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a(n) = 110*a(n-1) - a(n-2) for n >= 2, a(0)=1, a(1)=111.
16

%I #34 Feb 09 2018 04:29:10

%S 1,111,12209,1342879,147704481,16246150031,1786928798929,

%T 196545921732159,21618264461738561,2377812544869509551,

%U 261537761671184312049,28766775971285404815839,3164083819079723345430241,348020453322798282592510671,38279085781688731361830743569,4210351415532437651518789281919

%N a(n) = 110*a(n-1) - a(n-2) for n >= 2, a(0)=1, a(1)=111.

%C Sequence {s_k(110)} of A269254.

%C The sequence contains no primes; see A269254 for a proof by _N. J. A. Sloane_.

%C For detailed theory, see [Hone]. - _L. Edson Jeffery_, Feb 09 2018

%H Colin Barker, <a href="/A298677/b298677.txt">Table of n, a(n) for n = 0..450</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (110,-1).

%H Andrew N. W. Hone, et al., <a href="https://arxiv.org/abs/1802.01793">On a family of sequences related to Chebyshev polynomials</a>, arXiv:1802.01793 [math.NT], 2018.

%F G.f.: (1 + x)/(1 - 110*x + x^2).

%F a(n) = (1/18)*((55+12*sqrt(21))^(-n)*(9-2*sqrt(21) + (9+2*sqrt(21))*(55+12*sqrt(21))^(2*n))). - _Colin Barker_, Jan 25 2018

%t s[0, n_] := 1; s[1, n_] := n + 1; s[k_, n_] := n*s[k - 1, n] - s[k - 2, n]; Table[s[k, 110], {k, 0, 15}]

%t LinearRecurrence[{110, -1}, {1, 111}, 15]

%t CoefficientList[Series[(1 + x)/(1 - 110*x + x^2), {x, 0, 14}], x]

%o (PARI) Vec((1 + x)/(1 - 110*x + x^2) + O(x^20)) \\ _Colin Barker_, Jan 25 2018

%Y Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A269253, A269254, A294099, A298675, A298677, A298878, A299045, A299071.

%K nonn,easy

%O 0,2

%A _L. Edson Jeffery_, _Bob Selcoe_ and _Andrew Hone_, Jan 24 2018