)= A029547(n), T(n, 3)=A001541(n).
G.f.: (1-x)/(1-34*x+x^2).
a(n) = sqrt(8*A046176(n+1)^2 + 1)/3.
a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(2*n)-a(n-1), where k = (sqrt(2)+1)^4 = 17+12*sqrt(2) and a(0)=1. - Charles L. Hohn, Apr 05 2011
a(n) = a(-n-1) = A029547(n)-A029547(n-1) = ((1+sqrt(2))^(4n+2)+(1-sqrt(2))^(4n+2))/6. - Bruno Berselli, Nov 22 2011

A084128 a(n) = 4a(n-1) + 4a(n-2), a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 12, 56, 272, 1312, 6336, 30592, 147712, 713216, 3443712, 16627712, 80285696, 387653632, 1871757312, 9037643776, 43637604352, 210700992512, 1017354387456, 4912221519872, 23718303629312, 114522100596736, 552961616904192, 2669934870003712

Offset: 0

Paul Barry, May 16 2003

Original name was: Generalized Fibonacci sequence.

Binomial transform of A084058.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (4,4).

Cf. A001333 A001541, A002203, A057087, A084058, A084128, A201730.

Appears in A086346, A086347 and A086348. - Johannes W. Meijer, Aug 01 2010

[2^(n-1)*Evaluate(DicksonFirst(n,-1), 2): n in [0..40]]; // G. C. Greubel, Oct 13 2022

a:=proc(n) option remember; if n=0 then 1 elif n=1 then 2 else
4*a(n-1)+4*a(n-2); fi; end: seq(a(n), n=0..40); # Wesley Ivan Hurt, Jan 31 2017
a := n -> (2*I)^n*ChebyshevT(n, -I):
seq(simplify(a(n)), n = 0..23); # Peter Luschny, Dec 03 2023

CoefficientList[Series[(2 z - 1)/(4 z^2 + 4 z - 1), {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
Table[2^(n-1) LucasL[n, 2], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 07 2016 *)
LinearRecurrence[{4,4},{1,2},30] (* Harvey P. Dale, Mar 01 2018 *)

a(n)=if(n<0,0,polsym(4+4*x-x^2,n)[n+1]/2)

[lucas_number2(n,4,-4)/2 for n in range(0, 23)] # Zerinvary Lajos, May 14 2009

a(n) = 2^n * A001333(n).
G.f.: (1-2*x)/(1-4*x-4*x^2).
a(n) = 4*a(n-1) + 4*a(n-2), a(0)=1, a(1)=2.
a(n) = (2 + 2*sqrt(2))^n/2 + (2 - 2*sqrt(2))^n/2.
E.g.f.: exp(2*x)*cosh(2*x*sqrt(2)).
From Johannes W. Meijer, Aug 01 2010: (Start)
Lim_{k->infinity} a(n+k)/a(k) = A084128(n) + 2*A057087(n-1)*sqrt(2).
Lim_{n->infinity} A084128(n)/A057087(n-1) = sqrt(2). (End)
a(n) = Sum_{k=0..n} A201730(n,k)*7^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(4*k-2)/(x*(4*k+2) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) = 2^(n-1)*A002203(n). - Vladimir Reshetnikov, Oct 07 2016

A038725 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 11, 64, 373, 2174, 12671, 73852, 430441, 2508794, 14622323, 85225144, 496728541, 2895146102, 16874148071, 98349742324, 573224305873, 3340996092914, 19472752251611, 113495517416752, 661500352248901, 3855506596076654, 22471539224211023, 130973728749189484

Offset: 0

Barry E. Williams, May 02 2000

From Wolfdieter Lang, Feb 26 2015: (Start)
The sequence {2*a(n+1)}_{n >= 0}, gives all positive solutions y = y2(n) = 2*a(n+1) of the second class of the Pell equation x^2 - 2*y^2  = -7. For the corresponding terms x = x2(n) see A255236(n).
See A255236 for comments on the first class solutions and the relation to the Pell equation x^2 - 2*y^2 = 14. (End)

			n = 2: a(3) = sqrt((181^2 + 7)/2)/2 = 64.
a(3) = (53 + 75)/2 = 64. - _Wolfdieter Lang_, Mar 19 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Stefano Spezia, Table of n, a(n) for n = 0..1300
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See x'(n) in Lemma 7.9 p. 21.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001653 and A001541. Cf. A001109.

A038723(n) = a(-n).

a[0]:=1: a[1]:=2: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006

Union[Flatten[NestList[{#[[2]],#[[3]],6#[[3]]-#[[2]]}&,{1,2,11},25]]]  (* Harvey P. Dale, Mar 04 2011 *)
LinearRecurrence[{6,-1},{1,2},30] (* Harvey P. Dale, Jun 12 2017 *)

{a(n) = real((3 + 2*quadgen(8))^n * (1 - quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */

{a(n) = polchebyshev(n, 1, 3) - polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3); a(n) = ((4-sqrt(2))/8)*(3+2*sqrt(2))^(n-1)+((4+sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008
From Michael Somos, Sep 28 2008: (Start)
Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v.
G.f.: (1 - 4*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). (End)
From Wolfdieter Lang, Feb 26 2015: (Start)
a(n) = S(n, 6) - 4*S(n-1, 6), n>=0, with the Chebyshev polynomials S(n, x) (A049310), with S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell equation comment above.
a(n) = 6*a(n-1) - a(n-2), n >= 1, a(-1) = 4, a(0) = 1. (See the name.) (End)
From Wolfdieter Lang, Mar 19 2015: (Start)
a(n+1) = sqrt((A255236(n)^2 + 7)/2)/2, n >= 0.
a(n+1) = (A038761(n) + A038762(n))/2, n >= 0. See the Mar 19 2015 comment on A054490. (End)
E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) - sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, May 01 2020

A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

Original entry on oeis.org

1, 17, 577, 19601, 665857, 22619537, 768398401, 26102926097, 886731088897, 30122754096401, 1023286908188737, 34761632124320657, 1180872205318713601, 40114893348711941777, 1362725501650887306817, 46292552162781456490001

Offset: 0

Henry Bottomley, Aug 16 2000

The sequence satisfies the Pell equation a(n)^2 - 18 * A202299(n+1)^2 = 1. - Vincenzo Librandi, Dec 19 2011
Also numbers n such that n - 1 and 2*n + 2 are squares. - Arkadiusz Wesolowski, Mar 15 2015
And they, n - 1 and 2*n + 2, are the squares of A005319 and A003499. - Michel Marcus, Mar 15 2015
This sequence {a(n)} gives all the nonnegative integer solutions of the Pell equation a(n)^2 - 32*(3*A091761(n))^2 = +1. - Wolfdieter Lang, Mar 09 2019

			G.f. = 1 + 17*x + 577*x^2 + 19601*x^3 + 665857*x^4 + 22619537*x^5 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..600 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A001075, A001541, A001091, A001079, A023038, A011943, A001081, A023039, A001085 and note relationship with square triangular number sequences A001110 and A001109. A091761.

Row 3 of array A188644.

I:=[1, 17]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 18 2011

LinearRecurrence[{34,-1},{1,17},30] (* Vincenzo Librandi, Dec 18 2011 *)
a[ n_] := ChebyshevT[ 2 n, 3]; (* Michael Somos, May 28 2014 *)

makelist(expand(((17+sqrt(288))^n+(17-sqrt(288))^n))/2, n, 0, 15); /* Vincenzo Librandi, Dec 18 2011 */

{a(n) = polchebyshev( n, 1, 17)}; /* Michael Somos, Apr 05 2019 */

[lucas_number2(n,34,1)/2 for n in range(0,15)] # Zerinvary Lajos, Jun 27 2008

a(n) = (r^n + 1/r^n)/2 with r = 17 + sqrt(17^2-1).
a(n) = 16*A001110(n) + 1 = A001541(2n) = (4*A001109(n))^2 + 1 = 3*A001109(2n-1) - A001109(2n-2) = A001109(2n) - 3*A001109(2n-1).
a(n) = T(n, 17) = T(2*n, 3) with T(n, x) Chebyshev's polynomials of the first kind. See A053120. T(n, 3)= A001541(n).
G.f.: (1-17*x)/(1-34*x+x^2).
G.f.: (1 - 17*x / (1 - 288*x / (17 - x))). - Michael Somos, Apr 05 2019
a(n) = cosh(2n*arcsinh(sqrt(8))). - Herbert Kociemba, Apr 24 2008
a(n) = (a^n + b^n)/2 where a = 17 + 12*sqrt(2) and b = 17 - 12*sqrt(2); sqrt(a(n)-1)/4 = A001109(n). - James R. Buddenhagen, Dec 09 2011
a(-n) = a(n). - Michael Somos, May 28 2014
a(n) = sqrt(1 + 32*9*A091761(n)^2), n >= 0. See one of the Pell comments above. - Wolfdieter Lang, Mar 09 2019

More terms from James Sellers, Sep 07 2000

Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A188645 Array of ((k^n)+(k^(-n)))/2 where k=(sqrt(x^2+1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 3, 1, 17, 9, 1, 99, 161, 19, 1, 577, 2889, 721, 33, 1, 3363, 51841, 27379, 2177, 51, 1, 19601, 930249, 1039681, 143649, 5201, 73, 1, 114243, 16692641, 39480499, 9478657, 530451, 10657, 99, 1, 665857, 299537289, 1499219281, 625447713, 54100801, 1555849, 19601, 129, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; and constant k=f(x, y); then for all integers x>=1 and y=[+-]1, k may be irrational, but ((k^n)+(k^(-n)))/2 always produces integer sequences; y=1 results shown here; y=-1 results are A188644.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{k}(x), evaluated at x=2*n^2+1. - Seiichi Manyama, Jan 01 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   3,     17,        99,          577, ...
   2 | 1,   9,    161,      2889,        51841, ...
   3 | 1,  19,    721,     27379,      1039681, ...
   4 | 1,  33,   2177,    143649,      9478657, ...
   5 | 1,  51,   5201,    530451,     54100801, ...
   6 | 1,  73,  10657,   1555849,    227143297, ...
   7 | 1,  99,  19601,   3880899,    768398401, ...
   8 | 1, 129,  33281,   8586369,   2215249921, ...
   9 | 1, 163,  53137,  17322499,   5647081537, ...
  10 | 1, 201,  80801,  32481801,  13057603201, ...
  11 | 1, 243, 118097,  57394899,  27893802817, ...
  12 | 1, 289, 167041,  96549409,  55805391361, ...
  13 | 1, 339, 229841, 155831859, 105653770561, ...
  14 | 1, 393, 308897, 242792649, 190834713217, ...
  15 | 1, 451, 406801, 366934051, 330974107201, ...
  ...

Row 1 is A001541, row 2 is A023039, row 3 is A078986, row 4 is A099370, row 5 is A099397, row 6 is A174747, row 8 is A176368, (row 1)*2 is A003499, (row 2)*2 is A087215.
Column 1 is A058331, (column 1)*2 is A005899.
A188644 (f(x, y) as above with y=-1).
Diagonal gives A173128.
Cf. A188647.

max = 9; y = 1; t = Table[k = ((x^2 + y)^(1/2) + x)^2; ((k^n) + (k^(-n)))/2 // FullSimplify, {n, 0, max - 1}, {x, 1, max}]; Table[ t[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)

A(n,k) = (A188647(n,k-1) + A188647(n,k))/2.

A(n,k) = Sum_{j=0..k} binomial(2*k,2*j)*(n^2+1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited and extended by Seiichi Manyama, Jan 01 2019

A221328 T(n,k)=Number of nXk arrays of occupancy after each element stays put or moves to some horizontal or antidiagonal neighbor, with no occupancy greater than 2.

Original entry on oeis.org

1, 3, 1, 7, 17, 1, 17, 119, 99, 1, 41, 866, 2008, 577, 1, 99, 6328, 46105, 33873, 3363, 1, 239, 46211, 1010078, 2460824, 571358, 19601, 1, 577, 337274, 22181855, 162430981, 131347807, 9637322, 114243, 1, 1393, 2460918, 487335857, 10794156688

Offset: 1

R. H. Hardin Jan 11 2013

Table starts
.1........3............7................17...................41
.1.......17..........119...............866.................6328
.1.......99.........2008.............46105..............1010078
.1......577........33873...........2460824............162430981
.1.....3363.......571358.........131347807..........26135748494
.1....19601......9637322........7010554450........4205264399891
.1...114243....162555929......374176705928......676612123702617
.1...665857...2741882383....19970973695923...108863121975329470
.1..3880899..46248186229..1065911268786396.17515379016294357354
.1.22619537.780082534460.56890876705125765

			Some solutions for n=3 k=4
..1..2..0..1....2..1..2..0....2..1..0..2....2..1..1..1....2..2..1..2
..1..0..2..1....1..1..0..0....1..1..2..0....0..2..0..1....0..0..0..2
..1..1..2..0....1..2..1..1....1..0..2..0....1..1..0..2....2..0..1..0

R. H. Hardin, Table of n, a(n) for n = 1..96

Column 2 is A001541

Row 1 is A001333

A079496 a(0) = a(1) = 1; thereafter a(2n+1) = 2a(2n) - a(2n-1), a(2n) = 4a(2n-1) - a(2n-2).

Original entry on oeis.org

1, 1, 3, 5, 17, 29, 99, 169, 577, 985, 3363, 5741, 19601, 33461, 114243, 195025, 665857, 1136689, 3880899, 6625109, 22619537, 38613965, 131836323, 225058681, 768398401, 1311738121, 4478554083, 7645370045, 26102926097, 44560482149, 152139002499, 259717522849, 886731088897

Offset: 0

Benoit Cloitre, Jan 20 2003

a(1)=1, a(n) is the smallest integer > a(n-1) such that sqrt(2)*a(n) is closer and > to an integer than sqrt(2)*a(n-1) (i.e., a(n) is the smallest integer > a(n-1) such that frac(sqrt(2)*a(n)) < frac(sqrt(2)*a(n-1))).
n such that floor(sqrt(2)*n^2) = n*floor(sqrt(2)*n).
The sequence 1,1,3,5,17,... has g.f. (1+x-3x^2-x^3)/(1-6x^2+x^4); a(n) = Sum_{k=0..floor(n/2)} C(n,2k)*2^(n-k-floor((n+1)/2)); a(n) = -(sqrt(2)-1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4) - (sqrt(2)+1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4); a(2n) = A001541(n) = A001333(2n); a(2n+1) = A001653(n) = A000129(2n+1). - Paul Barry, Jan 22 2005
The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators=A143608 and denominators=A079496. - Clark Kimberling, Aug 27 2008
From Richard Choulet, May 09 2010: (Start)
This sequence is a particular case of the following situation: a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n) where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z.
The g.f is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4); so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n).
The general form of a(n) is given by:
a(2*m)=sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(b-Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)) and
a(2*m+1)=a*sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(a*b+q-a*Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)). (End)
The integer square roots of floor(n^2/2 + 1) or (A007590 + 1). - Richard R. Forberg, Aug 01 2013

			1 + x + 3*x^2 + 5*x^3 + 17*x^4 + 29*x^5 + 99*x^6 + 169*x^7 + 577*x^8 + ...

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Indranil Ghosh, Table of n, a(n) for n = 0..2608
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 47, 56.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv:1105.3399 [math.GM], 2011.
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Yujun Yang and Heping Zhang, Kirchhoff Index of linear hexagonal chains, Int. J. Quant. Chem. 108 (2008) 503-512, eq (3.3).
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A010914, A058580.

Cf. A000129, A084068.

H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -1):
a := n -> `if`(n < 3, [1, 1, 3][n+1], 2^(n - 1)*H(n, irem(n, 2), 1/2)):
seq(simplify(a(n)), n=0..26); # Peter Luschny, Sep 03 2019

a[1] = 1; a[2] = 3; a[3] = 5; a[n_] := a[n] = (a[n-1]*a[n-2] + 2) / a[n-3]; Table[a[n], {n, 1, 29}] (* Jean-François Alcover, Jul 17 2013, after Paul D. Hanna *)

{a(n) = n = abs(n); 2^((4-n)\2) * real( (10 + 7 * quadgen(8)) / 2 * (2 + quadgen(8))^(n-3) ) }  /* Michael Somos, Sep 03 2013 */

{a(n) = polcoeff( (1 + x - 3*x^2 - x^3) / (1 - 6*x^2 + x^4) + x * O(x^abs(n)), abs(n))} /* Michael Somos, Sep 03 2013 */

a(2n+1) - a(2n) = a(2n) - a(2n-1) = A001542(n).
a(2n+1) = ceiling((2+sqrt(2))/4*(3+2*sqrt(2))^n), a(2n) = ceiling(1/2*(3+2*sqrt(2))^n).
G.f.: (1 + x - 3*x^2 - x^3)/(1 - 6*x^2 + x^4).
a(n)*a(n+3) - a(n+1)*a(n+2) = 2. - Paul D. Hanna, Feb 22 2003
a(n) = 6*a(n-2) - a(n-4). - R. J. Mathar, Apr 04 2008
a(-n) = a(n) = A010914(n-3)*2^floor((4 - n)/2). - Michael Somos, Sep 03 2013
a(n) = (sqrt(2)*sqrt(2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n))/(2+sqrt(2)+(-1)^n*(-2+sqrt(2))). - Gerry Martens, Jun 06 2015
a(n) = 2^(n - 1)*H(n, n mod 2, 1/2) for n >= 3 where H(n, a, b) = hypergeom([a - n/2, b - n/2], [1 - n], -1). - Peter Luschny, Sep 03 2019
a(n) == Pell(n)^(-1) (mod Pell(n+1)) where Pell(n) = A000129(n), use the identity a(n)*Pell(n) - A084068(n-1)*Pell(n+1) = 1, taken modulo Pell(n+1). - Gary W. Adamson, Nov 21 2023
E.g.f.: cosh(x)*(cosh(sqrt(2)*x) + sinh(sqrt(2)*x)/sqrt(2)). - Stefano Spezia, Apr 21 2025

a(0)=1 added by Michael Somos, Sep 03 2013

A084703 Squares k such that 2*k+1 is also a square.

Original entry on oeis.org

0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844

Offset: 0

Amarnath Murthy, Jun 08 2003

With the exception of 0, a subsequence of A075114. - R. J. Mathar, Dec 15 2008
Consequently, A014105(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011
From M. F. Hasler, Jan 17 2012: (Start)
Bisection of A079291. The squares 2*k+1 are given in A055792.
A204576 is this sequence written in binary. (End)
a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012

D. W. Wilson, Table of n, a(n-1) for n = 1..100 (offset=1)
Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011), Article 11.5.6, table 3, k=2.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A001109, A001110, A001333, A001541, A001542, A001652, A001653.
Cf. A014105, A029549, A046090, A046176, A055792, A075114, A079291, A084702.
Cf. A089928, A204576.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).

[4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022

b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)
a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1,30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)

[4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = 4*A001110(n) = A001542(n)^2.
a(n+1) = A001652(n)*A001652(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2. - Charlie Marion, Jul 01 2003
a(n) = A001653(k+n)*A001653(k-n) - A001653(k)^2, for k >= n >= 0; e.g. 144 = 5741*5 - 169^2. - Charlie Marion, Jul 16 2003
G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - R. J. Mathar, Dec 15 2008
a(n) = A079291(2n). - M. F. Hasler, Jan 16 2012
From George F. Johnson, Nov 02 2012: (Start)
a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.
a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1)*a(n+1) = (a(n) - 4)^2.
2*a(n) + 1 = (A001541(n))^2.
a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.
a(n)*a(n+1) = (4*A029549(n))^2.
a(n+1) - a(n) = 4*A046176(n).
a(n) + a(n+1) = 4*(6*A029549(n) + 1).
a(n) = (2*A001333(n)*A000129(n))^2.
Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)
Empirical: a(n) = A089928(4*n-2), for n > 0. - Alex Ratushnyak, Apr 12 2013
a(n) = 4*A001109(n)^2. - G. C. Greubel, Aug 18 2022
Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Edited and extended by Robert G. Wilson v, Jun 15 2003

cp's OEIS Frontend

A090390 Repeatedly multiply (1,0,0) by ([1,2,2],[2,1,2],[2,2,3]); sequence gives leading entry.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

PARI

Perl

SageMath

Formula

A143608 A005319 and A002315 interleaved.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Formula

A077420 Bisection of Chebyshev sequence T(n,3) (odd part) with Diophantine property.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Formula

A084128 a(n) = 4*a(n-1) + 4*a(n-2), a(0)=1, a(1)=2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A038725 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

Views

Author

A084128 a(n) = 4a(n-1) + 4a(n-2), a(0)=1, a(1)=2.

A079496 a(0) = a(1) = 1; thereafter a(2n+1) = 2a(2n) - a(2n-1), a(2n) = 4a(2n-1) - a(2n-2).