cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Views

Author

Peter Bala, Jun 06 2014

Keywords

Comments

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.

Links

Crossrefs

Cf. A041111, A243470.

Programs

Formula

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

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

Views

Author

Kenneth J Ramsey, Jun 21 2011

Keywords

Comments

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.

Examples

			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 ...
.
.
.

Links

Crossrefs

Cf. A191579, A191971.

Formula

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

Extensions

Corrected and edited by Olivier Gérard, Jul 05 2011

A041110 Numerators of continued fraction convergents to sqrt(63).

Original entry on oeis.org

7, 8, 119, 127, 1897, 2024, 30233, 32257, 481831, 514088, 7679063, 8193151, 122383177, 130576328, 1950451769, 2081028097, 31084845127, 33165873224, 495407070263, 528572943487, 7895428279081, 8424001222568
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A010516, A041111.

Programs

  • Mathematica
    Numerator[Convergents[Sqrt[63], 30]] (* Vincenzo Librandi, Oct 26 2013 *)
LinearRecurrence[{0,16,0,-1},{7,8,119,127},30] (* Harvey P. Dale, Dec 17 2019 *)

Formula

a(n) = 16*a(n-2)-a(n-4). G.f.: (7+8*x+7*x^2-x^3)/(1-16*x^2+x^4). [Colin Barker, Jul 15 2012]

A243470 Numerators of the rational convergents to the periodic continued fraction 1/(2 + 1/(7 + 1/(2 + 1/(7 + ...)))).

Original entry on oeis.org

1, 7, 15, 112, 239, 1785, 3809, 28448, 60705, 453383, 967471, 7225680, 15418831, 115157497, 245733825, 1835294272, 3916322369, 29249550855, 62415424079, 466157519408, 994730462895, 7429270759673, 15853271982241, 118402174635360, 252657621252961, 1887005523406087
Offset: 1

Views

Author

Peter Bala, Jun 06 2014

Keywords

Comments

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, ...]. The present sequence is the sequence of numerators 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 A243469 for the sequence of denominators to the convergents.

Links

Crossrefs

Cf. A041111, A077412, A243469.

Programs

  • Magma
    I:=[1,7,15,112]; [n le 4 select I[n] else 16*Self(n-2) -Self(n-4): n in [1..31]]; // G. C. Greubel, May 21 2022
  • Mathematica
    LinearRecurrence[{0,16,0,-1},{1,7,15,112},30] (* Harvey P. Dale, Nov 06 2017 *)
  • PARI
    Vec(x*(1+7*x-x^2)/(1-16*x^2+x^4)+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2015
  • SageMath
    def b(n): return chebyshev_U(n,8) # b=A077412
def A243470(n): return 7*((n-1)%2)*b(n//2 -1) +(n%2)*(b((n-1)//2) -b((n-1)//2 -1))
[A243470(n) for n in (1..30)] # G. C. Greubel, May 21 2022

Formula

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) = 7*(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) = 7*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) = 7*a(2*n - 1) + a(2*n - 2) and a(2*n + 1) = 2*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 + 7*x - x^2)/(1 - 16*x^2 + x^4).
a(2n-1) = A157456(n), a(2n) = 7*A077412(n-1). - Ralf Stephan, Jun 13 2014
a(n) = (1/2)*( 7*(1+(-1)^n)*ChebyshevU((n-2)/2, 8) + (1-(-1)^n)*(ChebyshevU((n- 1)/2, 8) - ChebyshevU((n-3)/2, 8)) ). - G. C. Greubel, May 21 2022
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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