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A140344
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Catalan triangle A009766 prepended by n zeros in its n-th row.
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4
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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
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0,8
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COMMENTS
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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.
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
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LINKS
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EXAMPLE
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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;
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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