%I #11 Dec 16 2014 05:36:21

%S 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,

%T 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,

%U 165,297,429,429,0,0,0,0,0,0,0,0,1,8,35,110,275,572,1001,1430,1430

%N Catalan triangle A009766 prepended by n zeros in its n-th row.

%C The triangle's n-th row is also related to recurrences for sequences f(n) which p-th differences, p=n+2: The denominator of the generating function contains a factor 1-2x in these cases.

%C 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(n-1)-6f(n-2)+4f(n-3) from the 4th row in A135356, the denominator in the g.f. is 1-4x+6x^2-4x^3=(1-2x)(1-2x+2x^2), which yields the degenerate recurrence f(n)=2f(n-1)-2f(n-2) from the 2nd factor and leaves the first three coefficients of 1/(1-2x+2x^2)=1+2x+2x^2+.. in row 2.

%C 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.

%C In the case p=5, the recurrence is f(n)=5f(n-1)-10f(n-2)+10f(n-3)-5f(n-4)+2f(n-5), the denominator in the g.f. is 1-5x+10x^2-10x^3+5x^4-2x^5= (1-2x)(1-3x+4x^2-2x^3+x^4), where 1/(1-3x+4x^2-2x^3+x^4) = 1+3x+5x^2+5x^3+... and the 4 coefficients populate row 3.

%C A049016 obeys the main recurrence but not the degenerate recurrence f(n)=3f(n-1)-4f(n-2)+2f(n-3)-f(n-4), 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.

%C 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

%H D. Foata, G.-N. Han, <a href="http://dx.doi.org/10.1007/s11139-009-9194-9">The doubloon polynomial triangle</a>, Ramanujan J. 23 (2010), 107-126

%e Triangle starts

%e 1;

%e 0,1,1;

%e 0,0,1,2,2;

%e 0,0,0,1,3,5,5;

%e 0,0,0,0,1,4,9,14,14;

%t Table[Join[Array[0&, n], Table[Binomial[n+k, n]*(n-k+1)/(n+1), {k, 0, n}]], {n, 0, 8}] // Flatten (* _Jean-François Alcover_, Dec 16 2014 *)

%Y Cf. A135356, A130020, A139687, A140343 (p=6), A140342 (p=7).

%K nonn,tabf

%O 0,8

%A _Paul Curtz_, May 29 2008

%E Edited by _R. J. Mathar_, Jul 10 2008