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.

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A123168 Continued fraction for c = sqrt(2)*(exp(sqrt(2))-1)/(exp(sqrt(2))+1).

Original entry on oeis.org

0, 1, 6, 5, 14, 9, 22, 13, 30, 17, 38, 21, 46, 25, 54, 29, 62, 33, 70, 37, 78, 41, 86, 45, 94, 49, 102, 53, 110, 57, 118, 61, 126, 65, 134, 69, 142, 73, 150, 77, 158, 81, 166, 85, 174, 89, 182, 93, 190, 97, 198, 101, 206, 105, 214, 109, 222, 113
Offset: 1

Views

Author

Benoit Cloitre, Oct 02 2006

Keywords

Comments

This continued fraction shows exp(sqrt(2)) is irrational (see A274540).

References

  • J. Borwein and D. Bailey, Mathematics by experiment, plausible reasoning in the 21st Century, A. K. Peters, p. 77.

Links

Crossrefs

Cf. A000217, A000384, A212343.
Odd bisection of A062828 with 0 prepended.
Cf. A274540, A123251.

Programs

  • Mathematica
    $MinPrecision = 5 $MachinePrecision; ContinuedFraction[Sqrt[2]* (Exp[Sqrt[2]] - 1)/(Exp[Sqrt[2]] + 1), 100]  (* G. C. Greubel, Aug 17 2018 *) (* or *)
LinearRecurrence[{0, 2, 0, -1}, {0, 1, 6, 5, 14}, 100] (* Georg Fischer, Aug 26 2022 *)
  • PARI
    default(realprecision, 1000); contfrac(sqrt(2)*(exp(sqrt(2))-1)/ (exp(sqrt(2))+1)) \\ Michel Marcus, Oct 11 2016

Formula

a(2*n) = 4*n-3, a(2*n+1) = 8*n-2.
From Colin Barker, Apr 15 2012: (Start)
a(n) = 2*a(n-2) - a(n-4) for n>5.
G.f.: x^2*(1+6*x+3*x^2+2*x^3)/((1-x)^2*(1+x)^2). (End)
a(n) = (2*n-3)*(3-(-1)^n)/2 for n>1, with a(1) = 0. - Wesley Ivan Hurt, Apr 01 2022

A227168 a(n) = gcd(2*n, n*(n+1)/2)^2.

Original entry on oeis.org

1, 1, 36, 4, 25, 9, 196, 16, 81, 25, 484, 36, 169, 49, 900, 64, 289, 81, 1444, 100, 441, 121, 2116, 144, 625, 169, 2916, 196, 841, 225, 3844, 256, 1089, 289, 4900, 324, 1369, 361, 6084, 400
Offset: 1

Views

Author

Paul Curtz, Jul 03 2013

Keywords

Comments

a(n) is defined as A062828(n)^2 for n >= 1. If we extend the sequence to n=0 and negative n by use of the recurrence that relates a(n) to a(n+12), a(n+8) and a(n+4), we obtain a(0)=0, a(-1)=4 and a(-n) = A176743(n-2)^2 for n >= 2.
Define c(n) = a(n+2) - a(n-2) for c >= 0. Because a(n) is a shuffle of three interleaved 2nd-order polynomials, c(n) is a shuffle of three interleaved 1st-order polynomials: c(n) = 4* A062828(n)*(periodically repeated 1, 8, 1, 1).
The sequence a(n) is case p=0 of the family A062828(n)*A062828(n+p):
0, 1, 1, 36, 4, 25, 9, 196, ... = a(n).
0, 1, 6, 12, 10, 15, 42, 56, ... = A130658(n)*A000217(n) = A177002(n-1)*A064038(n+1).
0, 6, 2, 30, 6, 70, 12, 126, ... = 2*A198148(n)
0, 2, 5, 18, 28, 20, 27, 70, ... = A177002(n+2)*A160050(n+1) = A014695(n+2)*A000096(n).

Links

Crossrefs

Cf. A062828, A016814, A017138, A005565, A006752, A061037, A061038.

Programs

Formula

a(n) = A062828(n)^2.
a(4n) = (4*n+1)^2; a(2n+1) = (n+1)^2; a(4n+2) = 4*(4*n+3)^2.
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n) * (period 4: repeat 4, 1, 1, 4) = A061038(n).
A005565(n-3) = a(n+1) * A061037(n). - Corrected by R. J. Mathar, Jul 25 2013
a(n) = A130658(n-1)^2 * A181318(n). - Corrected by R. J. Mathar, Aug 01 2013
G.f.: -x*(1 + x + 36*x^2 + 4*x^3 + 22*x^4 + 6*x^5 + 88*x^6 + 4*x^7 + 9*x^8 + x^9 + 4*x^10) / ( (x-1)^3*(1+x)^3*(x^2+1)^3 ). - R. J. Mathar, Jul 20 2013
Sum_{n>=1} 1/a(n) = 47*Pi^2/192 + 3*G/8, where G is Catalan's constant (A006752). - Amiram Eldar, Aug 21 2022
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