

A140344


Catalan triangle A009766 prepended by n zeros in its nth row.


4



1, 0, 1, 1, 0, 0, 1, 2, 2, 0, 0, 0, 1, 3, 5, 5, 0, 0, 0, 0, 1, 4, 9, 14, 14, 0, 0, 0, 0, 0, 1, 5, 14, 28, 42, 42, 0, 0, 0, 0, 0, 0, 1, 6, 20, 48, 90, 132, 132, 0, 0, 0, 0, 0, 0, 0, 1, 7, 27, 75, 165, 297, 429, 429, 0, 0, 0, 0, 0, 0, 0, 0, 1, 8, 35, 110, 275, 572, 1001, 1430, 1430
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OFFSET

0,8


COMMENTS

The triangle's nth row is also related to recurrences for sequences f(n) which pth differences, p=n+2: The denominator of the generating function contains a factor 12x in these cases.
This factor may be "lifted" either by looking at auxiliary sequences f(n+1)2f(n) or by considering the corresponding "degenerate" shorter recurrences right away. In the case p=4, the recurrence is f(n)=4f(n1)6f(n2)+4f(n3) from the 4th row in A135356, the denominator in the g.f. is 14x+6x^24x^3=(12x)(12x+2x^2), which yields the degenerate recurrence f(n)=2f(n1)2f(n2) from the 2nd factor and leaves the first three coefficients of 1/(12x+2x^2)=1+2x+2x^2+.. in row 2.
A000749 is an example which follows the recurrence but not the degenerate recurrence, but still A000749(n+1)2A000749(n) = 0, 0, 1, 2, 2,.. starts with the 3 coefficients. A009545 follows both recurrences and starts with the three nonzero terms because there is only a power of x in the numerator of the g.f.
In the case p=5, the recurrence is f(n)=5f(n1)10f(n2)+10f(n3)5f(n4)+2f(n5), the denominator in the g.f. is 15x+10x^210x^3+5x^42x^5= (12x)(13x+4x^22x^3+x^4), where 1/(13x+4x^22x^3+x^4) = 1+3x+5x^2+5x^3+... and the 4 coefficients populate row 3.
A049016 obeys the main recurrence but not the degenerate recurrence f(n)=3f(n1)4f(n2)+2f(n3)f(n4), yet A049016(n+1)2A049016(n)=1, 3, 5, 5,.. starts with the 4 coefficients. A138112 obeys both recurrences and is constructed to start with the 4 coefficients themselves.
In the nomenclature of Foata and Han, this is the doubloon polynomial triangle d_{n,m}(0), up to index shifts.  R. J. Mathar, Jan 27 2011


LINKS

Table of n, a(n) for n=0..80.
D. Foata, G.N. Han, The doubloon polynomial triangle, Ramanujan J. 23 (2010), 107126


EXAMPLE

Triangle starts
1;
0,1,1;
0,0,1,2,2;
0,0,0,1,3,5,5;
0,0,0,0,1,4,9,14,14;


MATHEMATICA

Table[Join[Array[0&, n], Table[Binomial[n+k, n]*(nk+1)/(n+1), {k, 0, n}]], {n, 0, 8}] // Flatten (* JeanFrançois Alcover, Dec 16 2014 *)


CROSSREFS

Cf. A135356, A130020, A139687, A140343 (p=6), A140342 (p=7).
Sequence in context: A131018 A035395 A116856 * A279479 A177338 A024158
Adjacent sequences: A140341 A140342 A140343 * A140345 A140346 A140347


KEYWORD

nonn,tabf


AUTHOR

Paul Curtz, May 29 2008


EXTENSIONS

Edited by R. J. Mathar, Jul 10 2008


STATUS

approved



