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a(n+2) = 9*a(n+1) - a(n) + 1, with a(1)=1, a(2)=10.
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%I #26 Dec 31 2023 10:24:42

%S 1,10,90,801,7120,63280,562401,4998330,44422570,394804801,3508820640,

%T 31184580960,277152408001,2463187091050,21891531411450,

%U 194560595612001,1729153829096560,15367823866257040,136581260967216801

%N a(n+2) = 9*a(n+1) - a(n) + 1, with a(1)=1, a(2)=10.

%C Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 9/11.

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (10,-10,1).

%F G.f.: x/(1-10*x+10*x^2-x^3) = x/((1-x)*(1-9*x+x^2)).

%F a(n) = 10*a(n-1) - 10*a(n-2) + a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=10.

%F a(n) = (S(n,9) - S(n-1,9) - 1)/7, n >= 1.

%F a(n+1) = Sum_{k=0..n} S(n,9), n >= 0, with S(n,9) = U(n,9/2) = A018913(n+1). (Partial sums of Chebyshev sequence A018913.)

%t a[1] = 1; a[2] = 10; a[n_] := a[n] = 9a[n - 1] - a[n - 2] + 1; Table[ a[n], {n, 20}] (* _Robert G. Wilson v_, Apr 05 2004 *)

%t LinearRecurrence[{10,-10,1},{1,10,90},20] (* _Harvey P. Dale_, May 21 2023 *)

%Y Cf. A092521. Also cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

%Y Cf. A018913 (first differences).

%K nonn,easy

%O 1,2

%A M. N. Deshpande (dpratap_ngp(AT)sancharnet.in), Apr 04 2004

%E More terms from _Robert G. Wilson v_, Apr 05 2004

%E Chebyshev comments from _Wolfdieter Lang_, Aug 31 2004