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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A138749 a(n) = 2*a(n-1) - 5*a(n-2), with a(1) = -1, a(2) = -7.

Original entry on oeis.org

-1, -7, -9, 17, 79, 73, -249, -863, -481, 3353, 9111, 1457, -42641, -92567, 28071, 518977, 897599, -799687, -6087369, -8176303, 14084239, 69049993, 67678791, -209892383, -758178721, -466895527, 2857102551, 8048682737, 1811852719, -36619708247
Offset: 1

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Author

Gary W. Adamson, Mar 28 2008

Keywords

Examples

			a(5) = 79 = 2*a(4) - 5*a(3) = 2*17 - 5*(-9).
a(5) = 79 = left term in [1,-2, 2,1]^5.

Links

Programs

  • Mathematica
    Rest[CoefficientList[Series[-x*(1+5*x)/(1-2*x+5*x^2),{x,0,30}],x]] (* or *) LinearRecurrence[{2,-5},{-1,-7},30] (* James C. McMahon, Jun 21 2025 *)
  • PARI
    a(n)={local(v=Vec((1+2*I*x)^n)); sum(k=1,#v, real(v[k])-imag(v[k]));} /* cf. A116483 */ /* Joerg Arndt, Jul 06 2011 */

Formula

a(n) = 2*a(n-1) - 5*a(n-2), n>3.
a(n) = left term in [1,-2; 2,1]^n * [1,1].
O.g.f.: -x*(1+5*x)/(1-2*x+5*x^2). a(n)=-A045873(n)-5*A045873(n-1). - R. J. Mathar, Apr 03 2008
a(n) = (1/2)*(1+i)*((1+2*i)^n-i*(1-2*i)^n), where i=sqrt(-1). - Bruno Berselli, Jul 06 2011

A231958 Numbers n dividing the Lucas sequence u(n) defined by u(i) = 2*u(i-1) - 5*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 2, 4, 8, 12, 16, 24, 32, 36, 48, 56, 64, 72, 96, 108, 112, 128, 132, 144, 156, 168, 192, 216, 224, 256, 264, 272, 288, 312, 324, 336, 384, 392, 396, 432, 448, 468, 496, 504, 512, 528, 544, 552, 576, 624, 648, 672, 768, 784, 792, 816, 864, 896, 936, 972
Offset: 1

Views

Author

Thomas M. Bridge, Nov 15 2013

Keywords

Comments

All terms except 1 and 2 are divisible by 4. The sequence contains every nonnegative integer power of 2. There are infinitely many multiples of 12 in the sequence.

Links

Crossrefs

Cf. A000079 (powers of 2 (subsequence)).
Cf. A045873 (Lucas sequence).

Programs

  • Mathematica
    nn = 2000; s = LinearRecurrence[{2, -5}, {1, 2}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 20 2013 *)
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