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.

A157336 a(n) = 8*(8*n + 1).

Original entry on oeis.org

72, 136, 200, 264, 328, 392, 456, 520, 584, 648, 712, 776, 840, 904, 968, 1032, 1096, 1160, 1224, 1288, 1352, 1416, 1480, 1544, 1608, 1672, 1736, 1800, 1864, 1928, 1992, 2056, 2120, 2184, 2248, 2312, 2376, 2440, 2504, 2568, 2632, 2696, 2760, 2824, 2888, 2952
Offset: 1

Views

Author

Vincenzo Librandi, Feb 27 2009

Keywords

Comments

The identity (128*n^2+32*n+1)^2-(4*n^2+n)*(64*n+8)^2=1 can be written as A157337(n)^2-A007742(n)*a(n)^2=1. This is the case s=2 of the identity (8*n^2*s^4+8*n*s^2+1)^2-(n^2*s^2+n)*(8*n*s^3+4*s)^2=1. - Vincenzo Librandi, Jan 29 2012
Likewise, the immediate identity (a(n)^2+1)^2-(a(n)^2+2)*a(n)^2 = 1 can be rewritten as A158686(8*n+1)^2-(A158686(8*n+1)+1)*a(n)^2=1. - Bruno Berselli, Feb 13 2012

Links

Crossrefs

Cf. A007742, A157337.

Programs

Formula

From Vincenzo Librandi, Jan 29 2012: (Start)
G.f.: 8*(1+7*x)/(x-1)^2. [corrected by Georg Fischer, May 12 2019]
a(n) = 2*a(n-1)-a(n-2). (End)
E.g.f.: 8*(1+8*x)*exp(x). - G. C. Greubel, Feb 01 2018

A158685 a(n) = 32*(32*n^2 + 1).

Original entry on oeis.org

32, 1056, 4128, 9248, 16416, 25632, 36896, 50208, 65568, 82976, 102432, 123936, 147488, 173088, 200736, 230432, 262176, 295968, 331808, 369696, 409632, 451616, 495648, 541728, 589856, 640032, 692256, 746528, 802848, 861216, 921632, 984096, 1048608, 1115168, 1183776
Offset: 0

Views

Author

Vincenzo Librandi, Mar 24 2009

Keywords

Comments

The identity (64*n^2 + 1)^2 - (1024*n^2 + 32)*(2*n)^2 = 1 can be written as A158686(n)^2 - a(n)*A005843(n)^2 = 1.

Links

Crossrefs

Cf. A005843, A158575, A158686.

Programs

Formula

G.f.: -32*(1+30*x+33*x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 21 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/(4*sqrt(2)))*Pi/(4*sqrt(2)) + 1)/64.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(4*sqrt(2)))*Pi/(4*sqrt(2)) + 1)/64. (End)
From Elmo R. Oliveira, Jan 15 2025: (Start)
E.g.f.: 32*exp(x)*(1 + 32*x + 32*x^2).
a(n) = 32*A158575(n). (End)

Extensions

Comment rewritten, a(0) added and formula replaced by R. J. Mathar, Oct 22 2009

A295130 a(n) = 3*n*(64*n^2 + 1).

Original entry on oeis.org

195, 1542, 5193, 12300, 24015, 41490, 65877, 98328, 139995, 192030, 255585, 331812, 421863, 526890, 648045, 786480, 943347, 1119798, 1316985, 1536060, 1778175, 2044482, 2336133, 2654280, 3000075, 3374670, 3779217, 4214868, 4682775, 5184090, 5719965, 6291552, 6900003, 7546470, 8232105, 8958060, 9725487
Offset: 1

Views

Author

Michael H. Bischoff, Nov 15 2017

Keywords

Examples

			For examples see "Squares in a square" in the LINKS section.

References

  • Martin Gardner, Mathematical Carnival, 1975, Alfred A. Knopf Inc., New York.

Links

Crossrefs

Cf. A008585, A158686.

Programs

  • Mathematica
    f[n_] := 3n (64n^2 +1); Array[f, 33] (* or *)
CoefficientList[ Series[(3 (65 +254x +65x^2))/(-1 +x)^4, {x, 0, 33}], x] (* or *)
LinearRecurrence[{4, -6, 4, -1}, {195, 1542, 5193, 12300}, 34] (* Robert G. Wilson v, Dec 27 2017 *)
  • PARI
    Vec(3*x*(65 + 254*x + 65*x^2) / (1 - x)^4 + O(x^40)) \\ Colin Barker, Nov 23 2017
  • PARI
    a(n) = 192*n^3 + 3*n \\ Iain Fox, Dec 22 2017

Formula

a(n) = 3*n*(64*n^2 + 1).
From Colin Barker, Nov 23 2017: (Start)
G.f.: 3*x*(65 + 254*x + 65*x^2) / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>4.
(End)
a(n) = A008585(n) * A158686(n). - Omar E. Pol, Nov 24 2017
E.g.f.: 3*x*e^x * (65 + 192*x + 64*x^2). - Iain Fox, Dec 22 2017
Showing 1-3 of 3 results.

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