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A203976 a(n) = 3*a(n-2) - a(n-4), a(0)=0, a(1)=1, a(2)=5, a(3)=4.

%I A203976 #67 Mar 19 2023 14:08:22
%S A203976 0,1,5,4,15,11,40,29,105,76,275,199,720,521,1885,1364,4935,3571,12920,
%T A203976 9349,33825,24476,88555,64079,231840,167761,606965,439204,1589055,
%U A203976 1149851,4160200,3010349,10891545,7881196,28514435,20633239,74651760,54018521,195440845
%N A203976 a(n) = 3*a(n-2) - a(n-4), a(0)=0, a(1)=1, a(2)=5, a(3)=4.
%C A203976 a(n+1) = p(n+2) where p(x) is the unique degree-n polynomial such that p(k) = Lucas(k) for k = 1, ..., n+1.
%C A203976 This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m).
%C A203976 a(n) = row sums of triangle A226377(n), based on differences among Lucas Numbers.  - _Richard R. Forberg_, Aug 01 2013
%C A203976 A strong divisibility sequence, i.e., gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. The sequence of convergents of the 2-periodic continued fraction [0; 1, -5, 1, -5, ...] = 1/(1 - 1/(5 - 1/(1 - 1/(5 - ...)))) = 1/2*(5 - sqrt(5)) begins [0/1, 1/1, 5/4, 4/3, 15/11, 11/8, 40/29,...]. The present sequence is the sequence of numerators; the sequence of denominators [1, 1, 4, 3, 11, 8, 29,...] is A005013. - _Peter Bala_, May 19 2014
%C A203976 It appears that the first homology group of the branched n-th cyclic covering of the group of figure-eight knot is the direct sum of cyclic groups of orders a(n) and A005013(n), so the order of that group is the product of these numbers, i. e. A004146(n); see the table on p. 156 of the paper by Fox. - _Andrey Zabolotskiy_, Mar 16 2023
%H A203976 Reinhard Zumkeller, <a href="/A203976/b203976.txt">Table of n, a(n) for n = 0..1000</a>
%H A203976 R. H. Fox, <a href="https://www.maths.ed.ac.uk/~v1ranick/papers/quickfox.pdf">A quick trip through knot theory</a>, pages 120-167 in: Topology of 3-manifolds and related topics (Proceedings of The University of Georgia Institute, 1961), Prentice-Hall, 1962.
%H A203976 <a href="/index/Di#divseq">Index to divisibility sequences</a>
%H A203976 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-1).
%F A203976 a(1) = 1, a(2) = 5, a(3) = 4, a(n) * a(n-3) = a(n-1) * a(n-2) - 5. a(-n) = -a(n).
%F A203976 G.f.: x * (1 + 5*x + x^2) /  ( (x^2+x-1)*(x^2-x-1) ).
%F A203976 a(2*n) = 5 * A000045(2*n) (Fibonacci). a(2*n+1) = A000032(2*n+1) (Lucas).
%F A203976 a(A004277(n)) = A054888(n+1). - _Reinhard Zumkeller_, Jan 11 2012
%F A203976 a(n) = A000032(n+1) - A061084(n). - _R. J. Mathar_, Jun 23 2013
%F A203976 a(2n) = a(2n-1) + a(2n+1), for n>0. - _Richard R. Forberg_, Aug 01 2013
%F A203976 a(n) = (2^(-1-n)*((-5-sqrt(5)+(-1)^n*(-5+sqrt(5)))*((-1+sqrt(5))^n-(1+sqrt(5))^n)))/sqrt(5). - _Colin Barker_, Mar 28 2016
%F A203976 E.g.f.: exp(-phi*x)*(exp(x) - 1)*(phi*exp(sqrt(5)*x) - 1/phi), where phi = (1 + sqrt(5))/2. - _G. C. Greubel_, Mar 28 2016
%e A203976 a(3) = 4 since p(x) = (-x^2 + 7*x - 4) / 2 interpolates p(1) = 1, p(2) = 3, p(3) = 4, and p(4) = 4.
%t A203976 LinearRecurrence[{0,3,0,-1},{0,1,5,4},40] (* _Harvey P. Dale_, Apr 06 2013 *)
%o A203976 (PARI) {a(n) = if( n%2, fibonacci(n+1) + fibonacci(n-1), 5 * fibonacci(n))}
%o A203976 (PARI) {a(n) = if( n<0, -a(-n), polcoeff( x * (1 + 5*x + x^2) / (1 - 3*x^2 + x^4) + x * O(x^n), n))}
%o A203976 (PARI) {a(n) = if( n<0, -a(-n), subst( polinterpolate( vector( n, k, fibonacci(k-1) + fibonacci(k+1) )), x, n + 1))}
%o A203976 (Haskell)
%o A203976 a203976 n = a203976_list !! n
%o A203976 a203976_list = 0 : 1 : 5 : 4 : zipWith (-)
%o A203976    (map (* 3) $ drop 2 a203976_list) a203976_list
%o A203976 -- _Reinhard Zumkeller_, Jan 10 2012
%o A203976 (Magma) I:=[0,1,5,4]; [n le 4 select I[n] else 3*Self(n-2)-Self(n-4): n in [1..40]]; // _Vincenzo Librandi_, Mar 29 2016
%Y A203976 Cf. A000032, A000045, A201157 (bisection), A002878 (bisection). A005013.
%K A203976 nonn,easy
%O A203976 0,3
%A A203976 _Michael Somos_, Jan 08 2012