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.

A086088 Decimal expansion of the limit of the ratio of consecutive terms in the tetranacci sequence A000078.

Original entry on oeis.org

1, 9, 2, 7, 5, 6, 1, 9, 7, 5, 4, 8, 2, 9, 2, 5, 3, 0, 4, 2, 6, 1, 9, 0, 5, 8, 6, 1, 7, 3, 6, 6, 2, 2, 1, 6, 8, 6, 9, 8, 5, 5, 4, 2, 5, 5, 1, 6, 3, 3, 8, 4, 7, 2, 7, 1, 4, 6, 6, 4, 7, 0, 3, 8, 0, 0, 9, 6, 6, 6, 0, 6, 2, 2, 9, 7, 8, 1, 5, 5, 5, 9, 1, 4, 9, 8, 1, 8, 2, 5, 3, 4, 6, 1, 8, 9, 0, 6, 5, 3, 2, 5
Offset: 1

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Author

Eric W. Weisstein, Jul 08 2003

Keywords

Comments

The tetranacci constant corresponds to the Golden Section in a quadripartite division 1 = u_1 + u_2 + u_3 + u_4 of a unit line segment, i.e., if 1/u_1 = u_1/u_2 = u_2/u_3 = u_3/u_4 = c, c is the tetranacci constant. - Seppo Mustonen, Apr 19 2005
The other 3 polynomial roots of 1+x+x^2+x^3-x^4 are -0.77480411321543385... and the complex-conjugated pair -0.07637893113374572508475 +- i * 0.814703647170386526841... - R. J. Mathar, Oct 25 2008
The continued fraction expansion starts 1, 1, 12, 1, 4, 7, 1, 21, 1, 2, 1, 4, 6, 1, 10, 1, 2, 2, 1, 7, 1, 1,... - R. J. Mathar, Mar 09 2012
For n>=4, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. - Vladimir Shevelev, Mar 21 2014
Note that we have: c + c^(-4) = 2, and the k-nacci constant approaches 2 when k approaches infinity (Martin Gardner). - Bernard Schott, May 09 2022

Examples

			1.927561975...
		

References

  • Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, p. 101, Simon & Schuster, NY, 1961.

Crossrefs

Cf. A000078.
k-nacci constants: A001622 (Fibonacci), A058265 (tribonacci), this sequence (tetranacci), A103814 (pentanacci), A118427 (hexanacci), A118428 (heptanacci).

Programs

Formula

Equals 1/4 + sqrt(11/48 - s/72 + 7/s) + sqrt(11/24 + s/72 - 7/s + 1 / sqrt(704/507 - 128 * s/1521 + 7168 / (169 * s))) where s = (sqrt(177304464) + 7020)^(1/3). - Michal Paulovic, Oct 08 2022