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.

A157447 a(n) = 512*n - 16.

Original entry on oeis.org

496, 1008, 1520, 2032, 2544, 3056, 3568, 4080, 4592, 5104, 5616, 6128, 6640, 7152, 7664, 8176, 8688, 9200, 9712, 10224, 10736, 11248, 11760, 12272, 12784, 13296, 13808, 14320, 14832, 15344, 15856, 16368, 16880, 17392, 17904, 18416, 18928
Offset: 1

Views

Author

Vincenzo Librandi, Mar 01 2009

Keywords

Comments

The identity (2048*n^2 - 128*n + 1)^2 - (16*n^2 - n)*(512*n - 16)^2 = 1 can be written as A157448(n)^2 - A157446(n)*a(n)^2 = 1 (see also second comment at A157446). - Vincenzo Librandi, Jan 26 2012

Links

Crossrefs

Cf. A157446, A157448.

Programs

  • Magma
    I:=[496, 1008]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Jan 26 2012
  • Mathematica
    LinearRecurrence[{2,-1},{496,1008},40] (* Vincenzo Librandi, Jan 26 2012 *)
512*Range[50]-16 (* Harvey P. Dale, Apr 09 2019 *)
  • PARI
    for(n=1, 22, print1(512*n - 16", ")); \\ Vincenzo Librandi, Jan 26 2012

Formula

a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Jan 26 2012
G.f.: x*(16*x + 496)/(x-1)^2. - Vincenzo Librandi, Jan 26 2012

A157448 a(n) = 2048*n^2 - 128*n + 1.

Original entry on oeis.org

1921, 7937, 18049, 32257, 50561, 72961, 99457, 130049, 164737, 203521, 246401, 293377, 344449, 399617, 458881, 522241, 589697, 661249, 736897, 816641, 900481, 988417, 1080449, 1176577, 1276801, 1381121, 1489537, 1602049, 1718657, 1839361
Offset: 1

Views

Author

Vincenzo Librandi, Mar 01 2009

Keywords

Comments

The identity (2048*n^2 - 128*n + 1)^2 - (16*n^2 - n)*(512*n - 16)^2 = 1 can be written as a(n)^2 - A157446(n)*A157447(n)^2 = 1 (see also second comment at A157446). - Vincenzo Librandi, Jan 26 2012

Links

Crossrefs

Cf. A157446, A157447.

Programs

  • Magma
    I:=[1921, 7937, 18049]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jan 26 2012
  • Mathematica
    LinearRecurrence[{3,-3,1},{1921,7937,18049},40] (* Vincenzo Librandi, Jan 26 2012 *)
  • PARI
    for(n=1, 22, print1(2048*n^2 - 128*n + 1", ")); \\ Vincenzo Librandi, Jan 26 2012

Formula

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jan 26 2012
G.f.: x*(-1921 - 2174*x - x^2)/(x-1)^3. - Vincenzo Librandi, Jan 26 2012

A157716 One-eighth of triangular numbers (integers only).

Original entry on oeis.org

0, 15, 17, 62, 66, 141, 147, 252, 260, 395, 405, 570, 582, 777, 791, 1016, 1032, 1287, 1305, 1590, 1610, 1925, 1947, 2292, 2316, 2691, 2717, 3122, 3150, 3585, 3615, 4080, 4112, 4607, 4641, 5166, 5202, 5757, 5795, 6380, 6420, 7035, 7077, 7722, 7766, 8441
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Mar 04 2009

Keywords

Comments

From Lamine Ngom, Oct 27 2020: (Start)
Numbers of the form (4*k)^2-k (A157446) or (4*k)^2+k (A157474).
Also numbers k such that 1+64*k is a square. (End)
The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(32*n))*(1 - q^(32*n-15))*(1 - q^(32*n-17)) = 1 - q^15 - q^17 + q^62 + q^66 - q^141 - q^147 + + - - .... - Peter Bala, Dec 24 2024

Examples

			The first three members of A000217 that are divisible by 8 are A000217(0), A000217(15) and A000217(16), so a(1) = A000217(0)/8 = 0, a(2) = A000217(15)/8 = 15, a(3) = A000217(16)/8 = 17.

Links

Crossrefs

Cf. A000217, A101877, A219257.

Programs

  • Maple
    seq((2*n-1 + 7/8*(-1)^n)^2 - 1/64, n = 1 .. 1000); # Robert Israel, Apr 20 2014
  • Mathematica
    Array[(2 # - 1 + 7/8*(-1)^#)^2 - 1/64 &, 46] (* or *)
Rest@ CoefficientList[Series[x^2*(15 + 2 x + 15 x^2)/((1 + x)^2*(1 - x)^3), {x, 0, 46}], x] (* Michael De Vlieger, Nov 05 2020 *)

Formula

G.f.: x^2*(15+2*x+15*x^2)/((1+x)^2*(1-x)^3 ). [Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009; checked and corrected by R. J. Mathar, Sep 16 2009]
a(n) = (2*n-1 + 7/8*(-1)^n)^2 -1/64. - Robert Israel, Apr 20 2014
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Nov 10 2020
Sum_{n>=2} 1/a(n) = 16 - (sqrt(2*(2+sqrt(2))) + sqrt(2) + 1)*Pi. - Amiram Eldar, Mar 17 2022

Extensions

Definition edited by N. J. A. Sloane, Mar 08 2009
Showing 1-3 of 3 results.

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