A022100 Fibonacci sequence beginning 1, 10.

1, 10, 11, 21, 32, 53, 85, 138, 223, 361, 584, 945, 1529, 2474, 4003, 6477, 10480, 16957, 27437, 44394, 71831, 116225, 188056, 304281, 492337, 796618, 1288955, 2085573, 3374528, 5460101, 8834629, 14294730, 23129359, 37424089, 60553448, 97977537, 158530985

Offset

0,2

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(10; n-1-k, k), n >= 1, with a(-1)=9. These are the SW-NE diagonals in P(10; n, k), the (10,1) Pascal triangle A093645. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

In general, for b Fibonacci sequence beginning with 1, h, we have:

b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011

Links

Table of n, a(n) for n=0..36.

Tanya Khovanova, Recursive Sequences

Index entries for linear recurrences with constant coefficients, signature (1, 1).

Formula

a(n) = a(n-1) + a(n-2) for n >= 2, a(0)=1, a(1)=10, a(-1):=9.

G.f.: (1 + 9*x)/(1 - x - x^2).

a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*(9*binomial(1, k) - 8*binomial(0, k)). - Paul Barry, May 05 2005

a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (9/2)*((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009

From Bruno Berselli, Feb 20 2017: (Start)

a(n) = 9*A000045(n) + A000045(n+1).

a(n) = 11*A000045(n) - A000045(n-2). (End)

Mathematica

LinearRecurrence[{1, 1}, {1, 10}, 40] (* Harvey P. Dale, May 17 2017 *)

Prog

(Magma) a0:=1; a1:=10; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

Crossrefs

Cf. A000045.

a(n) = A109754(9, n+1) = A101220(9, 0, n+1).

Sequence in context: A298849 A277588 A339138 * A041475 A041204 A041202

Adjacent sequences: A022097 A022098 A022099 * A022101 A022102 A022103

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved