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Search: id:A059259
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| A059259 |
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Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-x-x*y-y^2) = 1/((1+y)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ... |
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+0
6
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| 1, 1, 0, 1, 1, 1, 1, 2, 2, 0, 1, 3, 4, 2, 1, 1, 4, 7, 6, 3, 0, 1, 5, 11, 13, 9, 3, 1, 1, 6, 16, 24, 22, 12, 4, 0, 1, 7, 22, 40, 46, 34, 16, 4, 1, 1, 8, 29, 62, 86, 80, 50, 20, 5, 0, 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1, 1, 10, 46, 128, 239, 314, 296, 200
(list; table; graph; listen)
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OFFSET
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0,8
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COMMENT
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This sequence provides the general solution to the recurrence a(n)=a(n-1)+k(k+1)a(n-2), a(0)=a(1)=1. The solution is (1,1,k^2+k+1,2k^2+2k+1, .....) whose coefficients can be read from the rows of the triangle. The row sums of the triangle are given by the case k=1. These are the Jacobsthal numbers, A001045. Viewed as a square array, its first row is (1,0,1,0,1,...) with E.g.f. cosh(x), G.f. 1/(1-x^2) and subsequent rows are successive partial sums given by 1/((1-x)^n)(1-x^2)). - Paul Barry (pbarry(AT)wit.ie), Mar 17 2003
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FORMULA
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G.f.: 1/(1-x-x*y-y^2)
As a square array read by antidiagonals, this is T(n, k)=sum{i=0..n, (-1)^(n-i)C(i+k, k)} - Paul Barry (pbarry(AT)wit.ie), Jul 01 2003
T(2*n,n)=A026641(n) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Mar 08 2007
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MAPLE
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read transforms; 1/(1-x-x*y-y^2); SERIES2(%, x, y, 12); SERIES2TOLIST(%, x, y, 12);
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CROSSREFS
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See A059260 for an explicit formula.
Diagonals of this triangle are given by A006498
Essentially the same as A035317 and A080242.
Adjacent sequences: A059256 A059257 A059258 this_sequence A059260 A059261 A059262
Sequence in context: A000209 A104245 A109754 this_sequence A124394 A086460 A136431
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KEYWORD
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nonn,tabl
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AUTHOR
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njas, Jan 23 2001
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