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A092420
a(n+2) = 9*a(n+1) - a(n) + 1, with a(1)=1, a(2)=10.
9
1, 10, 90, 801, 7120, 63280, 562401, 4998330, 44422570, 394804801, 3508820640, 31184580960, 277152408001, 2463187091050, 21891531411450, 194560595612001, 1729153829096560, 15367823866257040, 136581260967216801
OFFSET
1,2
COMMENTS
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.
FORMULA
G.f.: x/(1-10*x+10*x^2-x^3) = x/((1-x)*(1-9*x+x^2)).
a(n) = 10*a(n-1) - 10*a(n-2) + a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=10.
a(n) = (S(n,9) - S(n-1,9) - 1)/7, n >= 1.
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.)
MATHEMATICA
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 *)
LinearRecurrence[{10, -10, 1}, {1, 10, 90}, 20] (* Harvey P. Dale, May 21 2023 *)
CROSSREFS
Cf. A092521. Also cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).
Cf. A018913 (first differences).
Sequence in context: A276020 A164552 A057086 * A010579 A010576 A162983
KEYWORD
nonn,easy
AUTHOR
M. N. Deshpande (dpratap_ngp(AT)sancharnet.in), Apr 04 2004
EXTENSIONS
More terms from Robert G. Wilson v, Apr 05 2004
Chebyshev comments from Wolfdieter Lang, Aug 31 2004
STATUS
approved