A041023 -id:A041023 - cp's OEIS frontend

A070997 a(n) = 8*a(n-1) - a(n-2), a(0)=1, a(-1)=1.

1, 7, 55, 433, 3409, 26839, 211303, 1663585, 13097377, 103115431, 811826071, 6391493137, 50320119025, 396169459063, 3119035553479, 24556114968769, 193329884196673, 1522082958604615, 11983333784640247, 94344587318517361

Offset: 0

Joe Keane (jgk(AT)jgk.org), May 18 2002

A Pellian sequence.
In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,8), where L is defined as in A108299; see also A057080 for L(n,-8). - Reinhard Zumkeller, Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7} which do not end in 0. - Tanya Khovanova, Jan 10 2007
Hankel transform of A158197. - Paul Barry, Mar 13 2009
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(6)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Values of x (or y) in the solutions to x^2 - 8xy + y^2 + 6 = 0. - Colin Barker, Feb 05 2014
From Klaus Purath, May 06 2025: (Start)
Nonnegative solutions to the Diophantine equation 3*b(n)^2 - 5*a(n)^2 = -2. The corresponding b(n) are A057080(n). Note that (b(n)*b(n+2) - b(n+1)^2)/2 = -5 and (a(n)*a(n+2) - a(n+1)^2)/2 = 3.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A001090(n+1) = Lucas U(8,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n+1) + t(i))/(t(i+n+1) + t(i+n)) as long as t(i+n+1) + t(i+n) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 7*t(i+2) - 7*t(i+1) + t(i) or t(i+2) = 8*t(i+1) - t(i) regardless of initial values and including this sequence itself. (End)

			1 + 7*x + 55*x^2 + 433*x^3 + 3409*x^4 + 26839*x^5 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Fink, R. K. Guy and M. Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 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 (8,-1).

a(n) = sqrt((3*A057080(n)^2+2)/5) (cf. Richardson comment).
Cf. A057080, A001090, A001091.
Row 8 of array A094954.
Cf. A001090.
Cf. similar sequences listed in A238379.
Cf. A041023.

I:=[1, 7]; [n le 2 select I[n] else 8*Self(n-1) - Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 26 2013

CoefficientList[Series[(1 - x)/(1 - 8*x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jan 26 2013 *)
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 A041023 *)
a[15, 20] (* Gerry Martens, Jun 07 2015 *)
LinearRecurrence[{8,-1},{1,7},20] (* Harvey P. Dale, Dec 04 2021 *)

{a(n) = subst( 9*poltchebi(n) - poltchebi(n-1), x, 4) / 5} /* Michael Somos, Jun 07 2005 */

{a(n) = if( n<0, n=-1-n); polcoeff( (1 - x) / (1 - 8*x + x^2) + x * O(x^n), n)} /* Michael Somos, Jun 07 2005 */

[lucas_number1(n,8,1)-lucas_number1(n-1,8,1) for n in range(1, 21)] # Zerinvary Lajos, Nov 10 2009

For all members x of the sequence, 15*x^2 - 6 is a square. Lim_{n->infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 12 2002
a(n) = (5+sqrt(15))/10 * (4+sqrt(15))^n + (5-sqrt(15))/10 * (4-sqrt(15))^n.
a(n) ~ 1/10*sqrt(10)*(1/2*(sqrt(10)+sqrt(6)))^(2*n+1)
a(n) = U(n, 4)-U(n-1, 4) = T(2*n+1, sqrt(5/2))/sqrt(5/2), with Chebyshev's U and T polynomials and U(-1, x) := 0. U(n, 4)=A001090(n+1), n>=-1.
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 6) = a(n) - Benoit Cloitre, Nov 10 2002
a(n)*a(n+3) = 48 + a(n+1)a(n+2). - Ralf Stephan, May 29 2004
a(n) = (-1)^n*U(2n, i*sqrt(6)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
G.f.: (1-x)/(1-8*x+x^2).
a(n) = a(-1-n).
a(n) = Jacobi_P(n,-1/2,1/2,4)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006
[a(n), A001090(n+1)] = [1,6; 1,7]^(n+1) * [1,0]. - Gary W. Adamson, Mar 21 2008
For n>0, a(n) is the numerator of the continued fraction [2,3,2,3,...,2,3] with n repetitions of 2,3. For the denominators see A136325. - Greg Dresden, Sep 12 2019
From Peter Bala, Apr 30 2025: (Start)
a(n) = (1/sqrt(5)) * sqrt(1 - T(2*n+1, -4)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(n) divides a(3*n+1); a(n) divides a(5*n+2); in general, for k >= 0, a(n) divides a((2*k+1)*n + k).
The aerated sequence [b(n)]n>=1 = [1, 0, 7, 0, 55, 0, 433, 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 = -10, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/6 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A291033(n-1) - 1/A291033(n).) (End)
In addition to the first formula above: In general, the following applies to all recurrences (a(n)) of the form (8,-1) with a(0) = 1 and arbitrary a(1): 15*a(n)^2 + y = b^2 where y = x^2 + 8*x + 1 and x = a(1) - 8. Also y = a(k+1)^2 - a(k)*a(k+1) for any k >=0. - Klaus Purath, May 06 2025

A192062 Square Array T(ij) read by antidiagonals (from NE to SW) with columns 2j being the denominators of continued fraction convergents to square root of (j^2 + 2j).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 2, 0, 1, 1, 3, 3, 1, 0, 1, 1, 4, 4, 5, 3, 0, 1, 1, 5, 5, 11, 8, 1, 0, 1, 1, 6, 6, 19, 15, 13, 4, 0, 1, 1, 7, 7, 29, 24, 41, 21, 1, 0, 1, 1, 8, 8, 41, 35, 91, 56, 34, 5, 0, 1, 1, 9, 9, 55, 48, 169, 115, 153, 55, 1, 0, 1, 1, 10, 10, 71, 63, 281, 204, 436, 209, 89, 6

Offset: 0

Kenneth J Ramsey, Jun 21 2011

Column j=1 is the Fibonacci sequence A000045.  Column 2 is A002530; column 4 is A041011; column 6 is A041023; column 8 is A041039, column 10 is A041059, column 12 is A041083, column 14 is A041111 corresponding the denominators of continued fraction convergents to square root of 3,8,15,24,35,48 and 63.
  T(2*i-1,j)*T(2*i,j)^2*T(2*i+1,j)*j/2 appears to be always a triangular number, T(j*T(2*i,j)^2).
  T(2*i,j)*T(2*i+1,j)^2*T(2*i+2)*j/2 appears to always equal a triangular number, T(j*T(2*i,j)*T(2*i+2,j)).
Conjecture re relation of A192062 to the sequence of primes: T(2*n,j) = A(n,j)*T(n,j) where A(n,j) is from the square array A191971. There, A(3*n,j) = A(n,j)*B(n,j) where B(n,j) are integers. It appears further that B(5*n,j)=B(n,j)*C(n,j); C(7*n,j)= C(n,j)*D(n,j); D(11*n,j) = D(n,j)*E(n,j); E(13*n,j) = E(n,j)*F(n,j) and F(17*n,j) = F(n,j)*G(n,j) where C(n,j), D(n,j) etc. are all integers. My conjecture is that this property continues indefinitely and follows the sequence of primes.

			Array as meant by the definition
First column has index j=0
0  0  0   0   0   0   0 ...
1  1  1   1   1   1   1 ...
1  1  1   1   1   1   1 ...
1  2  3   4   5   6   7 ...
2  3  4   5   6   7   8 ...
1  5 11  19  29  41  55 ...
3  8 15  24  35  48  63 ...
1 13 41  91 169 281 433 ...
4 21 56 115 204 329 496 ...
.
.
.

Kenneth J Ramsey,Triangular and Fibonacci Numbers

Cf. A191579, A191971.

Each column j is a recursive sequence defined by T(0,j)=0, T(1,j) = 1, T(2i,j)= T(2i-2,j)+T(2i-1,j) and T(2i+1,j) = T(2i-1,j)+j*T(2i,j). Also, T(n+2,j) = (j+2)*T(n,j)-T(n-2,j).
T(2n,j) = Sum(k=1 to n) C(k)*T(2*k,j-1) where the C(k) are the n-th row of the triangle A191579.
T(2*i,j) = T(i,j)*A(i,j) where A(i,j) is from the table A(i,j) of A191971.
T(4*i,j) = (T(2*i+1)^2 - T(2*i-1)^2)/j
T(4*i+2,j) = T(2*i+2,j)^2 - T(2*i,j)^2

Corrected and edited by Olivier Gérard, Jul 05 2011

A041022 Numerators of continued fraction convergents to sqrt(15).

Original entry on oeis.org

3, 4, 27, 31, 213, 244, 1677, 1921, 13203, 15124, 103947, 119071, 818373, 937444, 6443037, 7380481, 50725923, 58106404, 399364347, 457470751, 3144188853, 3601659604, 24754146477, 28355806081

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-1).

Cf. A010472, A041023.

Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[15],n]]],{n,1,50}] (* Vladimir Joseph Stephan Orlovsky, Mar 17 2011 *)
Numerator[Convergents[Sqrt[15], 30]] (* Vincenzo Librandi, Oct 28 2013 *)
a0[n_] := (-((4-Sqrt[15])^n*(3+Sqrt[15]))+(-3+Sqrt[15])*(4+Sqrt[15])^n)/2 // Simplify
a1[n_] := ((4-Sqrt[15])^n+(4+Sqrt[15])^n)/2 // Simplify
Flatten[MapIndexed[{a0[#], a1[#]} &,Range[20]]] (* Gerry Martens, Jul 11 2015 *)

G.f.: (3+4*x+3*x^2-x^3)/(1-8*x^2+x^4).
From Gerry Martens, Jul 11 2015: (Start)
Interspersion of 2 sequences [a0(n),a1(n)] for n>0:
a0(n) = (-((4-sqrt(15))^n*(3+sqrt(15)))+(-3+sqrt(15))*(4+sqrt(15))^n)/2.
a1(n) = ((4-sqrt(15))^n+(4+sqrt(15))^n)/2. (End)

A145542 Numerators in continued fraction expansion of sqrt(3/5).

Original entry on oeis.org

1, 3, 7, 24, 55, 189, 433, 1488, 3409, 11715, 26839, 92232, 211303, 726141, 1663585, 5716896, 13097377, 45009027, 103115431, 354355320, 811826071, 2789833533, 6391493137, 21964312944, 50320119025, 172924670019, 396169459063, 1361433047208, 3119035553479

Offset: 1

Gary W. Adamson, Oct 12 2008

a(n)/A145543(n) tends to sqrt(3/5).

A strong divisibility sequence, that is gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. Related to the Lehmer sequence U_n(sqrt(R),Q) with parameters R = 6 and Q = -1. See A041023. - Peter Bala, Jun 06 2014

			[a(7), a(8)] = [433, 1488] X^4 * [1, 0] = [433, 1488].
a(5) = 55 = 2*a(4) + a(3) = 2*24 + 7.
G.f. = x + 3*x^2 + 7*x^3 + 24*x^4 + 55*x^5 + 189*x^6 + 433*x^7 + 1488*x^8 + ...

Peter Bala, Notes on 2-periodic continued fractions and Lehmer sequences
Eric Weisstein's World of Mathematics, Lehmer Number

Cf. A001090, A041203, A070997, A145543.

Numerator[Convergents[Sqrt[3/5], 30]] (* gives terms with 0 prepended *) (* Wesley Ivan Hurt, Jun 15 2014 *)

{a(n) = if( n<1, 0, polcoeff( x * (1 + 3*x - x^2) / (1 - 8*x^2 + x^4) + x * O(x^n), n))}; /* Michael Somos, Nov 14 2015 */

Numerators in continued fraction expansion of sqrt(3/5); i.e., of [1, 3, 2, 3, 2, 3, 2, 3, 2, ...].
[a(2*n - 1), a(2*n)] = X^n * [1,0], where X is the 2 X 2 matrix [1,2; 3,7].
Empirical G.f.: x*(1+3*x-x^2)/(1-8*x^2+x^4). - Colin Barker, Jan 04 2012
From Peter Bala, Jun 06 2014: (Start)
a(2*n + 1) = Product_{k=1..n} (6 + 4*cos^2(k*Pi/(2*n+1))).
a(2*n) = 3*Product_{k=1..n-1} (6 + 4*cos^2(k*Pi/(2*n))).
a(2*n + 1) = A070997(n); a(2*n) = 3*A001090(n). (End)

More terms from Wesley Ivan Hurt, Jun 15 2014

cp's OEIS Frontend

A070997 a(n) = 8*a(n-1) - a(n-2), a(0)=1, a(-1)=1.

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A192062 Square Array T(ij) read by antidiagonals (from NE to SW) with columns 2j being the denominators of continued fraction convergents to square root of (j^2 + 2j).

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A041022 Numerators of continued fraction convergents to sqrt(15).

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A145542 Numerators in continued fraction expansion of sqrt(3/5).

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