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.

A011922 a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3) with a(0)=1, a(1)=3, and a(2)=33.

Original entry on oeis.org

1, 3, 33, 451, 6273, 87363, 1216801, 16947843, 236052993, 3287794051, 45793063713, 637815097923, 8883618307201, 123732841202883, 1723376158533153, 24003533378261251, 334326091137124353, 4656561742541479683, 64857538304443591201, 903348974519668797123, 12582028104970919568513
Offset: 0

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Author

Mario Velucchi (mathchess(AT)velucchi.it)

Keywords

References

  • Mario Velucchi, Seeing couples, in Recreational and Educational Computing, to appear 1997.

Crossrefs

Programs

  • Magma
    I:=[1,3,33]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..17]]; // Bruno Berselli, Jul 09 2011
    
  • Maple
    a:= gfun:-rectoproc({a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3), a(0)=1,a(1)=3,a(2)=33},a(n),remember):
    map(a,[$0..100]); # Robert Israel, Jul 02 2015
  • Mathematica
    RecurrenceTable[{a[n] == 15 a[n - 1] - 15 a[n - 2] + a[n - 3], a[0] == 1, a[1] == 3, a[2] == 33}, a, {n, 0, 15}] (* Michael De Vlieger, Jul 02 2015 *)
    LinearRecurrence[{15,-15,1},{1,3,33},30] (* Harvey P. Dale, Dec 04 2018 *)
  • Maxima
    a[0]:1$ a[1]:3$ a[2]:33$ a[n]:=15*a[n-1]-15*a[n-2]+a[n-3]$ makelist(a[n], n, 0, 16); /* Bruno Berselli, Jul 09 2011 */
    
  • PARI
    a(n)=([0,1,0; 0,0,1; 1,-15,15]^n*[1;3;33])[1,1] \\ Charles R Greathouse IV, Jul 02 2015

Formula

a(n) = (2+sqrt(1+((((2+sqrt(3))^(2*n)-(2-sqrt(3))^(2*n))^2)/4)))/3. [corrected by Francisco Salinas (franciscodesalinas(AT)hotmail.com), Dec 30 2001]
a(n) = ((7+4*sqrt(3))^n+(7-4*sqrt(3))^n+4)/6. - Bruno Berselli, Jul 09 2011
G.f.: (1-12*x+3*x^2)/ ((1-x) * (x^2-14*x+1)). - R. J. Mathar, Apr 15 2010
Sqrt(3) = 1 + Sum_{n>=1} 2/a(n) = 1 + 2/3 + 2/33 + ... - Gary W. Adamson, Jun 12 2003
a(n)^2 = A103974(n+1)^2 - (4*A007655(n+1))^2. - Paul D. Hanna, Mar 06 2005
a(n) = (A011943(n+1) + 2)/3. - Ralf Stephan, Aug 13 2013
a(n) = A001075(n)^2 - A001353(n)^2. - Richard R. Forberg, Aug 24 2013
E.g.f.: exp(x)*(2 + exp(6*x)*cosh(4*sqrt(3)*x))/3. - Stefano Spezia, Dec 11 2022

Extensions

Recurrence in definition by R. J. Mathar, Apr 15 2010