

A104449


Fibonacci sequence with initial values a(0) = 3 and a(1) = 1.


10



3, 1, 4, 5, 9, 14, 23, 37, 60, 97, 157, 254, 411, 665, 1076, 1741, 2817, 4558, 7375, 11933, 19308, 31241, 50549, 81790, 132339, 214129, 346468, 560597, 907065, 1467662, 2374727, 3842389, 6217116, 10059505, 16276621, 26336126, 42612747, 68948873, 111561620
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OFFSET

0,1


COMMENTS

The old name was: The Pibonacci numbers (a Fibonaccitype sequence): each term is the sum of the two previous terms.
The 6th row in the Wythoff array begins with the 6th term of the sequence (14, 23, 37, 60, 97, 157, ...). a(n) = f(n3) + f(n+2) for the Fibonacci numbers f(n) = f(n1) + f(n2); f(0) = 0, f(1) = 1.
(a(2*k), a(2*k+1)) give for k >= 0 the proper positive solutions of one of two families (or classes) of solutions (x, y) of the indefinite binary quadratic form x^2 + x*y  y^2 of discriminant 5 representing 11. The other family of such solutions is given by (x2, y2) = (b(2*k), b(2*k+1)) with b = A013655. See the formula in terms of Chebyshev S polynomials S(n, 3) = A001906(n+1) below, which follows from the fundamental solution (3, 1) by applying positive powers of the automorphic matrix given in a comment in A013655. See also A089270 with the Alfred Brousseau link with D = 11.  Wolfdieter Lang, May 28 2019


REFERENCES

V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.


LINKS

G. C. Greubel, Table of n, a(n) for n = 0..1000
John Conway, Alex Ryba, The extra Fibonacci series and the Empire State Building, Math. Intelligencer 38 (2016), no. 1, 4148. (Uses the name Pibonacci.)
Tanya Khovanova, Recursive Sequences
Ron Knott, Fibonacci Numbers and the Golden Section .
Eric Weisstein's World of Mathematics, Fibonacci Number
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (1,1).
Index entries for sequences related to Chebyshev polynomials.


FORMULA

a(n) = a(n1) + a(n2) with a(0) = 3, a(1) = 1.
a(n) = 3*Fibonacci(n1) + Fibonacci(n).  Zerinvary Lajos, Oct 05 2007
G.f.: (32*x)/(1xx^2).  Philippe Deléham, Nov 19 2008
a(n) = ( (3*sqrt(5)1)*((1+sqrt(5))/2)^n + (3*sqrt(5)+1)*((1sqrt(5) )/2)^n )/(2*sqrt(5)).  Bogart B. Strauss, Jul 19 2013
Bisection: a(2*k) = 4*S(k1, 3)  3*S(k2, 3), a(2*k+1) = 2*S(k1, 3) + S(k, 3) for k >= 0, with the Chebyshev S(n, 3) polynomials from A001906(n+1) for n >= 1.  Wolfdieter Lang, May 28 2019
a(n) = Fibonacci(n1) + Lucas(n).  G. C. Greubel, May 29 2019
a(3n + 4)/a(3n + 1) = continued fraction 4,4,4,...,4,9 (that's n 4's followed by a single 9).  Greg Dresden and Shaoxiong Yuan, Jul 16 2019
E.g.f.: (exp((1/2)*(1  sqrt(5))*x)*(1 + 3*sqrt(5) + ( 1 + 3*sqrt(5))*exp(sqrt(5)*x)))/(2*sqrt(5)).  Stefano Spezia, Jul 18 2019


MAPLE

a:=n>3*fibonacci(n1)+fibonacci(n): seq(a(n), n=0..40); # Zerinvary Lajos, Oct 05 2007


MATHEMATICA

LinearRecurrence[{1, 1}, {3, 1}, 40] (* Harvey P. Dale, May 23 2014 *)


PROG

(PARI) a(n)=3*fibonacci(n1)+fibonacci(n) \\ Charles R Greathouse IV, Jun 05 2011
(MAGMA) [Fibonacci(n1) + Lucas(n): n in [0..40]]; // G. C. Greubel, May 29 2019
(Sage) ((32*x)/(1xx^2)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, May 29 2019
(GAP) a:=[3, 1];; for n in [3..40] do a[n]:=a[n1]+a[n2]; od; a; # G. C. Greubel, May 29 2019


CROSSREFS

Cf. Other Fibonaccitype sequences: A000045, A000032, A013655. Other related sequences: A001906, A013655, A089270, A103343, A103344.
Wythoff array: A035513.
Essentially the same as A000285.
Sequence in context: A183904 A105177 A050057 * A116416 A051203 A243553
Adjacent sequences: A104446 A104447 A104448 * A104450 A104451 A104452


KEYWORD

nonn,easy


AUTHOR

Casey Mongoven, Mar 08 2005


EXTENSIONS

Name changed by Wolfdieter Lang, Jun 17 2019


STATUS

approved



