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%I #76 Oct 03 2024 08:32:33
%S 1,7,8,15,23,38,61,99,160,259,419,678,1097,1775,2872,4647,7519,12166,
%T 19685,31851,51536,83387,134923,218310,353233,571543,924776,1496319,
%U 2421095,3917414,6338509,10255923,16594432,26850355,43444787,70295142,113739929
%N Fibonacci sequence beginning 1, 7.
%C a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(7;n-1-k,k) with n>=1, a(-1)=6. These are the SW-NE diagonals in P(7;n,k), the (7,1) Pascal triangle A093564. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.
%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 For n >= 1, a(n) is the number of edge covers of the tadpole graph T_{4,n-1} with T_{4,0} interpreted as just the cycle graph C_4. Example: If n=2, we have C_4 and path P_1 joined by a bridge. This is the cycle with a pendant and has 7 edge covers. - _Feryal Alayont_, Sep 22 2024
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/TadpoleGraph.html">Tadpole Graph</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) for n>=2, a(0)=1, a(1)=7, a(-1):=6.
%F G.f.: (1+6*x)/(1-x-x^2).
%F a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(-13 + sqrt(5)) + (1 + sqrt(5))^n*(13 + sqrt(5))))/sqrt(5). - _Herbert Kociemba_
%F a(n) = 6*A000045(n) + A000045(n+1). - _R. J. Mathar_, Aug 10 2012
%F a(n) = 8*A000045(n) - A000045(n-2). - _Bruno Berselli_, Feb 20 2017
%F From _Aamen Muharram_, Aug 05 2022: (Start)
%F a(n) = F(n-4) + F(n-1) + F(n+4),
%F a(n) = F(n) + F(n+4) - F(n-3),
%F where F(n) = A000045(n) is the Fibonacci numbers. (End)
%t First /@ NestList[{Last@ #, Total@ #} &, {1, 7}, 36] (* or *)
%t CoefficientList[Series[(1 + 6 x)/(1 - x - x^2), {x, 0, 36}], x] (* _Michael De Vlieger_, Feb 20 2017 *)
%t LinearRecurrence[{1,1},{1,7},40] (* _Harvey P. Dale_, May 17 2018 *)
%o (Magma) a0:=1; a1:=7; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // _Bruno Berselli_, Feb 12 2013
%o (PARI) a(n)=([0,1; 1,1]^n*[1;7])[1,1] \\ _Charles R Greathouse IV_, Oct 03 2016
%Y a(n) = A101220(6, 0, n+1) = A109754(6, n+1) = A118654(3, n).
%Y Cf. A000045, A131778.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_