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Recurrence a(n)*a(n-2) = a(n-1)*(a(n-1)+3) with a(0) = 1, a(1) = 4.
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%I #37 Sep 08 2022 08:45:55

%S 1,4,28,217,1705,13420,105652,831793,6548689,51557716,405913036,

%T 3195746569,25160059513,198084729532,1559517776740,12278057484385,

%U 96664942098337,761041479302308,5991666892320124,47172293659258681,371386682381749321

%N Recurrence a(n)*a(n-2) = a(n-1)*(a(n-1)+3) with a(0) = 1, a(1) = 4.

%C The non-linear recurrence equation a(n)*a(n-2) = a(n-1)*(a(n-1)+r) with initial conditions a(0) = 1, a(1) = 1+r has the solution a(n) = 1/2 + 1/2*sum_{k=0..n} (2*r)^k*binomial(n+k,2*k) = 1/2 + b(n,2*r)/2, where b(n,x) are the Morgan-Voyce polynomials of A085478. The recurrence produces sequences A101265 (r = 1), A011900 (r = 2) and A054318 (r = 4), as well as signed versions of A133872 (r = -1), A109613(r = -2), A146983 (r = -3) and A084159(r = -4).

%C Also the indices of centered pentagonal numbers (A005891) which are also centered triangular numbers (A005448). - _Colin Barker_, Jan 01 2015

%C Also positive integers y in the solutions to 3*x^2 - 5*y^2 - 3*x + 5*y = 0. - _Colin Barker_, Jan 01 2015

%H Seiichi Manyama, <a href="/A182432/b182432.txt">Table of n, a(n) for n = 0..1116</a> (first 201 terms from Vincenzo Librandi)

%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2016volume16/FG2016volume16.pdf#page=423">Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences</a>, Forum Geometricorum, Volume 16 (2016) 419-427.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Morgan-VoycePolynomials.html">Morgan-Voyce Polynomials</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-9,1).

%F a(n) = 1/2 + 1/2*sum_{k = 0..n} 6^k*binomial(n+k,2*k).

%F a(n) = R(n,3) where R(n,x) denotes the row polynomials of A211955.

%F a(n) = 1/u*T(n,u)*T(n+1,u) with u = sqrt(5/2) and T(n,x) the Chebyshev polynomial of the first kind.

%F Recurrence equation: a(n) = 8*a(n-1) - a(n-2)-3 with a(0) = 1, a(1) = 4.

%F O.g.f.: (1 - 5*x + x^2)/((1 - x)*(1 - 8*x + x^2)) = 1 + 4*x + 28*x^2 + ....

%F Sum_{n>=0} 1/a(n) = sqrt(5/3); 5 - 3*(sum_{n=0..2*n} 1/a(k))^2 = 2/A070997(n)^2.

%F a(0)=1, a(1)=4, a(2)=28, a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3). - _Harvey P. Dale_, May 14 2012

%t RecurrenceTable[{a[0]==1,a[1]==4,a[n]==(a[-1+n] (3+a[-1+n]))/a [-2+n]}, a[n],{n,30}] (* or *) LinearRecurrence[{9,-9,1},{1,4,28},30] (* _Harvey P. Dale_, May 14 2012 *)

%o (Magma) I:=[1, 4, 28]; [n le 3 select I[n] else 9*Self(n-1)-9*Self(n-2)+Self(n-3): n in [1..25]]; // _Vincenzo Librandi_, May 18 2012

%o (PARI) Vec((1-5*x+x^2)/((1-x)*(1-8*x+x^2)) + O(x^100)) \\ _Colin Barker_, Jan 01 2015

%Y Cf. A011900, A054318, A070997, A101265, A109613, A133872, A146983, A211955.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Apr 30 2012