A100059 First differences of A052911.
1, 5, 14, 45, 139, 434, 1351, 4209, 13110, 40837, 127203, 396226, 1234207, 3844441, 11975078, 37301261, 116189979, 361921042, 1127350583, 3511592833, 10938286998, 34071752661, 106130359315, 330586256610
Offset: 1
Keywords
Examples
a(5) = 139 = rightmost term in M^5 * [1 1 1] which is [434 205 139]. 434 = a(6), while 205 = A052911(5). a(6) = 434 = 3*a(5) + a(4) - 2*a(3) = 3*139 + 45 - 2*14.
References
- Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
Links
- Index entries for linear recurrences with constant coefficients, signature (3,1,-2).
- Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 27.
Programs
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Mathematica
LinearRecurrence[{3,1,-2},{1,5,14},30] (* Harvey P. Dale, Apr 21 2016 *)
Formula
G.f.: (2*x^2-2*x-1)*x / (-2*x^3+x^2+3*x-1).
Recurrence: a(n) = 3*a(n-1) + a(n-2) - 2*a(n-3).
a(n) = rightmost term in M^5 * [1 1 1], where M = the 3 X 3 upper triangular matrix [2 1 2 / 1 1 0 / 1 0 0].
INVERT transform of (1, 4, 5, 6, 7, 8, 9, ...) with offset 0.
Extensions
Edited by Ralf Stephan, Nov 02 2004
Comments