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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A254076 a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n>2, a(0)=-1, a(1)=-2, a(2)=-4.

Original entry on oeis.org

-1, -2, -4, 1, 2, 13, 26, 61, 122, 253, 506, 1021, 2042, 4093, 8186, 16381, 32762, 65533, 131066, 262141, 524282, 1048573, 2097146, 4194301, 8388602, 16777213, 33554426, 67108861, 134217722, 268435453, 536870906, 1073741821, 2147483642, 4294967293
Offset: 0

Views

Author

Paul Curtz, Jan 29 2015

Keywords

Comments

The main diagonal of the difference table is -A000079(n) = -2^n.
a(n) mod 9 is of period 6: repeat 8, 7, 5, 1, 2, 4.
a(n) + a(n+1) = -3, -6, -3, 3, 15, ...; all are multiples of 3.

Links

Crossrefs

Cf. A000079, A010704, A141725, A153130, A156067, A157823.

Programs

  • Mathematica
    a[0] = -1; a[n_] := 2^(n-1) + 3*Mod[n, 2] - 6; Table[a[n], {n, 0, 33}] (* Jean-François Alcover, Feb 04 2015 *)
  • PARI
    Vec((9*x^3+x^2-1)/((x-1)*(x+1)*(2*x-1)) + O(x^100)) \\ Colin Barker, Jan 30 2015

Formula

a(2n+1) = A141725(n-1), a(2n+2) = 2*a(2n+1).
a(n+1) = 2*a(n) + (period 2: repeat 0, 9), n>0.
a(n) = -A157823(n) - (period 2: repeat 6, 3).
a(n+1) = a(n) - A156067(n).
a(n+2) = a(n) + 3*2^(n-1), n>0.
a(n+4) = a(n) + 15*2^(n-1), n>0.
a(n+6) = a(n) + 63*2^(n-1), n>0.
a(n) = (2^n - 3*(-1)^n - 9)/2 for n>0. - Colin Barker, Jan 30 2015
G.f.: (9*x^3+x^2-1) / ((x-1)*(x+1)*(2*x-1)). - Colin Barker, Jan 30 2015

A158927 a(n) = -3a(n-1) - 3a(n-2) - 2a(n-3), n > 3.

Original entry on oeis.org

2, 2, 2, -7, 11, -16, 29, -61, 128, -259, 515, -1024, 2045, -4093, 8192, -16387, 32771, -65536, 131069, -262141, 524288, -1048579, 2097155, -4194304, 8388605, -16777213, 33554432, -67108867, 134217731, -268435456, 536870909, -1073741821, 2147483648
Offset: 0

Views

Author

Paul Curtz, Mar 31 2009

Keywords

Comments

The inverse binomial transform of A153130, after dropping A153130(0).
The inverse binomial transform of the full A153130 is A158916.
Dropping two initial terms of A153130 yields A158935, dropping three yields essentially a sign-reversed version of A158916, dropping 4 essentially the sequence here.

Links

Crossrefs

Same recurrence as A131562, A158916, A158926.

Programs

  • GAP
    a := [2,2,2,-7];; for n in [5..10^3] do a[n] := -3*a[n-1] - 3*a[n-2] - 2*a[n-3]; od; a; # Muniru A Asiru, Jan 27 2018
  • Maple
    a := proc(n) option remember: if n=0 then 2 elif n=1 then 2 elif n=2 then 2 elif n=3 then -7 elif n>=4 then -3*procname(n-1) - 3*procname(n-2) - 2*procname(n-3) fi; end:
seq(a(n), n=0..100); # Muniru A Asiru, Jan 27 2018

Formula

a(n) = -3a(n-1) - 3a(n-2) - 2a(n-3), with a(0)=a(1)=a(2)=2, a(3)=-7.
a(n) = (-1)^(n+1)*A157823(n) - A099838(n+3).
G.f.: (2+8*x+14*x^2+9*x^3)/((2*x+1)*(1+x+x^2)). - R. J. Mathar, Apr 09 2009
a(0)=2; a(n) = (1/2)*(-2)^n - 3*cos(2*Pi*n/3) + sqrt(3)*sin(2*Pi*n/3) for n >= 1. - Richard Choulet, Apr 23 2009

Extensions

Edited and extended by R. J. Mathar, Apr 09 2009
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