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 1-3 of 3 results.

A132596 X-values of solutions to the equation X*(X + 1) - 6*Y^2 = 0.

Original entry on oeis.org

0, 2, 24, 242, 2400, 23762, 235224, 2328482, 23049600, 228167522, 2258625624, 22358088722, 221322261600, 2190864527282, 21687323011224, 214682365584962, 2125136332838400, 21036680962799042, 208241673295152024
Offset: 0

Views

Author

Mohamed Bouhamida, Nov 14 2007

Keywords

Comments

Or, 3*A000217(X) is a square, (3*A004189(n))^2. - Zak Seidov, Apr 08 2009
"You can find an infinite number of [different] triangular numbers such that when multiplied together form a square number. For example, for every triangular number, T_n, there are an infinite number of other triangular numbers, T_m, such that T_n*T_m is a square. For example, T_2 * T_24 = 30^2." [Pickover] - Robert G. Wilson v, Apr 01 2010

References

  • Clifford A. Pickover, The Loom of God, Tapestries of Mathematics and Mysticism, Sterling, NY, 2009, page 33.

Links

Crossrefs

Cf. A007654, A001079, A000217, A098297, A108741 (Y^2), A004189 (Y).

Programs

Formula

a(n) = 10*a(n-1) - a(n-2) + 4, a(0)=0, a(1)=2.
a(n) = (A001079(n) - 1)/2. - Max Alekseyev, Nov 13 2009
From R. J. Mathar, Apr 20 2010: (Start)
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) = 2*A098297(n).
G.f.: -2*x*(1+x) / ( (x-1)*(x^2-10*x+1) ). (End)
a(n) = 2*A098297(n) = (1/2)*(T(2*n,sqrt(3)) - 1), where T(n,x) is the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 31 2012

A221074 Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 3.

Original entry on oeis.org

2, 8, 1, 16, 1, 96, 1, 176, 1, 968, 1, 1760, 1, 9600, 1, 17440, 1, 95048, 1, 172656, 1, 940896, 1, 1709136, 1, 9313928, 1, 16918720, 1, 92198400, 1, 167478080, 1, 912670088, 1, 1657862096, 1
Offset: 0

Views

Author

Peter Bala, Jan 06 2013

Keywords

Comments

Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 3. For other cases see A221073 (m = 2), A221075 (m = 4) and A221076 (m = 5).

Examples

			Product {n >= 0} {1 - sqrt(3)*(sqrt(3) - sqrt(2))^(4*n+3)}/{1 - sqrt(3)*(sqrt(3) - sqrt(2))^(4*n+1)} = 2.11180 16361 44098 52896 ...
= 2 + 1/(8 + 1/(1 + 1/(16 + 1/(1 + 1/(96 + 1/(1 + 1/(176 + ...))))))).
Since (sqrt(3) - sqrt(2))^3 = 9*sqrt(3) - 11*sqrt(2) we have the following simple continued fraction expansion:
product {n >= 0} {1 - sqrt(3)*(9*sqrt(3) - 11*sqrt(2))^(4*n+3)}/{1 - sqrt(3)*(9*sqrt(3) - 11*sqrt(2))^(4*n+1)} = 1 + 1/(16 + 1/(1 + 1/(968 + 1/(1 + 1/(17440 + 1/(1 + 1/(940896 + ...))))))).

Links

Crossrefs

Cf. A098297, A105438, A132596, A174500, A221073 (m = 2), A221075 (m = 4), A221076 (m = 5).

Formula

a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (sqrt(3) + sqrt(2))^(2*n) + (sqrt(3) - sqrt(2))^(2*n) - 2;
a(4*n - 1) = 1/sqrt(3)*{(sqrt(3) + sqrt(2))^(2*n + 1) + (sqrt(3) - sqrt(2))^(2*n + 1)} - 2.
a(4*n - 3) = 8*A098297(n) = 4*A132596(n); a(4*n - 1) = 4*A105038(n).
O.g.f.: 2 + x^2/(1 - x^2) + 8*x*(1 + x^2)^2/(1 - 11*x^4 + 11*x^8 - x^12) = 2 + 8*x + x^2 + 16*x^3 + x^4 + 96*x^5 + ....
If we denote the present sequence by [2; 8, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-sqrt(3)*{(sqrt(3)-sqrt(2))^(2*k+1)}^(4*n+3)]/[1 - sqrt(3)*{(sqrt(3)-sqrt(2))^(2*k+1)}^(4*n+1)]. An example is given below.
O.g.f.: (x^10-2*x^8-10*x^6+20*x^4-8*x^3+x^2-8*x-2) / ((x-1)*(x+1)*(x^8-10*x^4+1)). - Colin Barker, Jan 10 2014

A098298 Member r=13 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 13, 144, 1573, 17161, 187200, 2042041, 22275253, 242985744, 2650567933, 28913261521, 315395308800, 3440435135281, 37529391179293, 409382867836944, 4465682155027093, 48713120837461081, 531378647057044800, 5796451996790031721, 63229593317633304133
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Links

Crossrefs

Cf. A098296, A098297.

Programs

  • GAP
    a:=[0,1,13];; for n in [4..30] do a[n]:=12*a[n-1]-12*a[n-2]+ a[n-3]; od; a; # G. C. Greubel, May 24 2019
  • Magma
    [n le 2 select n-1 else 11*Self(n-1)- Self(n-2) + 2: n in [1..30]]; // Vincenzo Librandi, Mar 06 2016
  • Mathematica
    LinearRecurrence[{12,-12,1},{0,1,13},30] (* Harvey P. Dale, May 11 2012 *)
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == 11 a[n-1] - a[n-2] + 2}, a, {n, 30}] (* Vincenzo Librandi, Mar 06 2016 *)
  • PARI
    concat(0, Vec(x*(1+x)/((1-x)*(1-11*x+x^2)) + O(x^25))) \\ Colin Barker, Mar 06 2016
  • Sage
    (x*(1+x)/((1-x)*(1-11*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 24 2019

Formula

a(n) = 2*(T(n, 11/2) - 1)/9 with twice Chebyshev's polynomials of the first kind evaluated at x=11/2: 2*T(n, 11/2) = A057076(n) = ((11 + sqrt(117))^n + (11 - sqrt(117))^n)/2^n.
a(n) = 11*a(n-1) - a(n-2) + 2, n >= 2, a(0)=0, a(1)=1.
a(n) = 12*a(n-1) - 12*a(n-2) + a(n-3), n >= 3, a(0)=0, a(1)=1, a(2)=13.
G.f.: x*(1+x)/((1-x)*(1-11*x+x^2)) = x*(1+x)/(1-12*x+12*x^2-x^3) (from the Stephan link, see A092184).
Showing 1-3 of 3 results.

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