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A140993
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Triangle G(n, k) read by rows, for 1 <= k <= n, where G(n, n) = G(n+1, 1) = 1, G(n+2, 2) = 2, G(n+3, m) = G(n+1, m-1) + G(n+1, m-2) + G(n+2, m-1) for n >= 1 and m = 3..(n+2).
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24
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1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 5, 7, 1, 1, 2, 5, 11, 12, 1, 1, 2, 5, 12, 23, 20, 1, 1, 2, 5, 12, 28, 46, 33, 1, 1, 2, 5, 12, 29, 63, 89, 54, 1, 1, 2, 5, 12, 29, 69, 137, 168, 88, 1, 1, 2, 5, 12, 29, 70, 161, 289, 311, 143, 1, 1, 2, 5, 12, 29, 70, 168, 367, 594, 567, 232, 1, 1, 2, 5, 12, 29, 70, 169, 399, 817, 1194, 1021, 376, 1
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OFFSET
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1,5
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COMMENTS
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Let b(m) = lim_{n -> infinity} G(n, m) for each m >= 1. Then b(1) = 1, b(2) = 2, and b(m) = 2*b(m-1) + b(m-2) for m >= 3. (The existence of the limit can be proved by induction on m.) This means b(m) = A000129(m) for m >= 1 (known as the Pell numbers).
If we want to get the second main diagonal, we let c(n) = G(n+1, n) for n >= 1. Then c(n+2) = G(n+3, n+2) = G(n+1, n+1) + G(n+1, n) + G(n+2, n+1) = 1 + c(n) + c(n+1) with c(1) = G(2, 1) = 1 and c(2) = G(3, 2) = 2, which implies that c(n) = A000071(n+2) = Fibonacci(n+2) - 1 for n >= 1.
This array is the mirror image of A140998 (except for a shifting of the indices by 1). Thus, G(n, k) = A140998(n - 1, n - k) for 1 <= k <= n. This array has index of obliqueness e = 1, while array A140998 has index of obliqueness e = 0. Both arrays have the same index of asymmetry (s = 1). (End)
One of the two rectangular versions, say (RA(n,k): n,k >= 0), of this triangular array (G(n,k): 1 <= k <= n) is given by RA(n,k) = G(n+k-1,k) for n,k >= 1. Conversely, G(n,k) = RA(n-k+1, k) for 1 <= k <= n. (This assumes that the triangle G(n,k) is read from the array RA(n,k) by ascending antidiagonals.)
Note that [o.g.f of RA](x,y) = x*[o.g.f. of G](x, y/x) and [o.g.f of G](x,y) = x^(-1)*[o.g.f of RA](x,x*y).
The other rectangular version, say (RD(n,k): n,k >= 0), of this triangular array (G(n,k): 1 <= k <= n) is given by RD(n,k) = RA(k,n) = G(n+k-1,n) for n,k >= 1. Conversely, G(n,k) = RD(k,n-k+1) for 1 <= k <= n. (This assumes that the triangle G(n,k) is read from the array RD(n,k) by descending antidiagonals.)
Note that [o.g.f of RD](x,y) = y*[o.g.f. of G](y,x/y) and [o.g.f of G](x,y) = x^(-1)*[o.g.f of RD](x*y, x). (End)
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LINKS
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FORMULA
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G(n, k) = A140998(n - 1, n - k) for 1 <= k <= n.
Bivariate o.g.f.: Sum_{n >= 1, k >= 1} G(n, k)*x^n*y^k = x*y*(1 - x*y -x^2*y^2 + x^3*y^2)/((1 - x) * (1 - x*y) * (1 - x*y - x^2*y - x^2*y^2)). (Here, we let G(n, k) = 0 when 1 <= n < k.)
To get the row sums, we let y = 1 in the above bivariate g.f. and simplify. We get x/(1 - 2*x), which is the g.f. of sequence (A000079(n-1): n >= 1) = (2^(n-1): n >= 1). (End)
We give formulas about the rectangular array RA(n,k).
Initial conditions: RA(1,n) = RA(n+1,1) = 1 and RA(n+1,2) = 2 for n >= 1.
Recurrence: RA(n,k) = RA(n-1,k-1) + RA(n,k-2) + RA(n,k-1) for n >= 2 and k >= 3.
The main diagonal of the array is RA(n,n) = A000129(n) (Pell numbers).
Bivariate o.g.f: Sum_{n,k >= 1} RA(n,m)*x^n*y^k = x*y*(x*y^2 - y^2 - y + 1)/((1 - x)*(1 - y)*(-x*y - y^2 - y + 1)).
To obtain formulas about the other rectangular array, RD(n,k), we use the equations RD(n,k) = RA(k,n) for n,k >= 1 and [o.g.f. of RD](x,y) = [o.g.f. of RA](y,x). (End)
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EXAMPLE
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Triangle G(n,k) (with rows for n >= 1 and columns for 1 <= k <= n) begins:
1
1 1
1 2 1
1 2 4 1
1 2 5 7 1
1 2 5 11 12 1
1 2 5 12 23 20 1
1 2 5 12 28 46 33 1
1 2 5 12 29 63 89 54 1
1 2 5 12 29 69 137 168 88 1
1 2 5 12 29 70 161 289 311 143 1
1 2 5 12 29 70 168 367 594 567 232 1
1 2 5 12 29 70 169 399 817 1194 1021 376 1
1 2 5 12 29 70 169 407 934 1778 2355 1820 609 1
...
Rectangular array RA(n,k) (with rows for n >= 1 and columns for k >= 1) begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 2, 4, 7, 12, 20, 33, 54, 88, 143, ...
1, 2, 5, 11, 23, 46, 89, 168, 311, 567, ...
1, 2, 5, 12, 28, 63, 137, 289, 594, 1194, ...
1, 2, 5, 12, 29, 69, 161, 367, 817, 1778, ...
1, 2, 5, 12, 29, 70, 168, 399, 934, 2150, ...
1, 2, 5, 12, 29, 70, 169, 407, 975, 2316, ...
1, 2, 5, 12, 29, 70, 169, 408, 984, 2367, ...
1, 2, 5, 12, 29, 70, 169, 408, 985, 2377, ...
1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, ...
...
Reading the array RA(n,k) by ascending antidiagonals, we get triangle G(n,k) above. (End)
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MAPLE
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A140993 := proc(n, k) if k = n then 1; elif k = 1 then 1; elif k = 2 then 2; else procname(n-2, k-1)+procname(n-2, k-2)+procname(n-1, k-1) ; end if; end proc: seq(seq(A140993(n, k), k=1..n), n=1..15) ; # R. J. Mathar, Apr 28 2010
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MATHEMATICA
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t[n_, k_] := If[k == n, 1, If[k == 1, 1, If[k == 2, 2, t[n - 2, k - 1] + t[n - 2, k - 2] + t[n - 1, k - 1]]]]; Flatten[Table[ t[n, k], {n, 13}, {k, n}]] (* Robert G. Wilson v, Dec 22 2011 *)
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CROSSREFS
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Cf. A000071, A000079, A000129, A007318, A140994, A140995, A140996, A140997, A140998, A141020, A141021.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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