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A282745 Irregular triangle read by rows, giving coefficients arising when solving g(n) = f(n)+ f(n-1) + f(n-2) + f(n-3) + f(n-4) for f(n). 2

%I #8 Mar 04 2017 20:09:33

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

%T 11,10,6,5,1,0,12,11,7,6,2,1,13,12,8,7,3,2,14,13,9,8,4,3,15,14,10,9,5,

%U 4,0,16,15,11,10,6,5,1,0,17,16,12,11,7,6,2,1,18,17,13,12,8,7,3,2,19,18,14,13,9,8,4,3,20,19,15,14,10,9,5,4,0

%N Irregular triangle read by rows, giving coefficients arising when solving g(n) = f(n)+ f(n-1) + f(n-2) + f(n-3) + f(n-4) for f(n).

%H H. W. Gould, <a href="http://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. Note that there is a typo in line 2 of Table 4.

%e Triangle begins:

%e 0,

%e 1,0,

%e 2,1,

%e 3,2,

%e 4,3,

%e 5,4,0,

%e 6,5,1,0,

%e 7,6,2,1,

%e 8,7,3,2,

%e 9,8,4,3,

%e 10,9,5,4,0,

%e 11,10,6,5,1,0,

%e 12,11,7,6,2,1,

%e 13,12,8,7,3,2,

%e 14,13,9,8,4,3,

%e 15,14,10,9,5,4,0,

%e 16,15,11,10,6,5,1,0,

%e 17,16,12,11,7,6,2,1,

%e 18,17,13,12,8,7,3,2,

%e ...

%Y Cf. A282743, A282744.

%K nonn,tabf

%O 0,4

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

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Last modified April 25 01:35 EDT 2024. Contains 371964 sequences. (Running on oeis4.)