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A095660
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Pascal (1,3) triangle.
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15
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3, 1, 3, 1, 4, 3, 1, 5, 7, 3, 1, 6, 12, 10, 3, 1, 7, 18, 22, 13, 3, 1, 8, 25, 40, 35, 16, 3, 1, 9, 33, 65, 75, 51, 19, 3, 1, 10, 42, 98, 140, 126, 70, 22, 3, 1, 11, 52, 140, 238, 266, 196, 92, 25, 3, 1, 12, 63, 192, 378, 504, 462, 288, 117, 28, 3, 1, 13, 75, 255, 570, 882, 966, 750
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| This is the third member, q=3, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1), A029635 (q=2) (but with a(0,0)=2, not 1).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column nr. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x):=sum(a(n,m)*x^m,m=0..n) is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(3-2*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(3-2*z)/(1-(1+x)*z).
The SW-NE diagonals give sum(a(n-1-k,k),k=0..ceiling((n-1)/2)) = A000285(n-2), n>=2, with n=1 value 3. Observation by Paul Barry (pbarry(AT)wit.ie, Apr 29 2004. Proof via recursion relations and comparison of inputs.
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LINKS
| W. Lang, First 10 rows.
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FORMULA
| Recursion: a(n, m)=0 if m>n, a(0, 0)= 3; a(n, 0)=1 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (3-2*x)/(1-x)^(m+1), m>=0.
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EXAMPLE
| [3];[1,3];[1,4,3];[1,5,7,3];[1,6,12,10,3];...
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CROSSREFS
| Row sums: A000079(n+1), n>=1, 3 if n=0. Alternating row sums are [3, -2, followed by 0's].
Column sequences (without leading zeros) give for m=1..9 with n>=0: A000027(n+3), A055998(n+1), A006503(n+1), A095661, A000574, A095662-5.
Sequence in context: A023892 A085417 A201662 * A035648 A053575 A103790
Adjacent sequences: A095657 A095658 A095659 * A095661 A095662 A095663
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KEYWORD
| nonn,easy,tabl
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AUTHOR
| Wolfdieter Lang (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de), May 21 2004
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