This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A022099 #61 Feb 18 2024 01:58:46
%S A022099 1,9,10,19,29,48,77,125,202,327,529,856,1385,2241,3626,5867,9493,
%T A022099 15360,24853,40213,65066,105279,170345,275624,445969,721593,1167562,
%U A022099 1889155,3056717,4945872,8002589,12948461,20951050,33899511,54850561,88750072,143600633,232350705
%N A022099 Fibonacci sequence beginning 1, 9.
%C A022099 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 A022099 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 A022099 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 A022099 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 A022099 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A022099 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).
%F A022099 a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=9. a(-1):=8.
%F A022099 G.f.: (1+8*x)/(1-x-x^2).
%F A022099 a(n) = A109754(8, n+1) = A101220(8, 0, n+1).
%F A022099 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 A022099 a(n) = 8*A000045(n) + A000045(n+1). - _R. J. Mathar_, Aug 10 2012
%F A022099 a(n) = 10*A000045(n) - A000045(n-2). - _Bruno Berselli_, Feb 20 2017
%F A022099 a(n) = Lucas(n+3) + Fibonacci(n-4). - _Greg Dresden_ and Mary Beth Pittman, Mar 12 2022
%t A022099 LinearRecurrence[{1, 1}, {1, 9}, 36] (* _Robert G. Wilson v_, Apr 11 2014 *)
%o A022099 (Magma) a0:=1; a1:=9; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // _Bruno Berselli_, Feb 12 2013
%Y A022099 Cf. A000045, A001175, A093644, A101220, A109754.
%K A022099 nonn,easy
%O A022099 0,2
%A A022099 _N. J. A. Sloane_, Jun 14 1998