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 A140993 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). 24

%I

%S 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,

%T 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,

%U 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

%N 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).

%C From _Petros Hadjicostas_, Jun 10 2019: (Start)

%C 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).

%C 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.

%C 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).

%C (End)

%H Juri-Stepan Gerasimov, <a href="/A140998/a140998.jpg">Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...</a>

%F From _Petros Hadjicostas_, Jun 10 2019: (Start)

%F G(n, k) = A140998(n - 1, n - k) for 1 <= k <= n.

%F Bivariate 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.)

%F 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).

%F (End)

%e Triangle (with rows for n >= 1 and columns for k >= 1) begins as follows:

%e 1

%e 1 1

%e 1 2 1

%e 1 2 4 1

%e 1 2 5 7 1

%e 1 2 5 11 12 1

%e 1 2 5 12 23 20 1

%e 1 2 5 12 28 46 33 1

%e 1 2 5 12 29 63 89 54 1

%e 1 2 5 12 29 69 137 168 88 1

%e 1 2 5 12 29 70 161 289 311 143 1

%e 1 2 5 12 29 70 168 367 594 567 232 1

%e 1 2 5 12 29 70 169 399 817 1194 1021 376 1

%e 1 2 5 12 29 70 169 407 934 1778 2355 1820 609 1

%e ...

%p 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

%t 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 *)

%Y Cf. A000071, A000079, A000129, A007318, A140994, A140995, A140996, A140997, A140998, A141020, A141021.

%K nonn,tabl

%O 1,5

%A _Juri-Stepan Gerasimov_, Jul 08 2008

%E Entries checked by _R. J. Mathar_, Apr 28 2010

%E Name and offset edited by _Petros Hadjicostas_, Jun 10 2019

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Last modified July 15 02:31 EDT 2020. Contains 335762 sequences. (Running on oeis4.)