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Search: id:A099510
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| A099510 |
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Triangle, read by rows, of trinomial coefficients arranged so that there are n+1 terms in row n by setting T(n,k) equal to the coefficient of z^k in (1 + 2*z + z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2. |
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+0
6
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| 1, 1, 2, 1, 4, 1, 1, 6, 6, 4, 1, 8, 15, 20, 1, 1, 10, 28, 56, 15, 6, 1, 12, 45, 120, 70, 56, 1, 1, 14, 66, 220, 210, 252, 28, 8, 1, 16, 91, 364, 495, 792, 210, 120, 1, 1, 18, 120, 560, 1001, 2002, 924, 792, 45, 10, 1, 20, 153, 816, 1820, 4368, 3003, 3432, 495, 220, 1, 1, 22, 190
(list; table; graph; listen)
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OFFSET
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0,3
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COMMENT
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Row sums form A099511. In general if T(n,k) = coefficient of z^k in (a + b*z + c*z^2)^(n-[k/2]), then the resulting number triangle will have the o.g.f.: ((1-a*x-c*x^2*y^2) + b*x*y)/((1-a*x-c*x^2*y^2)^2 - x*(b*x*y)^2).
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FORMULA
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G.f.: (1-x+2*x*y-x^2*y^2)/((1-x)^2-2*x^2*y^2-2*x^3*y^2+x^4*y^4). T(n, k) = binomial(2*n-2*(k\2), k).
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EXAMPLE
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Rows begin:
[1],
[1,2],
[1,4,1],
[1,6,6,4],
[1,8,15,20,1],
[1,10,28,56,15,6],
[1,12,45,120,70,56,1],
[1,14,66,220,210,252,28,8],
[1,16,91,364,495,792,210,120,1],
[1,18,120,560,1001,2002,924,792,45,10],...
and can be derived from coefficients of (1+2*z+z^2)^n:
[1],
[1,2,1],
[1,4,6,4,1],
[1,6,15,20,15,6,1],
[1,8,28,56,70,56,28,8,1],
[1,10,45,120,210,252,210,120,45,10,1],...
by shifting each column k down by [k/2] rows.
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PROGRAM
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(PARI) T(n, k)=if(n<k|k<0, 0, polcoeff((1+2*z+z^2+z*O(z^k))^(n-k\2), k, z))
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CROSSREFS
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Cf. A034870, A099509, A099511, A099512.
Sequence in context: A133934 A055327 A105260 this_sequence A137633 A066633 A088443
Adjacent sequences: A099507 A099508 A099509 this_sequence A099511 A099512 A099513
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Oct 20 2004
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