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 A022095 Fibonacci sequence beginning 1, 5. 34
 1, 5, 6, 11, 17, 28, 45, 73, 118, 191, 309, 500, 809, 1309, 2118, 3427, 5545, 8972, 14517, 23489, 38006, 61495, 99501, 160996, 260497, 421493, 681990, 1103483, 1785473, 2888956, 4674429, 7563385, 12237814, 19801199, 32039013, 51840212, 83879225, 135719437 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(5; n-1-k, k), n >= 1, with a(-1)=4. These are the sums of the SW-NE diagonals in P(5; n, k), the (5,1) Pascal triangle A093562. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also sums of the SW-NE diagonals in the (1,4)-Pascal triangle A095666. Row sums of triangle A131776 starting (1, 5, 6, 11, 17, 28, ...). - Gary W. Adamson, Jul 14 2007 In general, for a Fibonacci sequence beginning with 1,b we have: a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2b) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2b)))/sqrt(5). - Herbert Kociemba, Dec 18 2011 Subsequence of primes: 5, 11, 17, 73, 191, 809, 421493, 1103483, ... . - R. J. Mathar, Aug 09 2012 Pisano periods: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 9, 60, ... (differs from A001175). - R. J. Mathar, Aug 10 2012 LINKS Muniru A Asiru, Table of n, a(n) for n = 0..960 Tanya Khovanova, Recursive Sequences José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368 [cs.DM], 2012. José L. Ramírez and Gustavo N. Rubiano, Properties and Generalizations of the Fibonacci Word Fractal, The Mathematica Journal, Vol. 16 (2014). Matty van Son, Uniqueness conjectures for extended Markov numbers, arXiv:1911.00746 [math.NT], 2019. Index entries for linear recurrences with constant coefficients, signature (1,1). FORMULA a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=5. G.f.: (1+4*x)/(1-x-x^2). a(n) = 4*Fibonacci(n) + Fibonacci(n+1), n >= 1. - Zerinvary Lajos, Oct 05 2007, corrected by R. J. Mathar, Apr 07 2011 a(n-1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5)) + 2*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009 a(n) = 4*Fibonacci(n+2) - 3*Fibonacci(n+1). - Gary Detlefs, Dec 21 2010 a(n) = (L(n-2) + 8*L(n-1) + 4*L(n) + 2*L(n+1))/5 for the Lucas numbers L(n). - J. M. Bergot, Oct 22 2012 a(n) = ((2*sqrt(5) - 1)*(((1 + sqrt(5))/2)^(n+1)) + (2*sqrt(5) + 1)*(((1 - sqrt(5))/2)^(n+1)))/(sqrt(5)). - Bogart B. Strauss, Jul 19 2013 a(n) = Lucas(n-1) + Fibonacci(n+3) = Lucas(n+2) - Fibonacci(n-3). - Greg Dresden and Griffin Donaldson, Mar 03 2022 MAPLE with(combinat): a:= n-> 4*fibonacci(n)+fibonacci(n+1): seq(a(n), n=0..32); # Zerinvary Lajos, Oct 05 2007 MATHEMATICA f[n_] := (LucasL[n - 2] + 8*LucasL[n - 1] + 4*LucasL[n] + 2*LucasL[n + 1])/5; Array[f, 38, 0] (* or *) LinearRecurrence[{1, 1}, {1, 5}, 38] (* Robert G. Wilson v, Oct 22 2012 *) PROG (PARI) a(n)=fibonacci(n-1)+5*fibonacci(n) \\ Charles R Greathouse IV, Jun 05 2011 (Magma) a0:=1; a1:=5; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013 (GAP) List([0..40], n->4*Fibonacci(n)+Fibonacci(n+1)); # Muniru A Asiru, Mar 04 2018 CROSSREFS a(n) = A101220(4, 0, n+1). a(n) = A109754(4, n+1). Cf. A000032, A000045, A131776. Sequence in context: A070373 A231000 A274283 * A042531 A042839 A041373 Adjacent sequences: A022092 A022093 A022094 * A022096 A022097 A022098 KEYWORD nonn,easy AUTHOR N. J. A. Sloane STATUS approved

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Last modified September 17 15:47 EDT 2024. Contains 375987 sequences. (Running on oeis4.)