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%I #61 Feb 18 2024 01:58:46
%S 1,9,10,19,29,48,77,125,202,327,529,856,1385,2241,3626,5867,9493,
%T 15360,24853,40213,65066,105279,170345,275624,445969,721593,1167562,
%U 1889155,3056717,4945872,8002589,12948461,20951050,33899511,54850561,88750072,143600633,232350705
%N Fibonacci sequence beginning 1, 9.
%C a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(9;n-1-k,k) with n>=1, a(-1)=8. These are the SW-NE diagonals in P(9;n,k), the (9,1) Pascal triangle A093644. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.
%C 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
%C Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (perhaps the same as A001175). - _R. J. Mathar_, Aug 10 2012
%C No, it is not the same as in A001175. The Pisano periods are different for moduli 71 and 142, where they are 35 and 105 instead of 70 and 210. Otherwise they coincide with those of the Fibonacci sequence. - _Klaus Purath_, Jun 26 2022
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).
%F a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=9. a(-1):=8.
%F G.f.: (1+8*x)/(1-x-x^2).
%F a(n) = A109754(8, n+1) = A101220(8, 0, n+1).
%F a(n+1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5))+ 4*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
%F a(n) = 8*A000045(n) + A000045(n+1). - _R. J. Mathar_, Aug 10 2012
%F a(n) = 10*A000045(n) - A000045(n-2). - _Bruno Berselli_, Feb 20 2017
%F a(n) = Lucas(n+3) + Fibonacci(n-4). - _Greg Dresden_ and Mary Beth Pittman, Mar 12 2022
%t LinearRecurrence[{1, 1}, {1, 9}, 36] (* _Robert G. Wilson v_, Apr 11 2014 *)
%o (Magma) a0:=1; a1:=9; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // _Bruno Berselli_, Feb 12 2013
%Y Cf. A000045, A001175, A093644, A101220, A109754.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Jun 14 1998
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