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%I A022098 #71 Feb 16 2024 10:22:50
%S A022098 1,8,9,17,26,43,69,112,181,293,474,767,1241,2008,3249,5257,8506,13763,
%T A022098 22269,36032,58301,94333,152634,246967,399601,646568,1046169,1692737,
%U A022098 2738906,4431643,7170549,11602192,18772741,30374933,49147674,79522607,128670281
%N A022098 Fibonacci sequence beginning 1, 8.
%C A022098 a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(8; n-1-k, k) with n >= 1, a(-1) = 7. These are the SW-NE diagonals in P(8; n, k), the (8, 1) Pascal triangle A093565. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.
%C A022098 Pisano period lengths: 1, 3, 8, 6, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, ... (is this the same as A106291?). - _R. J. Mathar_, Aug 10 2012
%C A022098 Also the sum of five consecutive Lucas numbers starting with L(-3). - _Alonso del Arte_, Sep 26 2013
%C A022098 The Pisano period lengths of this sequence are exactly the same as those of the Lucas sequence (A000032), given in A106291. - _Klaus Purath_, Apr 20 2019
%H A022098 Vincenzo Librandi, <a href="/A022098/b022098.txt">Table of n, a(n) for n = 0..1000</a>
%H A022098 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A022098 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).
%F A022098 a(n) = a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=8, and a(-1):=7.
%F A022098 G.f.: (1 + 7*x)/(1 - x - x^2).
%F A022098 a(n) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5)) + 3.5*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)) for n>0. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
%F A022098 a(n) = 2^(-1-n)*((1 - sqrt(5))^n*(-15 + sqrt(5)) + (1 + sqrt(5))^n*(15 + sqrt(5)))/sqrt(5). - _Herbert Kociemba_, Dec 18 2011
%F A022098 a(n) = 7*A000045(n) + A000045(n+1). - _R. J. Mathar_, Aug 10 2012
%F A022098 a(n) = 9*A000045(n) - A000045(n-2). - _Bruno Berselli_, Feb 20 2017
%F A022098 a(n) = Lucas(n+5) + Lucas(n-1) - 3*Lucas(n+2). - _Bruno Berselli_, Dec 29 2016, corrected by _Greg Dresden_, Mar 02 2022
%F A022098 a(n) = Lucas(n+3) - Lucas(n-2). - _Greg Dresden_ and Michael Nyc, Mar 02 2022
%t A022098 LinearRecurrence[{1, 1}, {1, 8}, 40] (* _Alonso del Arte_, Sep 26 2013 *)
%t A022098 CoefficientList[Series[(1 + 7 x)/(1 - x - x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Sep 27 2013 *)
%t A022098 Table[LucasL[n + 3] + LucasL[n - 3] - 3 LucasL[n], {n, 2, 40}] (* _Bruno Berselli_, Dec 30 2016 *)
%o A022098 (Magma) a0:=1; a1:=8; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // _Bruno Berselli_, Feb 12 2013
%o A022098 (PARI) a(n)=([0,1; 1,1]^n*[1;8])[1,1] \\ _Charles R Greathouse IV_, Oct 07 2016
%Y A022098 Cf. A000032, A000045.
%Y A022098 a(n) = A109754(7, n+1) = A101220(7, 0, n+1).
%K A022098 nonn,easy
%O A022098 0,2
%A A022098 _N. J. A. Sloane_

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