

A093406


A sequence converging to 1 + 2^(1/4).


2



1, 3, 11, 31, 71, 145, 289, 601, 1321, 2979, 6683, 14743, 32111, 69697, 151777, 332113, 728689, 1598883, 3503627, 7668079, 16774775, 36704017, 80343361, 175916521, 385196761, 843365379, 1846290395, 4041672871, 8847607391, 19368919297, 42403014721, 92830645537
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OFFSET

1,2


COMMENTS

a(n)/a(n1) tends to 2.189207115... = 1 + 2^(1/4). Example: a(18)/a(17) = 1598883/728689 = 2.19419... A052101 is the series derived from analogous 3rd order operations, with lim a(n)/a(n1) as n approaches inf. = 1 + 2^(1/3).


REFERENCES

E. J. Barbeau, Polynomials, SpringerVerlag NY Inc, 1989, p. 136.


LINKS

Table of n, a(n) for n=1..32.
Index entries for linear recurrences with constant coefficients, signature (4,6,4,1).


FORMULA

We use a 4 X 4 matrix corresponding to the characteristic polynomial (x  1)^4  2 = 0 = x^4  4x^3 + 6x^2  4x  1 = 0, being [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / 1 4 6 4]. Let the matrix = M. Perform M^n * [1, 1, 1, 1]. a(n) = the third term from the left, (the other 3 terms being offset members of the series).
a(n) = 4*a(n1)6*a(n2)+4*a(n3)+a(n4). G.f.: x*(x^3+5*x^2x+1)/ (x^4+4*x^36*x^2+4*x1). [Colin Barker, Oct 21 2012]


EXAMPLE

a(4) = 31, since M^4 * [1,1,1,1] = [3, 11, 31, 71].


MATHEMATICA

LinearRecurrence[{4, 6, 4, 1}, {1, 3, 11, 31}, 40] (* Harvey P. Dale, Jul 22 2013 *)


CROSSREFS

Cf. A052101.
Sequence in context: A261148 A071568 A097081 * A107587 A245931 A190590
Adjacent sequences: A093403 A093404 A093405 * A093407 A093408 A093409


KEYWORD

nonn,easy


AUTHOR

Gary W. Adamson, Mar 28 2004


EXTENSIONS

Corrected by T. D. Noe, Nov 08 2006


STATUS

approved



