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.

A228954 Bisection of A195240(n).

Original entry on oeis.org

1, 7, 11, 7, 19, 337, 5, -1681, 22133, -87223, 427291, -118181363, 4276553, -11874730297, 4307920641583, -3854660520481, 1288843929185, -13157635776526258889, 1464996956920781, -130541359248224557643
Offset: 0

Views

Author

Paul Curtz, Sep 09 2013

Keywords

Comments

The first bisection is b(n) = 0, 1, 8, 10, 8, 14, 1028, -2, 1936, -21734,... .
a(n) and b(n) are twice linked to Bernoulli numbers (A027641(n+4) or A164555(n+4))/A027642(n+4).

Programs

  • Mathematica
    evb = Join[{0, 1, 0}, Table[BernoulliB[n], {n, 2, 42}]]; ievb = Table[ Sum[Binomial[n, k]*evb[[k + 1]], {k, 0, n}], {n, 0, Length[evb] - 3}]; A195240 = Differences[ievb, 2] // Numerator; Partition[A195240, 2][[All, 2]]
(* or *)
A000367[n_] := BernoulliB[2*n] // Numerator; A001897[n_] := -2*(2^(2*n - 1) - 1)*BernoulliB[2*n] // Denominator; a[0] = 1; a[n_] := (A000367[n + 1] + A001897[n + 1])/2; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Sep 09 2013, after R. J. Mathar *)

Formula

A195240(2n+1).
a(n+1) = b(n+2) + A000367(n+2).
a(n+1) = A001897(n+2) - b(n+2).
2*a(n+1) = A000367(n+2) + A001897(n+2).

Extensions

More terms from Jean-François Alcover, Sep 09 2013

A206012 Modular recursion: a(0)=a(1)=a(2)=a(3)=1, thereafter: a(n) equals a(n - 2) + a(n - 3) when n = 0 mod 5, a(n - 1) + a(n - 3) when n = 1 mod 5, a(n - 1) + a(n - 2) when n = 2 mod 5, a(n - 1) + a(n - 4) when n = 3 mod 5, and a(n - 1) + a(n - 2) + a(n - 3) otherwise.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 3, 5, 8, 16, 13, 21, 34, 50, 105, 84, 134, 218, 323, 675, 541, 864, 1405, 2080, 4349, 3485, 5565, 9050, 13399, 28014, 22449, 35848, 58297, 86311, 180456, 144608, 230919, 375527, 555983, 1162429, 931510, 1487493, 2419003, 3581432, 7487928
Offset: 0

Views

Author

Roger L. Bagula, Mar 19 2012

Keywords

Comments

This sequence was inspired by the work of Paul Curtz on three part sequences. I did a three part version of this that gave a generating polynomial and got even more variance by adding two more modulo sequences.

Links

Crossrefs

Cf. A194880, A195240, A185138, A206568, A110897, A110898.

Programs

  • Magma
    m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((x^15-x^13+x^12-2*x^10-2*x^9+2*x^8-x^7-3*x^6-4*x^5+3*x^4+x^3+x^2+x+1)/(x^15-3*x^10-6*x^5+1))); // Bruno Berselli, Mar 20 2012
  • Mathematica
    a[0] = 1; a[1] = 1; a[2] = 1; a[3] = 1;a[n_Integer] := a[n]=If[Mod[n, 5] == 0, a[n - 2] + a[n - 3], If[Mod[n, 5] == 1, a[n - 1] + a[n - 3], If[Mod[n, 5] == 2, a[n - 1] + a[n - 2], If[Mod[n, 5] == 3, a[n - 1] + a[n - 4], a[n - 1] + a[n - 2] + a[n - 3]]]]];b = Table[a[n], {n, 0, 50}];(* FindSequenceFunction gives*);Table[c[n] = b[[n]], {n, 1, 16}];c[n_Integer] := c[n] = -c[-15 + n] + c[-10 + n] + 6 c[-5 + n];d = Table[c[n], {n, 1, Length[b]}]
CoefficientList[Series[(x^15-x^13+x^12-2*x^10-2*x^9+2*x^8-x^7-3*x^6-4*x^5+3*x^4+x^3+x^2+x+1)/(x^15-3*x^10-6*x^5+1),{x,0,1001}],x] (* Vincenzo Librandi, Apr 01 2012 *)
  • PARI
    Vec((x^15-x^13+x^12-2*x^10-2*x^9+2*x^8-x^7-3*x^6-4*x^5+3*x^4+x^3+x^2+x+1)/(x^15-3*x^10-6*x^5+1)+O(x^99)) \\ Charles R Greathouse IV, Mar 19 2012

Formula

G.f.: (x^15 - x^13 + x^12 - 2x^10 - 2x^9 + 2x^8 - x^7 - 3x^6 - 4x^5 + 3x^4 + x^3 + x^2 + x + 1) / (x^15 - 3x^10 - 6x^5 + 1). - Alois P. Heinz, Mar 19 2012

A256003 a(n) = 0 followed by numerators of 2*A176327(n)/A176289(n).

Original entry on oeis.org

0, 2, 0, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 7, 0, -3617, 0, 43867, 0, -174611, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -7709321041217, 0, 2577687858367, 0, -26315271553053477373, 0
Offset: 0

Views

Author

Paul Curtz, May 06 2015

Keywords

Comments

Offset 0 is chosen instead of -1. (The offset 0 corresponds to A176327(n), -1 to 0 followed by A176327(n).)
Denominators: b(n) = 1 followed by A141459(n).
Difference table of a(n)/b(n):
0, 2, 0, 1/3, 0, -1/15, 0, ...
2, -2, 1/3, -1/3, -1/15, 1/15, ...
-4, 7/3, -2/3, 4/15, 2/15, ...
19/3, -3, 14/15, -2/15, ...
-28/3, 59/15, -16/15, ...
199/15, -5, ...
-274/15, ...
etc.
Without the first column, the antidiagonal sums are (-1)^n * A254667(n+1).
The Bernoulli numbers A027641(n)/A027642(n) or A164555(n)/A027642(n) come from A000027. 0 followed by the Bernoulli numbers comes from A001477. a(0)=0 is a choice.

Crossrefs

Cf. A176327/A176289, A172086/A172087, A141459, A176618, A195240, A254667.

Formula

a(2n) = 0. a(2n+1) = A172086(2n), from the main pure Bernoulli twin numbers.
Showing 1-3 of 3 results.

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