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

%I #14 Jan 05 2025 19:51:41

%S 0,1,0,2,1,3,2,4,3,0,5,4,1,0,6,5,2,1,7,6,3,2,8,7,4,3,0,9,8,5,4,1,0,10,

%T 9,6,5,2,1,11,10,7,6,3,2,12,11,8,7,4,3,0,13,12,9,8,5,4,1,0,14,13,10,9,

%U 6,5,2,1,15,14,11,10,7,6,3,2,16,15,12,11,8

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

%H Lars Blomberg, <a href="/A282744/b282744.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,

%e 4,3,0,

%e 5,4,1,0,

%e 6,5,2,1,

%e 7,6,3,2,

%e 8,7,4,3,0,

%e 9,8,5,4,1,0,

%e 10,9,6,5,2,1,

%e 11,10,7,6,3,2,

%e 12,11,8,7,4,3,0,

%e 13,12,9,8,5,4,1,0,

%e 14,13,10,9,6,5,2,1

%e 15,14,11,10,7,6,3,2

%e 16,15,12,11,8,7,4,3,0

%e ...

%Y Cf. A282743, A282745.

%K nonn,tabf

%O 0,4

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

%E More terms from _Lars Blomberg_, Mar 05 2017