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%I #67 Oct 21 2022 22:04:54
%S 1,1,2,6,21,77,286,1066,3977,14841,55386,206702,771421,2878981,
%T 10744502,40099026,149651601,558507377,2084377906,7779004246,
%U 29031639077,108347552061,404358569166,1509086724602,5631988329241,21018866592361,78443478040202,292755045568446
%N a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).
%C Consider the matrix M=[1,1,0; 1,3,1; 0,1,1]; characteristic polynomial of M is x^3 - 5*x^2 + 5*x - 1. Use (M^n)[1,1] to define the recursion a(0) = 1, a(1) = 1, a(2) = 2, for n>2 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).
%C a(n+1)/a(n) converges to 2 + sqrt(3) as n goes to infinity, the largest root of the characteristic polynomial. a(n) = A061278(n) + 1; (M^n)[1,2] = A001353(n); (M^n)[1,3] = A061278(n-1) for n>0; all with the same recursive properties.
%C Consecutive terms of this sequence and consecutive terms of A032908 provide all positive integer pairs for which K=(a+1)/b+(b+1)/a is integer. For this sequence K=4. - _Andrey Vyshnevyy_, Sep 18 2015
%C The two-page Reid Barton article was sent to me around 2002, but for some reason it was not included in the OEIS at that time. I recently rediscovered it in my files. - _N. J. A. Sloane_, Sep 08 2018
%H Seiichi Manyama, <a href="/A101879/b101879.txt">Table of n, a(n) for n = 0..1750</a>
%H Reid Barton, <a href="https://faculty.uml.edu/jpropp/reach/Barton/b-interp.html">A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...</a>
%H Reid Barton, <a href="/A101879/a101879.pdf">A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...</a>, [Annotated scanned copy]
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-5,1).
%F a(n) = A101265(n), n>0. - _R. J. Mathar_, Aug 30 2008
%F a(n) = A079935(n+1) - A001571(n). - _Gerry Martens_, Jun 05 2015
%F a(0) = a(1) = 1, for n>1 a(n) = (a(n-1) + a(n-1)^2) / a(n-2). - _Seiichi Manyama_, Aug 11 2016
%F From _Ilya Gutkovskiy_, Aug 11 2016: (Start)
%F G.f.: (1 - 4*x + 2*x^2)/((1 - x)*(1 - 4*x + x^2)).
%F a(n) = (6+(3-sqrt(3))*(2+sqrt(3))^n + (2-sqrt(3))^n*(3+sqrt(3)))/12. (End)
%F a(n) = 4*a(n-1) - a(n-2) - 1. - _Seiichi Manyama_, Aug 26 2016
%F From _Seiichi Manyama_, Sep 03 2016: (Start)
%F a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).
%F a(n) = A005246(n)*A005246(n+1). (End)
%F From _Michael Somos_, Jul 09 2017: (Start)
%F 0 = +a(n)*(+1 +a(n) -4*a(n+1)) +a(n+1)*(+1 +a(n+1)) for all n in Z.
%F a(n) = a(1 - n) = (1 + A001835(n)) / 2 for all n in Z. (End)
%t LinearRecurrence[{5, -5, 1}, {1, 1, 2}, 30] (* _Vincenzo Librandi_, Sep 18 2015 *)
%t CoefficientList[Series[(1 - 4 x + 2 x^2)/((1 - x) (1 - 4 x + x^2)), {x, 0, 27}], x] (* _Michael De Vlieger_, Aug 11 2016 *)
%t a[ n_] := If[ n < 1, a[1 - n], SeriesCoefficient[ (1/(1 - x) + (1 - 3 x)/(1 - 4 x + x^2)) / 2, {x, 0, n}]]; (* _Michael Somos_, Jul 09 2017 *)
%o (PARI) M=[1,1,0; 1,3,1; 0,1,1]; for(i=0,40,print1((M^i)[1,1],","))
%o (Magma) I:=[1,1,2]; [n le 3 select I[n] else 5*Self(n-1)-5*Self(n-2)+Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Sep 18 2015
%o (PARI) {a(n) = if( n<1, a(1-n), polcoeff( (1/(1 - x) + (1 - 3*x)/(1 - 4*x + x^2)) / 2 + x * O(x^n), n))}; /* _Michael Somos_, Jul 09 2017 */
%Y Cf. A061278, A001353, A001835, A005246, A276122, A276271.
%K nonn,easy
%O 0,3
%A Lambert Klasen (lambert.klasen(AT)gmx.net) and _Gary W. Adamson_, Jan 28 2005
%E a(26)-a(27) from _Vincenzo Librandi_, Sep 18 2015
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