%I
%S 1,1,1,1,2,1,1,2,4,1,1,2,4,8,1,1,2,4,9,15,1,1,2,4,9,19,28,1,1,2,4,9,
%T 19,40,52,1,1,2,4,9,19,41,83,96,1,1,2,4,9,19,41,88,170,177,1,1,2,4,9,
%U 19,41,88,188,345,326,1,1,2,4,9,19,41,88,189,400,694,600,1,1,2,4,9,19,41,88,189,406,846,1386,1104,1
%N Triangle G(n, k), for 0 <= k <= n, read by rows, where G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, m) = G(n+1, m2) + G(n+1, m3) + G(n+2, m2) + G(n+3, m1) for n >= 0 and m = 3..(n+3).
%C From _Petros Hadjicostas_, Jun 12 2019: (Start)
%C This is a mirror of image of triangular array A140997. The current array has index of asymmetry s = 2 and index of obliqueness (obliquity) e = 1. Array A140997 has the same index of asymmetry, but has index of obliqueness e = 0. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but the stone slab that appears over a tomb in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.)
%C In general, if the index of asymmetry (from the Pascal triangle A007318) is s, then the order of the recurrence is s + 2 (because the recurrence of the Pascal triangle has order 2). There are also s + 2 infinite sets of initial conditions (as opposed to the Pascal triangle that has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).
%C Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), and arrays A140996 and A140995 have s = 3 (with e = 0 and e = 1, respectively).
%C If A(x,y) = Sum_{n,k >= 0} G(n, k)*x^n*y^k is the bivariate g.f. for this array (with G(n, k) = 0 for 0 <= n < k) and B(x, y) = Sum_{n, k} A140997(n, k)*x^n*y^k, then A(x, y) = B(x*y, y^(1)). This can be proved using formal manipulation of double series expansions and the fact G(n, k) = A140997(n, nk) for 0 <= k <= n.
%C If we let b(k) = lim_{n > infinity} G(n, k) for k >= 0, then b(0) = 1, b(1) = 2, b(2) = 4, and b(k) = b(k1) + 2*b(k2) + b(k3) for k >= 3. (The existence of the limit can be proved by induction on k.) It follows that b(k) = A141015(k) for k >= 0.
%C (End)
%H Robert Price, <a href="/A140994/b140994.txt">Table of n, a(n) for n = 0..1325</a>
%H JuriStepan 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 12 2019: (Start)
%F G(n, k) = A140997(n, nk) for 0 <= k <= n.
%F Bivariate g.f.: Sum_{n,k >= 0} G(n, k)*x^n*y^k = (x^4*y^3  x^3*y^3  x^2*y^2 + x^2*y  x*y + 1)/((1 x*y)*(1  x)*(1 x*y  x^2*y^2  x^3*y^3  x^3*y^2)).
%F (End)
%e Triangle begins:
%e 1
%e 1 1
%e 1 2 1
%e 1 2 4 1
%e 1 2 4 8 1
%e 1 2 4 9 15 1
%e 1 2 4 9 19 28 1
%e 1 2 4 9 19 40 52 1
%e 1 2 4 9 19 41 83 96 1
%e 1 2 4 9 19 41 88 170 177 1
%e 1 2 4 9 19 41 88 188 345 326 1
%e 1 2 4 9 19 41 88 189 400 694 600 1
%e 1 2 4 9 19 41 88 189 406 846 1386 1104 1
%e ... [corrected by _Petros Hadjicostas_, Jun 12 2019]
%e E.g., G(12, 9) = G(9, 7) + G(9, 6) + G(10, 7) + G(11, 8) = 170 + 88 + 188 + 400 = 846.
%p G := proc(n,k) if k=0 or n =k then 1; elif k= 1 then 2 ; elif k =2 then 4; elif k > n or k < 0 then 0 ; else procname(n3,k2)+procname(n3,k3)+procname(n2,k2)+procname(n1,k1) ; end if; end proc: seq(seq(G(n,k),k=0..n),n=0..15) ; # _R. J. Mathar_, Apr 14 2010
%t nlim = 50;
%t Do[G[n, 0] = 1, {n, 0, nlim}];
%t Do[G[n, n] = 1, {n, 1, nlim}];
%t Do[G[n + 2, 1] = 2, {n, 0, nlim}];
%t Do[G[n + 3, 2] = 4, {n, 0, nlim}];
%t Do[G[n + 4, m] =
%t G[n + 1, m  2] + G[n + 1, m  3] + G[n + 2, m  2] +
%t G[n + 3, m  1], {n, 0, nlim}, {m, 3, n + 3}];
%t A140994 = {}; For[n = 0, n <= nlim, n++,
%t For[k = 0, k <= n, k++, AppendTo[A140994, G[n, k]]]];
%t A140994 (* _Robert Price_, Aug 19 2019 *)
%Y Cf. A007318, A008937, A140993, A140995, A140996, A140997, A140998, A141015, A141018, A141020, A141021, A141031, A141065, A141066, A141067.
%K nonn,tabl,changed
%O 0,5
%A _JuriStepan Gerasimov_, Jul 08 2008
%E Entries checked by _R. J. Mathar_, Apr 14 2010
