proposed

approved

editing

proposed

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 *)

approved

editing

_Paul Curtz (bpcrtz(AT)free.fr), _, May 29 2008

Edited by _R. J. Mathar (mathar(AT)strw.leidenuniv.nl), _, Jul 10 2008

reviewed

approved

proposed

reviewed

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

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

approved

proposed

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.

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.

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.

nonn,tabf,new

Sequences identical to their p-th differences and natural Catalan's triangle A009766.From two methods.The first considers every principal sequence from A131577 and,after, A024495, A000749, A049016 with 4 leading 0's.First 2n+1 terms of a(n+1)-2a(n) are 1 0, 1, 1 0, 0, 1, 2, 2 or on line triangle 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; The second considers principal sequences for degenerate case.See 0, A009545, A138112, today 0, 0, 0, 0, 1, 4, 9 and 0, 0, 0, 0, 0, 1, 5, 14.

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

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.

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.

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(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.

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.

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;

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

nonn,uned,newtabf

Edited by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 10 2008