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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A213968 Probably an erroneous version of A213967.

Original entry on oeis.org

1, 2, 3, 7, 13, 24, 45, 83, 152, 273, 471, 768, 1167, 1641
Offset: 1

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Author

N. J. A. Sloane, Jun 30 2012

Keywords

References

  • Atanassov, K. T.; Atanassova, V.; Shannon, A. G.; Turner, J. C. New visual perspectives on Fibonacci numbers. With a foreword by A. F. Horadam. World Scientific Publishing Co., Inc., River Edge, NJ, 2002. xvi+313 pp. ISBN: 981-238-134-1 MR1932564 (2003h:11015). See p. 68.

A248098 a(n) = 1 + a(n-1) + a(n-2) + a(n-3) if n>=4; a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

1, 1, 1, 4, 7, 13, 25, 46, 85, 157, 289, 532, 979, 1801, 3313, 6094, 11209, 20617, 37921, 69748, 128287, 235957, 433993, 798238, 1468189, 2700421, 4966849, 9135460, 16802731, 30905041, 56843233, 104551006, 192299281, 353693521
Offset: 1

Views

Author

Emeric Deutsch, Dec 28 2014

Keywords

Comments

Number of vertices in the Fibonacci ternary tree T(n). The Fibonacci ternary tree T(n) is defined inductively: T(1), T(2), T(3) consist of only a root node, while for n>=4, T(n) consists of a root node with 3 ordered children T(n-1), T(n-2), T(n-3) from left to right. See the Chang reference.
The number of leaves in the Fibonacci ternary tree T(n) is the tribonacci number A000213(n-1).
In general, adding a constant to each successive term of a third-order linear recurrence with signature (x,y,z) results in a fourth-order recurrence with signature (x+1, y-x, z-y, -z). - Gary Detlefs, Jul 20 2023

Links

Crossrefs

Cf. A000213.
Cf. A213967.

Programs

  • Haskell
    a248098 n = a248098_list !! (n-1)
a248098_list = 1 : 1 : 1 : map (+ 1) (zipWith3 (((+) .) . (+))
               a248098_list (tail a248098_list) (drop 2 a248098_list))
-- Reinhard Zumkeller, Dec 29 2014
  • Maple
    a[1]:=1: a[2]:=1: a[3]:=1: for n from 4 to 40 do a[n] := 1+a[n-1]+a[n-2]+a[n-3] end do: seq(a[n], n=1..40);
g:=z*(1-z^2+2*z^3-z)/((1-z)*(1-z-z^2-z^3)): gser:=series(g,z=0,45): seq(coeff(gser,z,n), n=1..40);

Formula

G.f. = z*(1-z-z^2+2*z^3)/((1-z)*(1-z-z^2-z^3)).
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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