A141065 -id:A141065 - cp's OEIS frontend

A140994 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, m-2) + G(n+1, m-3) + G(n+2, m-2) + G(n+3, m-1) for n >= 0 and m = 3..(n+3).

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, 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, 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

Offset: 0

Juri-Stepan Gerasimov, Jul 08 2008

From Petros Hadjicostas, Jun 12 2019: (Start)
This is a mirror image of the 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 on 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.)
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, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).
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).
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, n-k) for 0 <= k <= n.
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(k-1) + 2*b(k-2) + b(k-3) 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.
(End)

			Triangle begins:
  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 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 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
... [corrected by _Petros Hadjicostas_, Jun 12 2019]
E.g., G(12, 9) = G(9, 7) + G(9, 6) + G(10, 7) + G(11, 8) = 170 + 88 + 188 + 400 = 846.

Robert Price, Table of n, a(n) for n = 0..1325
Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A007318, A008937, A140993, A140995, A140996, A140997, A140998, A141015, A141018, A141020, A141021, A141031, A141065, A141066, A141067.

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(n-3,k-2)+procname(n-3,k-3)+procname(n-2,k-2)+procname(n-1,k-1) ; end if; end proc: seq(seq(G(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Apr 14 2010

nlim = 50;
Do[G[n, 0] = 1, {n, 0, nlim}];
Do[G[n, n] = 1, {n, 1, nlim}];
Do[G[n + 2, 1] = 2, {n, 0, nlim}];
Do[G[n + 3, 2] = 4, {n, 0, nlim}];
Do[G[n + 4, m] =
   G[n + 1, m - 2] + G[n + 1, m - 3] + G[n + 2, m - 2] +
    G[n + 3, m - 1], {n, 0, nlim}, {m, 3, n + 3}];
A140994 = {}; For[n = 0, n <= nlim, n++,
 For[k = 0, k <= n, k++, AppendTo[A140994, G[n, k]]]];
A140994 (* Robert Price, Aug 19 2019 *)

From Petros Hadjicostas, Jun 12 2019: (Start)
G(n, k) = A140997(n, n-k) for 0 <= k <= n.
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)).
(End)

Entries checked by R. J. Mathar, Apr 14 2010

A140997 Triangle G(n,k) read by rows, for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, and G(n+4, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) for n >= 0 and m = 1..n+1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 15, 9, 4, 2, 1, 1, 28, 19, 9, 4, 2, 1, 1, 52, 40, 19, 9, 4, 2, 1, 1, 96, 83, 41, 19, 9, 4, 2, 1, 1, 177, 170, 88, 41, 19, 9, 4, 2, 1, 1, 326, 345, 188, 88, 41, 19, 9, 4, 2, 1, 1, 600, 694, 400, 189, 88, 41, 19, 9, 4, 2, 1, 1, 1104, 1386, 846, 406, 189, 88, 41, 19, 9, 4, 2, 1, 1, 2031, 2751, 1779, 871, 406, 189, 88, 41, 19, 9, 4, 2, 1, 1, 3736, 5431, 3719, 1866, 872, 406, 189, 88, 41, 19, 9, 4, 2, 1

Offset: 0

Juri-Stepan Gerasimov, Jul 08 2008

From Petros Hadjicostas, Jun 12 2019: (Start)
This is a mirror image of the triangular array A140994. The current array has index of asymmetry s = 2 and index of obliqueness (obliquity) e = 0. Array A140994 has the same index of asymmetry, but has index of obliqueness e = 1. (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 on 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.)
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, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).
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).
(End)

			Triangle begins:
  1
  1   1
  1   2   1
  1   4   2   1
  1   8   4   2   1
  1  15   9   4   2  1
  1  28  19   9   4  2  1
  1  52  40  19   9  4  2  1
  1  96  83  41  19  9  4  2 1
  1 177 170  88  41 19  9  4 2 1
  1 326 345 188  88 41 19  9 4 2 1
  1 600 694 400 189 88 41 19 9 4 2 1
  ...
E.g., G(14, 2) = G(11, 1) + G(11, 2) + G(12, 2) + G(13, 2) = 600 + 694 + 1386 + 2751 = 5431.

Robert Price, Table of n, a(n) for n = 0..1325
Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A007318, A008937, A140993, A140994, A140995, A140996, A140998, A141015, A141018, A141020, A141021, A141065, A141066, A141067.

nlim = 50;
Do[G[n, 0] = 1, {n, 0, nlim}];
Do[G[n + 1, n + 1] = 1, {n, 0, nlim}];
Do[G[n + 2, n + 1] = 2, {n, 0, nlim}];
Do[G[n + 3, n + 1] = 4, {n, 0, nlim}];
Do[G[n + 4, m] =
   G[n + 1, m - 1] + G[n + 1, m] + G[n + 2, m] + G[n + 3, m], {n, 0,
   nlim}, {m, 1, n + 1}];
A140997 = {}; For[n = 0, n <= nlim, n++,
 For[k = 0, k <= n, k++, AppendTo[A140997, G[n, k]]]];
A140997 (* Robert Price, Aug 25 2019 *)

From Petros Hadjicostas, Jun 12 2019: (Start)
G(n, k) = A140994(n, n-k) for 0 <= k <= n.
Bivariate g.f.: Sum_{n,k >= 0} G(n,k)*x^n*y^k = (1 - x - x^2 - x^3 + x^2*y + x^4*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^3 - x^3*y)).
Differentiating once w.r.t. y and setting y = 0, we get the g.f. of column k = 1: x/((1 - x) * (1 - x - x^2 - x^3)). This is the g.f. of sequence A008937.
(End)

Typo in definition corrected by R. J. Mathar, Sep 19 2008
Name edited by and more terms from Petros Hadjicostas, Jun 12 2019
Deleted extraneous term at a(29) by Robert Price, Aug 25 2019
Added 13 missing terms at a(79) by Robert Price, Aug 25 2019

A140996 Triangle G(n, k) read by rows for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) + G(n+4, m) for n >= 0 for m = 1..(n+1).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 16, 8, 4, 2, 1, 1, 31, 17, 8, 4, 2, 1, 1, 60, 35, 17, 8, 4, 2, 1, 1, 116, 72, 35, 17, 8, 4, 2, 1, 1, 224, 148, 72, 35, 17, 8, 4, 2, 1, 1, 432, 303, 149, 72, 35, 17, 8, 4, 2, 1, 1, 833, 618, 308, 149, 72, 35, 17, 8, 4, 2, 1, 1, 1606, 1257, 636, 308, 149, 72, 35, 17, 8, 4, 2, 1

Offset: 0

Juri-Stepan Gerasimov, Jul 08 2008

From Petros Hadjicostas, Jun 12 2019: (Start)
This is a mirror image of the triangular array A140995. The current array has index of asymmetry s = 3 and index of obliqueness (obliquity) e = 0. Array A140995 has the same index of asymmetry, but has index of obliqueness e = 1. (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 on 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.)
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, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).
Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), arrays A140997 and A140994 have s = 2 (with e = 0 and e = 1, respectively), and arrays A141020 and A141021 have s = 4 (with e = 0 and e = 1, respectively).
(End)

			Triangle (with rows n >= 0 and columns k >= 0) begins as follows:
  1
  1   1
  1   2   1
  1   4   2   1
  1   8   4   2   1
  1  16   8   4   2  1
  1  31  17   8   4  2  1
  1  60  35  17   8  4  2  1
  1 116  72  35  17  8  4  2 1
  1 224 148  72  35 17  8  4 2 1
  1 432 303 149  72 35 17  8 4 2 1
  1 833 618 308 149 72 35 17 8 4 2 1
  ...

Robert Price, Table of n, a(n) for n = 0..5150
Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A007318, A107066, A140993, A140994, A140995, A140997, A140998, A141020, A141021, A141031, A141065, A141066, A141067, A141068, A141069, A141070, A141072, A141073, A309462.

nlim = 100;
For[n = 0, n <= nlim, n++, G[n, 0] = 1];
For[n = 1, n <= nlim, n++, G[n, n] = 1];
For[n = 2, n <= nlim, n++, G[n, n-1] = 2];
For[n = 3, n <= nlim, n++, G[n, n-2] = 4];
For[n = 4, n <= nlim, n++, G[n, n-3] = 8];
For[n = 5, n <= nlim, n++, For[k = 1, k < n - 3, k++,
   G[n, k] = G[n-4, k-1] + G[n-4, k] + G[n-3, k] + G[n-2, k] + G[n-1, k]]];
A140996 = {}; For[n = 0, n <= nlim, n++,
For[k = 0, k <= n, k++, AppendTo[A140996, G[n, k]]]];
A140996 (* Robert Price, Jul 03 2019 *)
G[n_, k_] := G[n, k] = Which[k < 0 || k > n, 0, k == 0 || k == n, 1, k == n - 1, 2, k == n - 2, 4, k == n - 3, 8, True, G[n - 1, k] + G[n - 2, k] + G[n - 3, k] + G[n - 4, k] + G[n - 4, k - 1]];
Table[G[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 28 2024 *)

From Petros Hadjicostas, Jun 12 2019: (Start)
G(n, k) = A140995(n, n - k) for 0 <= k <= n.
Bivariate g.f.: Sum_{n,k >= 0} G(n, k)*x^n*y^k = (1 - x - x^2 - x^3 - x^4 + x^2*y + x^3*y + x^5*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^3 - x^4 - x^4*y)).
If we take the first derivative of the bivariate g.f. w.r.t. y and set y = 0, we get the g.f. of column k = 1: x/((1 - x) * (1 - x - x^2 - x^3 - x^4)). This is the g.f. of a shifted version of sequence A107066.
Substituting y = 1 in the above bivariate function and simplifying, we get the g.f. of row sums: 1/(1 - 2*x). Hence, the row sums are powers of 2; i.e., A000079.
(End)

Name edited by Petros Hadjicostas, Jun 12 2019

A140995 Triangle G(n, k) read by rows, for 0 <= k <= n, where G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, m) = G(n+1, m-3) + G(n+1, m-4) + G(n+2, m-3) + G(n+3, m-2) + G(n+4, m-1) for n >= 0 and m = 4..(n+4).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 4, 8, 1, 1, 2, 4, 8, 16, 1, 1, 2, 4, 8, 17, 31, 1, 1, 2, 4, 8, 17, 35, 60, 1, 1, 2, 4, 8, 17, 35, 72, 116, 1, 1, 2, 4, 8, 17, 35, 72, 148, 224, 1, 1, 2, 4, 8, 17, 35, 72, 149, 303, 432, 1, 1, 2, 4, 8, 17, 35, 72, 149, 308, 618, 833, 1, 1, 2, 4, 8, 17, 35, 72, 149, 308, 636, 1257, 1606, 1

Offset: 0

Juri-Stepan Gerasimov, Jul 08 2008

From Petros Hadjicostas, Jun 13 2019: (Start)
This is a mirror image of the triangular array A140996. The current array has index of asymmetry s = 3 and index of obliqueness (obliquity) e = 1. Array A140996 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 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.)
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 A140997 and A140994 have s = 2 (with e = 0 and e = 1, respectively).
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} A140996(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) = A140996(n, n-k) for 0 <= k <= n.
If we let b(k) = lim_{n -> infinity} G(n, k) for k >= 0, then b(0) = 1, b(1) = 2, b(2) = 4, b(3) = 8, and b(k) = b(k-1) + b(k-2) + 2*b(k-3) + b(k-4) for k >= 4. (The existence of the limit can be proved by induction on k.) Thus, the limiting sequence is 1, 2, 4, 8, 17, 35, 72, 149, 308, 636, 1314, 2715, 5609, 11588, 23941, 49462, 102188, 211120, 436173, ... (sequence A309462). (End)

			Triangle begins:
  1
  1 1
  1 2 1
  1 2 4 1
  1 2 4 8  1
  1 2 4 8 16  1
  1 2 4 8 17 31  1
  1 2 4 8 17 35 60   1
  1 2 4 8 17 35 72 116   1
  1 2 4 8 17 35 72 148 224   1
  1 2 4 8 17 35 72 149 303 432   1
  1 2 4 8 17 35 72 149 308 618 833 1
  ...

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A000079, A007318, A140993, A140994, A140996, A140997, A140998, A141020, A141021, A141031, A141065, A141066, A141067, A141068, A141069, A141070, A141072, A141073, A309462.

From Petros Hadjicostas, Jun 13 2019: (Start)
G(n, k) = A140996(n, n-k) for 0 <= k <= n.
Bivariate g.f.: Sum_{n,k >= 0} G(n, k)*x^n*y^k = (x^5*y^4 - x^4*y^4 - x^3*y^3 + x^3*y^2 - 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^4*y^4 - x^4*y^3)).
Substituting y = 1 in the above bivariate function and simplifying, we get the g.f. of row sums: 1/(1 - 2*x). Hence, the row sums are powers of 2; i.e., A000079.
(End)

Entries checked by R. J. Mathar, Apr 14 2010

Name edited by and more terms from Petros Hadjicostas, Jun 13 2019

A141020 Pascal-like triangle with index of asymmetry y = 4 and index of obliqueness z = 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 16, 8, 4, 2, 1, 1, 32, 16, 8, 4, 2, 1, 1, 63, 33, 16, 8, 4, 2, 1, 1, 124, 67, 33, 16, 8, 4, 2, 1, 1, 244, 136, 67, 33, 16, 8, 4, 2, 1, 1, 480, 276, 136, 67, 33, 16, 8, 4, 2, 1, 1, 944, 560, 276, 136, 67, 33, 16, 8, 4, 2, 1

Offset: 0

Juri-Stepan Gerasimov, Jul 11 2008

The left column is set to 1. The four rightmost columns start with powers of 2:
T(n, 0) = T(n, n)=1; T(n, n-1)=2; T(n, n-2)=4; T(n, n-3)=8; T(n, n-4)=16.
Recurrence: T(n, k) = T(n-1, k) + T(n-2, k) + T(n-3, k) + T(n-4, k) + T(n-5, k) + T(n-5,k-1), k = 1..n-5.
From Petros Hadjicostas, Jun 14 2019: (Start)
In the attached photograph we see that the index of asymmetry is denoted by s (rather than y) and the index of obliqueness by e (rather than z).
The general recurrence is G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{1 <= m <= s+1} G(n+m, k-e*s+m*e-2*e) for n >= 0 with k = 1..(n+1) when e = 0 and k = (s+1)..(n+s+1) when e = 1. The initial conditions are G(n+x+1, n-e*n+e*x-e+1) = 2^x for x=0..s and n >= 0. There is one more initial condition, namely, G(n, e*n) = 1 for n >= 0.
For s = 0, we get Pascal's triangle A007318. For s = 1, we get A140998 (e = 0) and A140993 (e = 1). For s = 2, we get A140997 (e = 0) and A140994 (e = 1). For s = 3, we get A140996 (e = 0) and A140995 (e = 1). For s = 4, we have the current array (with e = 0) and array A141021 (with e = 1). In some of these arrays, the indices n and k are sometimes shifted.
(End)

			Pascal-like triangle with y = 4 and z = 0 begins as follows:
  1
  1   1
  1   2   1
  1   4   2   1
  1   8   4   2   1
  1  16   8   4   2  1
  1  32  16   8   4  2  1
  1  63  33  16   8  4  2  1
  1 124  67  33  16  8  4  2 1
  1 244 136  67  33 16  8  4 2 1
  1 480 276 136  67 33 16  8 4 2 1
  1 944 560 276 136 67 33 16 8 4 2 1
  ...

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A001949, A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141021, A141031, A141064, A141065, A141066, A141067, A141069, A141070, A141072, A141073.

A141020 := proc(n,k) option remember ; if k<0 or k>n then 0 ; elif k=0 or k=n then 1 ; elif k=n-1 then 2 ; elif k=n-2 then 4 ; elif k=n-3 then 8 ; elif k=n-4 then 16 ; else procname(n-1,k) +procname(n-2,k)+procname(n-3,k)+procname(n-4,k) +procname(n-5,k)+procname(n-5,k-1) ; fi; end:
for n from 0 to 20 do for k from 0 to n do printf("%d,",A141020(n,k)) ; od: od: # R. J. Mathar, Sep 19 2008

T[n_, k_] := T[n, k] = Which[k < 0 || k > n, 0, k == 0 || k == n, 1, k == n-1, 2, k == n-2, 4, k == n-3, 8, k == n-4, 16, True, T[n-1, k] + T[n-2, k] + T[n-3, k] + T[n-4, k] + T[n-5, k] + T[n-5, k-1]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 18 2019, after R. J. Mathar *)

From Petros Hadjicostas, Jun 14 2019: (Start)
T(n, k) = A141021(n, n-k) for 0 <= k <= n.
Bivariate g.f.: Sum_{n,k >= 0} T(n, k)*x^n*y^k = (1 - x - x^2 - x^3 - x^4 - x^5 + y*x^2*(1 + x + x^2 + x^4)) / ((1 - x) * (1 - x*y) * (1 - x - x^2 - x^3 - x^4 - x^5 - x^5*y)).
Differentiating the bivariate w.r.t. y and setting y = 0, we get the g.f. of the column k = 1: x/((-1 + x)*(x^5 + x^4 + x^3 + x^2 + x - 1)). This is the g.f. of a shifted version of sequence A001949.
(End)

Partially edited by N. J. A. Sloane, Jul 18 2008

Recurrence rewritten by R. J. Mathar, Sep 19 2008

A141066 List of different composites in Pascal-like triangles with index of asymmetry y = 2 and index of obliquity z = 0 or z = 1.

Original entry on oeis.org

4, 8, 9, 15, 28, 40, 52, 96, 88, 170, 177, 188, 326, 345, 189, 400, 600, 694, 406, 846, 1104, 1386, 871, 1779, 2031, 2751, 872, 1866, 3736, 6872, 7730, 10672, 4022, 8505, 12640, 15979, 20885, 4023, 8633, 18079, 23249, 32859, 40724, 42762, 67240, 18559, 39677, 78652, 80866, 153402

Offset: 1

Juri-Stepan Gerasimov, Jul 14 2008

nonn

For the Pascal-like triangle G(n, k) with index of asymmetry y = 2 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) for k = 1..(n+1). (This is array A140997.)
For the Pascal-like triangle with index of asymmetry y = 1 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, k) = G(n+1, k-2) + G(n+1, k-3) + G(n+2, k-2) + G(n+3, k-1) for k = 3..(n+3). (This is array A140994.)
From Petros Hadjicostas, Jun 12 2019: (Start)
The two triangular arrays A140997 and A140994, which are described above, are mirror images of each other.
To make the current sequence uniquely defined, we follow the suggestion of R. J. Mathar for sequence A141064. For each row of array A140997, the composites not appearing in earlier rows are collected, sorted, and added to the sequence. We get exactly the same sequence by working with array A140994 instead.
Finally, we mention that in the attached picture about the connection between Stepan's triangles and the Pascal triangle, the letter s is used to describe the index of asymmetry and the letter e is used to describe the index of obliqueness (instead of the letters y and z, respectively). The Pascal triangle A007318 has index of asymmetry s = y = 0 (and it does not matter whether we use e = 0 or e = 1 in the general formulas in the attached photograph).
(End)

			Pascal-like triangle with y = 2 and z = 0 (i.e., A140997) begins as follows:
  1, so no composite.
  1 1, so no composite.
  1 2 1, so no composite.
  1 4 2 1, so a(1) = 4.
  1 8 4 2 1, so a(2) = 8.
  1 15 9 4 2 1, so a(3) = 9 and a(4) = 15.
  1 28 19 9 4 2 1, so a(5) = 28.
  1 52 40 19 9 4 2 1, so a(6) = 40 and a(7) = 52.
  1 96 83 41 19 9 4 2 1, so a(8) = 96.
  1 177 170 88 41 19 9 4 2 1, so a(9) = 88, a(10) = 170, and a(11) = 177.
  1 326 345 188 88 41 19 9 4 2 1, so a(12) = 188, a(13) = 326, and a(14) = 345.
  1 600 694 400 189 88 41 19 9 4 2 1, so a(15) = 189, a(16) = 400, a(17) = 600, and a(18) = 694.
... [example edited by _Petros Hadjicostas_, Jun 11 2019]

Petros Hadjicostas, Table of n, a(n) for n = 1..191 [b-file format amended by _Georg Fischer_, Jun 22 2019]
Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A007318 (y = 0), A140993 (y = 1 and z = 1), A140994 (y = 2 and z = 1), A140995 (y = 3 and z = 1), A140996 (y = 3 and z = 0), A140997 (y = 2 and z = 0), A140998 (y = 1 and z = 0), A141020 (y = 4 and z = 0), A141021 (y = 4 and z = 1), A141064 (has primes when y = 1), A141065 (has composites when y = 1), A141067 (has primes when y = 2), A141068 (has primes when y = 3), A141069 (has composites when y = 3).

# This is a modification of R. J. Mathar's program for A141031 (for the case y = 4 and z = 0).
# Construction of array A140997 (y = 2 and z = 0):
A140997 := proc(n, k) option remember; if k < 0 or n < k then 0; elif k = 0 or k = n then 1; elif k = n - 1 then 2; elif k = n - 2 then 4; else procname(n - 1, k) + procname(n - 2, k) + procname(n - 3, k) + procname(n - 3, k - 1); end if; end proc;
# Construction of the current sequence:
A141066 := proc(nmax) local a, b, n, k, new; a := []; for n from 0 to nmax do b := []; for k from 0 to n do new := A140997(n, k); if not (new = 1 or isprime(new) or new in a or new in b) then b := [op(b), new]; end if; end do; a := [op(a), op(sort(b))]; end do; RETURN(a); end proc;
# Generation of numbers in the current sequence:
A141066(19);
# If one wishes to sort the numbers, then replace RETURN(a) with RETURN(sort(a)) in the above Maple code. In this case, however, the sequence is not uniquely defined because it depends on the maximum n. - Petros Hadjicostas, Jun 15 2019

Partially edited by N. J. A. Sloane, Jul 18 2008
Comments and Example edited by Petros Hadjicostas, Jun 12 2019
More terms from Petros Hadjicostas, Jun 12 2019

A141031 Nonprimes in the triangle A141020.

Original entry on oeis.org

1, 4, 8, 16, 32, 33, 63, 124, 136, 244, 276, 480, 560, 561, 944, 1135, 1140, 1856, 2298, 2316, 3649, 4705, 7174, 9398, 9558, 9559, 14104, 18984, 19415, 27728, 38320, 39432, 39457, 54512, 77298, 80075, 80163, 107168, 155823, 162583, 162863, 162864, 210687, 313927, 330878, 414200, 632080, 669872, 814296, 1271960, 1600864

Offset: 1

Juri-Stepan Gerasimov, Jul 12 2008

nonn

From Petros Hadjicostas, Jun 22 2019: (Start)
This is a dynamically defined sequence. Since the nonprimes from each row are mixed with the nonprimes of previous rows and then sorted, the value of a(n) may change each time we add a new row.
For a modification of R. J. Mathar's program below so that nonprimes are sorted only within each row (so as to get a uniquely defined sequence) see the documentation of sequences A141064, A141065, A141066, A141067, A141068, and A141069.
(End)

			Scanning rows of A141020 or A141021 and sorting new nonprimes into the list we get:
  1 yields a(1) = 1.
  1 1 yields no new member.
  1 2 1 yields no new member.
  1 4 2 1 yields a(2) = 4.
  1 8 4 2 1 yields a(3) = 8.
  1 16 8 4 2 1 yields a(4) = 16.
  1 32 16 8 4 2 1 yields a(5) = 32.
  1 63 33 16 8 4 2 1 yields a(6) = 33 and a(7) = 63.
  1 124 67 33 16 8 4 2 1 yields a(8) = 124.
  1 244 136 67 33 16 8 4 2 1 yields a(9) = 136 and a(10) = 244.
  1 480 276 136 67 33 16 8 4 2 1 yields a(11) = 276 and a(12) = 480.
  1 944 560 276 136 67 33 16 8 4 2 1 yields a(13) = 560 and a(14) = 944.
  ...
From _Petros Hadjicostas_, Jun 22 2019: (Start)
In the above example, we only sort the nonprimes up to row 11; we get the same output from _R. J. Mathar_'s program below if we say A141031(11). If, however, we include more rows in the program, the indexing of the nonprimes changes.
For example, the nonprimes in the data above come from the nonprimes of 22 rows. If we include more rows, then the indexing again changes and the value of each a(n) may not stay the same.
(End)

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141021, A141064, A141065, A141066, A141067, A141068, A141069.

A141020 := proc(n,k) option remember ; if k<0 or k>n then 0 ; elif k=0 or k=n then 1 ; elif k=n-1 then 2 ; elif k=n-2 then 4 ; elif k=n-3 then 8 ; elif k=n-4 then 16 ; else procname(n-1,k)+procname(n-2,k)+procname(n-3,k)+procname(n-4,k) +procname(n-5,k)+procname(n-5,k-1) ; fi; end:
A141031 := proc(nmax) local a,n,k ; a := [] ; for n from 0 to nmax do for k from 0 to n do a141020 := A141020(n,k) ; if not isprime(a141020) and not a141020 in a then a := [op(a),a141020] ; fi; od: od: RETURN(sort(a)) ; end: A141031(30) ; # R. J. Mathar, Sep 19 2008

Partially edited by N. J. A. Sloane, Jul 18 2008

Simplified definition, corrected values by R. J. Mathar, Sep 19 2008

A141069 List of different composites in Pascal-like triangles with index of asymmetry y = 3 and index of obliqueness z = 0 or z = 1.

Original entry on oeis.org

4, 8, 16, 35, 60, 72, 116, 148, 224, 303, 432, 308, 618, 833, 636, 1257, 1606, 1313, 2550, 3096, 1314, 2709, 5160, 5968, 2715, 5584, 10418, 11504, 5609, 11499, 20991, 22175, 23655, 42215, 42744, 11588, 23934, 48607, 82392, 84752, 23941, 99763, 158816, 169880

Offset: 1

Juri-Stepan Gerasimov, Jul 16 2008

nonn

For the Pascal-like triangle with index of asymmetry y = 3 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) + G(n+4, k) for n >= 0 and k = 1..(n+1). (This is array A140996.)
For the Pascal-like triangle with index of asymmetry y = 3 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and  G(n+5, k) = G(n+1, k-3) + G(n+1, k-4) + G(n+2, k-3) + G(n+3, k-2) + G(n+4, k-1) for n >= 0 and k = 4..(n+4). (This is array A140995.)
From Petros Hadjicostas, Jun 13 2019: (Start)
The arrays A140995 and A140996, which are described above, are mirror images of one another.
To make the current sequence uniquely defined, we follow the suggestion of R. J. Mathar for sequence A141064. For each row of array A140996, the composites not appearing in earlier rows are collected, sorted, and added to the sequence. We get exactly the same sequence by working with array A140995 instead.
Finally, we explain the meaning of the double recurrence in the attached photograph. It concerns the connection between Stepan's triangles and Pascal's triangles. The creator of the stone slab uses the notation G_n^k to denote the double array G(n, k), where 0 <= k <= n.
On the stone slab, the letter s is used to denote the "index of asymmetry" (denoted by y here) and the letter e is used to denote the 0-1 "index of obliqueness" (denoted by z here). Thus, as described above, there are two kinds of Stepan-Pascal triangles depending on whether e is equal to 0 or 1. (The case s = 0 corresponds to Pascal's triangle A007318.)
If e = 0, the value of k goes from 1 to n + 1, whereas if e = 1 the value of k goes from s + 1 to n + s + 1.
The "index of asymmetry" s = y can take any (fixed) integer value from 0 to infinity. The fixed value of s = y determines the number of initial conditions: G(n + x + 1, n - e*n + e*x - e + 1) = 2^x for x = 0, 1, ..., s. In addition, there is one more initial condition: G(n, e*n) = 1.
The "index of asymmetry" s = y also determines the order of the recurrence (which is probably s + 2 = y + 2): G(n + s + 2, k) = G(n + 1, k - e*s + e - 1) + Sum_{1 <= m <= s + 1} G(n + m, k - e*s + m*e - 2*e).
(End)

			Pascal-like triangle with y = 3 and z = 0 (i.e., A140996) begins as follows:
  1, so no composites.
  1 1, so no composites.
  1 2 1, so no composites.
  1 4 2 1, so a(1) = 4.
  1 8 4 2 1, so a(2) = 8.
  1 16 8 4 2 1, so a(3) = 16.
  1 31 17 8 4 2 1, so no new composites.
  1 60 35 17 8 4 2 1, so a(4) = 35 and a(5) = 60.
  1 116 72 35 17 8 4 2 1, so a(6) = 72 and a(7) = 116.
  1 224 148 72 35 17 8 4 2 1, so a(8) = 148 and a(9) = 224.
  1 432 303 149 72 35 17 8 4 2 1, so a(10) = 303 and a(11) = 432.
... [edited by _Petros Hadjicostas_, Jun 13 2019]

Petros Hadjicostas, Table of n, a(n) for n = 1..110
Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141021, A141031, A141064, A141065, A141066, A141067, A141069, A141070, A141072, A141073.

# This is a modification of R. J. Mathar's program from sequence A141031 (for the case y = 4 and z = 0).
# Definition of sequence A140996 (y = 3 and z = 0):
A140996 := proc(n, k) option remember; if k < 0 or n < k then 0; elif k = 0 or k = n then 1; elif k = n - 1 then 2; elif k = n - 2 then 4; elif k = n - 3 then 8; else procname(n - 1, k) + procname(n - 2, k) + procname(n - 3, k) + procname(n - 4, k) + procname(n - 4, k - 1); end if; end proc;
# Definition of current sequence:
A141069 := proc(nmax) local a, b, n, k, new; a := []; for n from 0 to nmax do b := []; for k from 0 to n do new := A140996(n, k); if not (new = 1 or isprime(new) or new in a or new in b) then b := [op(b), new]; end if; end do; a := [op(a), op(sort(b))]; end do; RETURN(a); end proc;
# Generation of current sequence until row n = 30:
A141069(30);
# If one wishes the composites to be sorted, then replace RETURN(a) with RETURN(sort(a)) in the above Maple code. In such a case, however, the output may not necessarily be uniquely defined (because it changes with the value of n). - Petros Hadjicostas, Jun 15 2019

Partially edited by N. J. A. Sloane, Jul 18 2008

More terms from Petros Hadjicostas, Jun 13 2019

A141068 List of different primes in Pascal-like triangles with index of asymmetry y = 3 and index of obliquity z = 0 or z = 1.

Original entry on oeis.org

2, 17, 31, 149, 11587, 49429, 15701951, 21304973, 3846277, 251375273, 5449276159, 296410704409, 750391353973, 205109154121, 875366796349, 72210869205443, 139884035510017, 79014319582741129, 94461530406533783, 2562508045902551

Offset: 1

Juri-Stepan Gerasimov, Jul 16 2008

nonn

For the Pascal-like triangle with index of asymmetry y = 3 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) + G(n+4, k) for k = 1..(n+1). (This is array A140996.)
For the Pascal-like triangle with index of asymmetry y = 3 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, k) = G(n+1, k-3) + G(n+1, k-4) + G(n+2, k-3) + G(n+3, k-2) + G(n+4, k-1) for k = 4..(n+4). (This is array A140995.)
From Petros Hadjicostas, Jun 13 2019: (Start)
The two triangular arrays A140995 and A140996, which are described above, are mirror images of each other.
To make the current sequence uniquely defined, we follow the suggestion of R. J. Mathar for sequence A141064. For each row of array A140996, the primes not appearing in earlier rows are collected, sorted, and added to the sequence. We get exactly the same sequence by working with array A140995 instead.
Finally, we mention that in the attached picture about the connection between Stepan's triangles and the Pascal triangle, the letter s is used to describe the index of asymmetry and the letter e is used to describe the index of obliqueness (instead of the letters y and z, respectively). The Pascal triangle A007318 has index of asymmetry s = y = 0.
(End)

			Pascal-like triangle with y = 3 and z = 0 (i.e., A140996) begins as follows:
  1, so no primes.
  1 1, so no primes
  1 2 1, so a(1) = 2.
  1 4 2 1, so no new primes.
  1 8 4 2 1, so no new primes.
  1 16 8 4 2 1, so new primes.
  1 31 17 8 4 2 1, so a(2) = 17 and a(3) = 31.
  1 60 35 17 8 4 2 1, so no new primes.
  1 116 72 35 17 8 4 2 1, so no new primes.
  1 224 148 72 35 17 8 4 2 1, so new primes.
  1 432 303 149 72 35 17 8 4 2 1, so a(4) = 149.
  ...

Juri-Stepan Gerasimov, Pascal-like triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141021, A141031, A141064, A141065, A141066, A141067, A141069, A141070, A141072, A141073.

# This is a modification of R. J. Mathar's program for A141031 (for the case y = 4 and z = 0).
# Definition of sequence A140996 (y = 3 and z = 0):
A140996 := proc(n, k) option remember; if k < 0 or n < k then 0; elif k = 0 or k = n then 1; elif k = n - 1 then 2; elif k = n - 2 then 4; elif k = n - 3 then 8; else procname(n - 1, k) + procname(n - 2, k) + procname(n - 3, k) + procname(n - 4, k) + procname(n - 4, k - 1); end if; end proc;
# Definition of the current sequence:
A141068 := proc(nmax) local a, b, n, k, new; a := []; for n from 0 to nmax do b := []; for k from 0 to n do new := A140996(n, k); if not (new = 1 or not isprime(new) or new in a or new in b) then b := [op(b), new]; end if; end do; a := [op(a), op(sort(b))]; end do; RETURN(a); end proc;
# Generation of the current sequence:
A141068(80);
# If one wishes to get the primes sorted (as R. J. Mathar does in A141031), then replace RETURN(a) in the code above with RETURN(sort(a)). In such a case, however, the output sequence is not uniquely defined because it depends on the maximum n. - Petros Hadjicostas, Jun 15 2019

Partially edited by N. J. A. Sloane, Jul 18 2008

More terms from Petros Hadjicostas, Jun 13 2019

A141073 List of central integer pairs in Pascal-like triangles with index of asymmetry y = 3 and index of obliqueness z = 0 or z = 1.

Original entry on oeis.org

1, 1, 4, 2, 8, 4, 17, 8, 35, 17, 72, 35, 149, 72, 308, 149, 636, 308, 1314, 636, 2715, 1314, 5609, 2715, 11588, 5609, 23941, 11588, 49462, 23941, 102188, 49462, 211120, 102188, 436173, 211120, 901131, 436173, 1861732, 901131, 3846329, 1861732, 7946496, 3846329

Offset: 1

Juri-Stepan Gerasimov, Jul 16 2008

For the Pascal-like triangle G(n, k) with index of asymmetry y = 3 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) + G(n+4, k) for n >= 0 and k = 1..(n+1). (This is array A140996.)
For the Pascal-like triangle G(n, k) with index of asymmetry y = 3 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, k) = G(n+1, k-3) + G(n+1, k-4) + G(n+2, k-3) + G(n+3, k-2) + G(n+4, k-1) for n > = 0 and k = 4..(n+4). (This is array A140995.)
Arrays A140995 and A140996 are mirror images of each other. For discussion about their properties and their connection to Stepan's triangles, see their documentation. See also the documentation of the sequences in the CROSSREFS. - Petros Hadjicostas, Jun 13 2019

			Pascal-like triangle with y = 3 and z = 0 (i.e., A140996) begins as follows:
  1, so no central pair.
  1 1, so a(1) = 1 and a(2) = 1.
  1 2 1, so no central pair.
  1 4 2 1, so a(3) = 4 and a(4) = 2.
  1 8 4 2 1, so no central pair.
  1 16 8 4 2 1, so a(5) = 8 and a(6) = 4.
  1 31 17 8 4 2 1, so no central pair.
  1 60 35 17 8 4 2 1, so a(7) = 17 and a(8) = 8.
  1 116 72 35 17 8 4 2 1, so no central pair.
  1 224 148 72 35 17 8 4 2 1, so a(9) = 35 and a(10) = 17.
  1 432 303 149 72 35 17 8 4 2 1, so no central pair.
  1 833 618 308 149 72 35 17 8 4 2 1, so a(11) = 72 and a(12) = 35.
... [edited by _Petros Hadjicostas_, Jun 13 2019]

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A129847, A140993, A140994, A140995, A140996, A140997, A140998, A141065, A141066, A141067, A141068, A141069, A141070, A141072.

Rest[CoefficientList[Series[x*(x^8 + 3*x^6 + x^5 + 3*x^4 + x^3 + 3*x^2 + x + 1)/(1 - x^2 - x^4 - 2*x^6 -x^8),{x,0,44}],x]] (* James C. McMahon, Jul 16 2025 *)

From Petros Hadjicostas, Jun 13 2019: (Start)
a(2*n - 1) = A140996(2*n - 1, n - 1) = A140995(2*n - 1, n) and a(2*n) = A140996(2*n - 1, n) = A140995(2*n - 1, n - 1) for n >= 1.
a(2*n) = a(2*n - 3) for n >= 3.
a(n) = 2*a(n-2) + A129847(floor(n/2) - (4 + (-1)^n)) for n >= 9.
G.f.: x*(x^8 + 3*x^6 + x^5 + 3*x^4 + x^3 + 3*x^2 + x + 1)/(1 - x^2 - x^4 - 2*x^6 -x^8). (End)

Partially edited by N. J. A. Sloane, Jul 18 2008

More terms from Petros Hadjicostas, Jun 13 2019

cp's OEIS Frontend

A140994 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, m-2) + G(n+1, m-3) + G(n+2, m-2) + G(n+3, m-1) for n >= 0 and m = 3..(n+3).

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A140997 Triangle G(n,k) read by rows, for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, and G(n+4, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) for n >= 0 and m = 1..n+1.

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A140996 Triangle G(n, k) read by rows for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) + G(n+4, m) for n >= 0 for m = 1..(n+1).

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A140995 Triangle G(n, k) read by rows, for 0 <= k <= n, where G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, m) = G(n+1, m-3) + G(n+1, m-4) + G(n+2, m-3) + G(n+3, m-2) + G(n+4, m-1) for n >= 0 and m = 4..(n+4).

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A141020 Pascal-like triangle with index of asymmetry y = 4 and index of obliqueness z = 0.

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A141066 List of different composites in Pascal-like triangles with index of asymmetry y = 2 and index of obliquity z = 0 or z = 1.

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A141031 Nonprimes in the triangle A141020.

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