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Irregular triangle read by rows, giving coefficients arising when solving g(n) = f(n)+ f(n-1) + f(n-2) for f(n).
3

%I #11 Jan 05 2025 19:51:41

%S 0,1,0,2,1,3,2,0,4,3,1,0,5,4,2,1,6,5,3,2,0,7,6,4,3,1,0,8,7,5,4,2,1,9,

%T 8,6,5,3,2,0,10,9,7,6,4,3,1,0,11,10,8,7,5,4,2,1,12,11,9,8,6,5,3,2,0,

%U 13,12,10,9,7,6,4,3,1,0,14,13,11,10,8,7,5,4

%N Irregular triangle read by rows, giving coefficients arising when solving g(n) = f(n)+ f(n-1) + f(n-2) for f(n).

%H Lars Blomberg, <a href="/A282743/b282743.txt">Table of n, a(n) for n = 0..10000</a>

%H H. W. Gould, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/44-4/quartgould04_2006.pdf">The inverse of a finite series and a third-order recurrent sequence</a>, Fibonacci Quart. 44 (2006), no. 4, 302-315.

%e Triangle begins:

%e 0,

%e 1,0,

%e 2,1,

%e 3,2,0,

%e 4,3,1,0,

%e 5,4,2,1,

%e 6,5,3,2,0,

%e 7,6,4,3,1,0,

%e 8,7,5,4,2,1,

%e 9,8,6,5,3,2,0,

%e 10,9,7,6,4,3,1,0,

%e 11,10,8,7,5,4,2,1

%e 12,11,9,8,6,5,3,2,0

%e 13,12,10,9,7,6,4,3,1,0

%e 14,13,11,10,8,7,5,4,2,1

%e ...

%Y Cf. A282744, A282745.

%K nonn,tabf

%O 0,4

%A _N. J. A. Sloane_, Mar 04 2017

%E More terms from _Lars Blomberg_, Mar 05 2017