A004254 -id:A004254 - cp's OEIS frontend

A179943 Triangle read by rows, antidiagonals of an array (r,k), r=(0,1,2,...), generated from 2 X 2 matrices of the form [1,r; 1,(r+1)].

1, 1, 2, 1, 3, 3, 1, 4, 8, 4, 1, 5, 15, 21, 5, 1, 6, 24, 56, 55, 6, 1, 7, 35, 115, 209, 144, 7, 1, 8, 48, 204, 551, 780, 377, 8, 1, 9, 63, 329, 1189, 2640, 2911, 987, 9, 1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10, 1, 11, 99, 711, 3905, 15456, 40391, 60605, 40545, 6765, 11

Offset: 0

Gary W. Adamson, Aug 07 2010

Row sums = A179944: (1, 3, 7, 17, 47, 148, 518,...)
Row 1 = A001906, row 2 = A001353, row 3 = A004254, row 4 = A001109, row 5 = A004187, row 6 = A001090, row 7 = A018913, row 9 = A004189.
Let S_m(x) be the m-th Chebyshev S-polynomial, described by Wolfdieter Lang in his draft [Lang], defined by S_0(x)=1, S_1(x)=x and S_m(x)=x*S_{m-1}(x)-S_{m-2}(x) (m>1). Let A = (A(r,c)) denote the rectangular array (not the triangle). Then A(r,c) = S_c(r+2), r,c=0,1,2,.... - L. Edson Jeffery, Aug 14 2011
As to the array, (n+1)-th row is the INVERT transform of n-th row. - Gary W. Adamson, Jun 30 2013
If the array sequences are labeled (2,3,4,...) for the n-th sequence, convergence tends to (n + sqrt(n^2 - 4))/2. - Gary W. Adamson, Aug 20 2013

			First few rows of the array:
  1, 2,  3,   4,    5,    6,     7,...
  1, 3,  8,  21,   55,  144,   377,...
  1, 4, 15,  56,  209,  780,  2911,...
  1, 5, 24, 115,  551, 2640, 12649,...
  1, 6, 35, 204, 1189, 6930, 40391,...
Taking antidiagonals, we obtain triangle A179943:
  1;
  1, 2;
  1, 3, 3;
  1, 4, 8, 4;
  1, 5, 15, 21, 5;
  1, 6, 24, 56, 55, 6;
  1, 7, 35, 115, 209, 144, 7;
  1, 8, 48, 204, 551, 780, 377, 8;
  1, 9, 63, 329, 1189, 2640, 2911, 987, 9;
  1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10;
  1, 11, 99, 711, 3905, 15456, 40391, 60605, 40545, 6765, 11;
  1, 12, 120, 980, 6319, 30744, 105937, 235416, 290376, 151316, 17711, 12;
  ...
Examples: Row 1 of the array: (1, 3, 8, 21, 55, 144,...); 144 = term (1,5) of the array = term (2,1) of M^6; where M = the 2 X 2 matrix [1,1; 1,2] and M^6 = [89,144; 144,233].
Term (1,5) of the array = 144 = (r+2)*(term (1,4)) - (term (1,3)) = 3*55 - 21.

Alois P. Heinz, Rows n = 0..140, flattened
Robert G. Donnelly, Molly W. Dunkum, Sasha V. Malone, and Alexandra Nance, Symmetric Fibonaccian distributive lattices and representations of the special linear Lie algebras, arXiv:2012.14991 [math.CO], 2020.
Wolfdieter Lang, Chebyshev S-polynomials: ten applications.

Cf. A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.

invtr:= proc(b) local a;
          a:= proc(n) option remember; local i;
          `if`(n<1, 1, add(a(n-i) *b(i-1), i=1..n+1)) end
        end:
A:= proc(n) A(n):= `if`(n=0, k->k+1, invtr(A(n-1))) end:
seq(seq(A(d-k)(k), k=0..d), d=0..10);  # Alois P. Heinz, Jul 17 2013
# using observation by Gary W. Adamson

a[, 0] = 0; a[, 1] = 1; a[r_, k_] := a[r, k] = (r+1)*a[r, k-1] - a[r, k-2]; Table[a[r-k+2, k], {r, 0, 10}, {k, 1, r+1}] // Flatten (* Jean-François Alcover, Feb 23 2015 *)

Antidiagonals of an array, (r,k), a(k) = (r+2)*a(k-1) - a*(k-2), r=0,1,2,... where (r,k) = term (2,1) in the 2 X 2 matrix [1,r; 1,r+1]^(k+1).

G.f. for row r of array: 1/(1 - (r+2)*x + x^2). - L. Edson Jeffery, Oct 26 2012

A290902 p-INVERT of the positive integers, where p(S) = 1 - 3*S.

Original entry on oeis.org

3, 15, 72, 345, 1653, 7920, 37947, 181815, 871128, 4173825, 19997997, 95816160, 459082803, 2199597855, 10538906472, 50494934505, 241935766053, 1159183895760, 5553983712747, 26610734667975, 127499689627128, 610887713467665, 2926938877711197, 14023806675088320

Offset: 0

Clark Kimberling, Aug 17 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A290890 for a guide to related sequences.

			s = (1,2,3,4,...), p(S) = 1-3*S;
S(x) = x + 2 x^2 + 3 x^3 + ... ;
p(S(x)) = 1 - 3(x + 2 x^2 + 3 x^3 + ...);
1/p(S(x)) = 1 + 3 x + 15 x^2 + 72 x^3 + ... ;
(-p(0) + 1/p(S(x)))/x = 3 + 15 x + 72 x^2 + ... ;
t(s) = (3, 15, 72, ...), with offset 0.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, -1)

Cf. A000027, A004254, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 3 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290902 *)
u/3 (* A004254 shifted *)

G.f.: 3/(1 - 5 x + x^2).
a(n) = 5*a(n-1) - a(n-2).
a(n) = 3*A004254(n+1) for n >= 0.

A372817 Table read by antidiagonals: T(m,n) = number of 1-metered (m,n)-parking functions.

Original entry on oeis.org

1, 0, 2, 0, 3, 3, 0, 4, 8, 4, 0, 6, 21, 15, 5, 0, 8, 55, 56, 24, 6, 0, 12, 145, 209, 115, 35, 7, 0, 16, 380, 780, 551, 204, 48, 8, 0, 24, 1000, 2912, 2640, 1189, 329, 63, 9, 0, 32, 2625, 10868, 12649, 6930, 2255, 496, 80, 10, 0, 48, 6900, 40569, 60606, 40391, 15456, 3905, 711, 99, 11

Offset: 1

Spencer Daugherty, May 13 2024

			For T(3,2) the 1-metered (3,2)-parking functions are 111, 121, 211, 212.
Table begins:
  1,  2,    3,     4,     5,      6,      7, ...
  0,  3,    8,    15,    24,     35,     48, ...
  0,  4,   21,    56,   115,    204,    329, ...
  0,  6,   55,   209,   551,   1189,   2255, ...
  0,  8,  145,   780,  2640,   6930,  15456, ...
  0, 12,  380,  2912, 12649,  40391, 105937, ...
  0, 16, 1000, 10868, 60606, 235416, 726103, ...
  ...

Spencer Daugherty, Pamela E. Harris, Ian Klein, and Matt McClinton, Metered Parking Functions, arXiv:2406.12941 [math.CO], 2024.

Main diagonal is A097690 and first row of A372816.
First, second, and third diagonals above main are A097691, A342167, A342168.
Second column A029744. Second row A005563. Third row A242135.
Cf. A001353, A004254, A001109, A004187, A372818, A372821.

T(m,n) = (n*(n+sqrt(n^2 - 4))-2)/(n*(n+sqrt(n^2 - 4))-4)*((n+sqrt(n^2-4))/2)^m + (n*(n-sqrt(n^2 - 4))-2)/(n*(n-sqrt(n^2 - 4))-4)*((n-sqrt(n^2-4))/2)^m.

T(m,n) = n*T(m-1,n) - T(m-2,n) with T(0,n) = 1.

A073134 Table by antidiagonals of T(n,k)=n*T(n,k-1)-T(n,k-2) starting with T(n,1)=1.

Original entry on oeis.org

1, 1, 1, 0, 2, 1, -1, 3, 3, 1, -1, 4, 8, 4, 1, 0, 5, 21, 15, 5, 1, 1, 6, 55, 56, 24, 6, 1, 1, 7, 144, 209, 115, 35, 7, 1, 0, 8, 377, 780, 551, 204, 48, 8, 1, -1, 9, 987, 2911, 2640, 1189, 329, 63, 9, 1, -1, 10, 2584, 10864, 12649, 6930, 2255, 496, 80, 10, 1, 0, 11, 6765, 40545, 60605, 40391, 15456, 3905, 711, 99, 11, 1, 1, 12

Offset: 1

Henry Bottomley, Jul 16 2002

			Rows start:
  1, 1,  0, -1,  -1,   0,    1, ...;
  1, 2,  3,  4,   5,   6,    7, ...;
  1, 3,  8, 21,  55, 144,  377, ...;
  1, 4, 15, 56, 209, 780, 2911, ...;
  ...

Shmuel T. Klein, Combinatorial Representation of Generalized Fibonacci Numbers, Fib. Quarterly 29 (2) (1991) 124-131, variable U_n^m. [From _R. J. Mathar_, Feb 19 2010]

Rows include A010892, A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190. Columns include (with some gaps) A000012, A000027, A005563, A057722.

Cf. A094954.

T(n,k) = sum(j=0,k-1,A049310(k-1,j)*n^j) \\ Jason Yuen, Aug 20 2024

T(n, k) = A073133(n, k)-2*A073135(n, k-2).

T(n, k) = Sum_{j=0..k-1} A049310(k-1, j)*n^j.

A092499 Chebyshev polynomials S(n-1,21) with Diophantine property.

Original entry on oeis.org

0, 1, 21, 440, 9219, 193159, 4047120, 84796361, 1776676461, 37225409320, 779956919259, 16341869895119, 342399310878240, 7174043658547921, 150312517518628101, 3149388824232642200, 65986852791366858099

Offset: 0

Rainer Rosenthal, Apr 05 2004

Sequence R_21: Starts with 0,1,21 and satisfies A*C=B^2-1 for successive A,B,C.
The natural numbers a(n)=n satisfy the recurrence a(n-1)*a(n+1)=a(n)^2-1. Let R_r denote the sequence starting with 0,1,r and with this recurrence. We see that R_2 = "the natural numbers" and we find R_3 = A001906. These R_r form a "family" of sequences, which coincides with the m-family (r=m-2, n -> n+1) provided by Wolfdieter Lang (see A078368). This sequence R_21 is strongly related to A041833, which gives the denominators in the continued fraction of sqrt(437).
All positive integer solutions of Pell equation b(n)^2 - 437*a(n)^2 = +4 together with b(n)=A097777(n), n>=0.
For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 21's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,20}. - Milan Janjic, Jan 25 2015

			a(3)=440 because a(1)*440 = a(2)^2-1.

Indranil Ghosh, Table of n, a(n) for n = 0..756
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (21,-1).

Cf. R_3=A001906, R_4=A001353, R_5=A004254, R_6=A001109, R_7=A004187, R_8=A001090, R_9=A018913, R_10=A004189, R_11=A004190, R_12=A004191, R_13=A078362, R_14=A007655, R_15=A078364, R_16=A077412, R_17=A078366, R_18=A049660, R_19=A078368, R_20=A075843, R_21=this, sequence, R_22=A077421. See also A041219 and A041917.

LinearRecurrence[{21,-1},{0,1},30] (* Harvey P. Dale, Apr 23 2015 *)

[lucas_number1(n,21,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008

a(0)=0, a(1)=1, a(2)=21 and a(n-1)*a(n+1) = a(n)^2-1
a(n) = S(n-1, 21)=U(n-1, 21/2) with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x)= 0 = U(-1, x).
a(n) = S(2*n-1, sqrt(23))/sqrt(23), n>=1.
a(n) = 21*a(n-1)-a(n-2), n >= 1; a(0)=0, a(1)=1.
a(n) = (ap^n-am^n)/(ap-am) with ap := (21+sqrt(437))/2 and am := (21-sqrt(437))/2.
G.f.: x/(1-21*x+x^2).
a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*20^k. - Philippe Deléham, Feb 10 2012
Product {n >= 1} (1 + 1/a(n)) = 1/19*(19 + sqrt(437)). - Peter Bala, Dec 23 2012
Product {n >= 2} (1 - 1/a(n)) = 1/42*(19 + sqrt(437)). - Peter Bala, Dec 23 2012

Extension, Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

Corrected by T. D. Noe, Nov 07 2006

A125078 Fifth in an infinite set of generalized Pascal's triangles, with trigonometric properties.

Original entry on oeis.org

1, 1, 4, 1, 5, 19, 1, 9, 24, 91, 1, 10, 63, 115, 436, 1, 14, 73, 397, 551, 2089, 1, 15, 132, 470, 2358, 2640, 10009, 1, 19, 147, 1043, 2828, 13482, 12649, 47956, 1, 20, 226, 1190, 7441, 16310, 75061, 60605, 229771

Offset: 1

Gary W. Adamson, Nov 18 2006

The triangle is the fifth in an infinite set of generalized Pascal's triangles constrained by two properties: row sums = powers of N and upward sloping diagonals solve for N + 2*Cos 2Pi/Q. Row sums are powers of 5. Right border (1, 4, 19, 91, 436...) = A004253. Next to right border (1, 5, 24, 115...) = A004254.

			First few rows of the triangle are:
1;
1, 4;
1, 5, 19;
1, 9, 24, 91;
1, 10, 63, 115, 436;
1, 14, 73, 397, 551, 2089;
1, 15, 132, 470, 2358, 2640, 10009;
...
The upward sloping diagonal (1, 14, 63, 91) is derived from the characteristic polynomial x^3 - 14x^2 + 63x - 91 and relates to the Heptagon (Q=7) since a root = 6.24697960...= 5 + 2*Cos 2Pi/7. The corresponding matrix is [4, 1, 0; 1, 5, 1; 0, 1, 5]. The next upward sloping diagonal (1, 15, 73, 115) relates to the Octagon (Q=8) since a root = 6.41421356... = 5 + 2*Cos 2Pi/8. The corresponding matrix is [5, 1, 0; 1, 5, 1; 0, 1, 5].

Cf. A125076, A125078, A004253, A004254.

Upward sloping diagonals are derived from interleaved characteristic polynomials of two types of matrices, relating to odd and even polygons. Matrices with an eigenvalue 5 + 2*Cos 2Pi/Q, Q is odd, are of the form: all 1's in the super and subdiagonals and 4,5,5,5... in the main diagonal. Matrices (Q is even) are of the form: all 1's in the super and subdiagonals and 5,5,5... in the main diagonal.

A136210 Numerators in continued fraction [0; 1, 3, 1, 3, 1, 3, ...].

Original entry on oeis.org

1, 3, 4, 15, 19, 72, 91, 345, 436, 1653, 2089, 7920, 10009, 37947, 47956, 181815, 229771, 871128, 1100899, 4173825, 5274724, 19997997, 25272721, 95816160, 121088881, 459082803, 580171684, 2199597855, 2779769539, 10538906472

Offset: 1

Gary W. Adamson, Dec 21 2007

A136210(n)/A136211(n) tends to 0.7912878474... = (sqrt(21) - 3)/2 = continued fraction [0; 1, 3, 1, 3, 1, 3, ...] = the inradius of a right triangle with hypotenuse 5, legs 2 and sqrt(21).

This is a strong divisibility sequence, that is, GCD(a(n),a(m)) = a(GCD(n,m)) for all natural numbers n and m. - Peter Bala, May 14 2014

			a(4) = 15 = 3*a(3) + a(2) = 3*4 + 3.
a(5) = 19 = a(4) + a(3) = 15 + 4.
T^3 = [19, 72; 24, 91], where [19, 72] = [a(5), a(6)]. [24, 91] = [A136211(5), A136211(6)].
G.f. = x + 3*x^2 + 4*x^3 + 15*x^4 + 19*x^5 + 72*x^6 + 91*x^7 + 345*x^8 + ...

J. L. Ramirez, F. Sirvent, A q-Analogue of the Bi-Periodic Fibonacci Sequence, J. Int. Seq. 19 (2016) # 16.4.6, t_n at a=3, b=1.
Index entries for linear recurrences with constant coefficients, signature (0,5,0,-1).

Cf. A004253, A004254, A136211.

a = {1, 3}; Do[If[EvenQ[n], AppendTo[a, 3*a[[ -1]] + a[[ -2]]], AppendTo[a, a[[ -1]] + a[[ -2]]]], {n, 3, 30}]; a (* Stefan Steinerberger, Dec 31 2007 *)
a[n_] := FromContinuedFraction[ Join[{0}, 3 - 2*Array[Mod[#, 2]&, n]]] // Numerator; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, May 15 2014 *)

{a(n) = (-1)^((n+1) * (n<0)) * polcoeff( x * (1 + 3*x - x^2) / (1 - 5*x^2 + x^4) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, May 15 2014 */

a(0) = 0, a(1) = 1, a(2n) = 3*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). Given the 2 X 2 matrix [1, 3; 1, 4] = T, [a(2n-1), a(2n)] = top row of T^n.
g.f.: x*(1+3*x-x^2)/(1-5*x^2+x^4). - Colin Barker, Jan 04 2012
a(-n) = -(-1)^n * a(n). a(2*n - 1) = A004253(n). a(2*n) = 3 * A004254(n). - Michael Somos, May 15 2014
a(n+1) - a(n-1) = a(n) * (2 - (-1)^n) for all n in Z. - Michael Somos, May 15 2014

More terms from Stefan Steinerberger, Dec 31 2007

A221365 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(5 - sqrt(21)).

Original entry on oeis.org

1, 3, 1, 21, 1, 108, 1, 525, 1, 2523, 1, 12096, 1, 57963, 1, 277725, 1, 1330668, 1, 6375621, 1, 30547443, 1, 146361600, 1, 701260563, 1, 3359941221, 1, 16098445548, 1, 77132286525, 1, 369562987083, 1, 1770682648896, 1

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 5. See also A221364 (N = 3), A221366 (N = 7) and A221367 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(5 - sqrt(21)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...].

			F(1/2*(5 - sqrt(21))) = 1.25274 83510 08359 27965 ... = 1 + 1/(3 + 1/(1 + 1/(21 + 1/(1 + 1/(108 + 1/(1 + 1/(525 + ...))))))).
F((1/2*(5 - sqrt(21)))^2) = 1.04545 84663 16495 30047 ... = 1 + 1/(21 + 1/(1 + 1/(525 + 1/(1 + 1/(12096 + 1/(1 + 1/(277725 + ...))))))).
F((1/2*(5 - sqrt(21)))^3) = 1.00917 43188 83793 73068 ... = 1 + 1/(108 + 1/(1 + 1/(12096 + 1/(1 + 1/(1330668 + 1/(1 + 1/(146361600 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-6,0,1).

Cf. A004254, A030221, A054493, A174500 (N = 4), A221364 (N = 3), A221366 (N = 7), A221369 (N = 9).

LinearRecurrence[{0,6,0,-6,0,1},{1,3,1,21,1,108},40] (* Harvey P. Dale, Jun 06 2023 *)

a(2*n-1) = (1/2*(5 + sqrt(21)))^n + (1/2*(5 - sqrt(21)))^n - 2 = 3*A054493(n); a(2*n) = 1.
a(4*n+1) = 3*(A030221(n))^2; a(4*n-1) = 21*(A004254(n))^2.
a(n) = 6*a(n-2)-6*a(n-4)+a(n-6). G.f.: -(x^4+3*x^3-5*x^2+3*x+1) / ((x-1)*(x+1)*(x^4-5*x^2+1)). - Colin Barker, Jan 20 2013

A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)sqrt(21a(n-1)^2 + 4) for n >= 1.

Original entry on oeis.org

1, 5, 115, 60605, 16831644835, 1298263252133919638045, 7723873922612696850892381990249713732303715, 273388347343560518533856033712658350781293745092679040607342582493129736504927611387805

Offset: 0

Peter Bala, Oct 30 2013

For integer N, the recurrence equation a(n) = a(n-1)*sqrt((N^2 - 4)*a(n-1)^2 + 4) for n >= 1, with starting value a(0) = 1, produces an integer sequence. The present sequence is the case N = 5. Cf. A000079 (case N = 2), A058635 (case N = 3) and A071579 (case N = 4).
Sequence of numerators in the Engel series representation of 1/2*(7 - sqrt(21)) = 1 + 1/5 + 1 /115 + 1/60605 + .... The corresponding Engel expansion is A003487.
The sequence also has a description as a Pierce expansion of the quadratic irrational 1/2*(5 - sqrt(21)) to the base b := 1/sqrt(21) (see A058635 for a definition of this term).
The associated Pierce series representation of 1/2*(5 - sqrt(21)) to the base b begins 1/2*(5 - sqrt(21)) = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + ....
More generally, for n >= 0, the sequence [a(n), a(n+1), a(n+2), ...] gives a Pierce expansion of ( 1/2*(5 - sqrt(21)) )^(2^n) to the base b = 1/sqrt(21). Some examples are given below.

			Let b = 1/sqrt(21) and x = 1/2*(5 - sqrt(21)). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + b^5/(1*5*115*60605*16831644835) - ....
x^2 = b/5 - b^2/(5*115) + b^3/(5*115*60605) - b^4/(5*115*60605*16831644835) + ....
x^4 = b/115 - b^2/(115*60605) + b^3/(115*60605*16831644835) - ....
x^8 = b/60605 - b^2/(60605*16831644835) + ....

Seiichi Manyama, Table of n, a(n) for n = 0..10

Cf. A003487, A004254, A058635, A071579.

a[n_] := a[n - 1]*Sqrt[21 a[n - 1]^2 + 4]; a[0] = 1; Array[a, 8, 0] (* Robert G. Wilson v, Mar 19 2014 *)
a[ n_] := If[ n < 0, 0, ChebyshevU[2^n - 1, 5/2]]; (* Michael Somos, Dec 06 2016 *)

a(n) = if( n<0, 0, imag( (5 + quadgen(84))^2^n) / 2^(2^n - 1)); /* Michael Somos, Dec 06 2016 */

a(n) = 1/sqrt(21)*( alpha^(2^n) - (1/alpha)^(2^n) ), where alpha = 1/2*(5 + sqrt(21)).
a(n) = product {k = 0..n-1} A003487(k).
Defining recurrence equation:
a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.
Other recurrence equations:
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = 21*a(n-2)^2 + 2 for n >= 2.
a(n) = A004254(2^n). - Michael Somos, Dec 06 2016

A334673 a(n) = 23*a(n-1) - a(n-2) + 1 for n > 1, a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 24, 552, 12673, 290928, 6678672, 153318529, 3519647496, 80798573880, 1854847551745, 42580695116256, 977501140122144, 22439945527693057, 515141245996818168, 11825808712399124808, 271478459139183052417, 6232178751488811080784, 143068632825103471805616

Offset: 0

Francesca Arici, Sep 11 2020

Michael De Vlieger, Table of n, a(n) for n = 0..735
Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
Index entries for linear recurrences with constant coefficients, signature (24,-24,1).

Cf. A004253, A004254, A030221, A097778 (first differences).

Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

CoefficientList[Series[x/((1 - x) (x^2 - 23 x + 1)), {x, 0, 18}], x] (* Michael De Vlieger, Apr 07 2021 *)

a(n) = A004254(n)*A004254(n+1)/5 = A160695(n+1)/5.
G.f.: x/((1-x)*(x^2-23*x+1)). - Alois P. Heinz, Sep 11 2020
From Klaus Purath, Jun 18 2025: (Start)
a(n) = (A004253(n+1)^2 - 1) / 15.
a(n) = (A030221(n)^2 - 1) / 35.
a(n) + a(n+1) = A004253(n+1)^2. (End)

a(13)-a(14) corrected and more terms added by Alois P. Heinz, Sep 11 2020

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