cp's OEIS Frontend

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)

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)

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)

The triangle here is A141020 with each row reversed.

From Petros Hadjicostas, Jun 16 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 array A141020 (with e = 0) and the current array (with e = 1). In some of these arrays, the indices n and k are sometimes shifted.

Putting k = 1 in Stepan's triangles with index of asymmetry s and index of obliqueness e = 0, we get G(n + s + 2, 1) = 1 + Sum_{1 <= m <= s+1} G(n+m, 1) for n >= 0 and k = 1..(n+1) with initial conditions G(x+1, 1) = 2^x for x = 0..s. Thus, we get a shifted version of column s+1 in array A172119. These sequences were first studied by Dunkel (1925).

Thus, the second main diagonal of Stepan's triangles with index of asymmetry s and index of obliqueness e = 1 is equal to a shifted version of column s + 1 in array A172119.

It follows from Eq. (20) on p. 360 in Dunkel (1925) that, for Stepan's triangles with index of asymmetry s and index of obliqueness e = 0, we have G(n, 1) = Sum_{t = 1..floor((n + s + 1)/(s + 2))} (-1)^(t + 1) * binomial(n + s - t*(s + 1), t - 1) * 2^(n + s - t*(s + 2) + 1) for n >= 0.

In a similar way, for Stepan's triangles with index of asymmetry s and index of obliqueness e = 1, we have G(n, n - 1) = Sum_{t = 1..floor((n + s + 1)/(s + 2))} (-1)^(t + 1) * binomial(n + s - t*(s + 1), t - 1) * 2^(n + s - t*(s + 2) + 1) for n >= 1.

Let A_s(x, y) be the bivariate g.f. of G(n, k) with index of asymmetry s and index of obliqueness e = 0 and let B_s(x, y) be the bivariate g.f. of the other G(n, k) with index of asymmetry s and index of obliqueness e = 1. Because the two triangular arrays are mirror images of each other, we have B_s(x, y) = A_s(x*y, y^(-1)).

(End)

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)

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)

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, and G(n+4, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) for n >= 0 and k = 1..(n+1). (This is array A140997.)

For the Pascal-like triangle G(n, k) 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, and 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 n >= 0 and 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 primes 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)

For the Pascal-like triangle G(n, k) with index of asymmetry y = 1 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, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) for n >= 0 and k = 1..(n+1).

For the Pascal-like triangle G(n, k) 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, k) = G(n+1, k-1) + G(n+1, k-2) + G(n+2, k-1) for n >= 0 and k = 2..(n+2).

From Petros Hadjicostas, Jun 09 2019: (Start)

For the triangle with index of asymmetry y = 1 and index of obliqueness z = 0, read by rows, we have G(n, k) = A140998(n, k) for 0 <= k <= n.

For the triangle with index of asymmetry y = 1 and index of obliqueness z = 1, read by rows, we have G(n, k) = A140993(n+1, k+1) for 0 <= k <= n.

Thus, except for the unfortunate shifting of the indices by 1, triangular arrays A140998 and A140993 are mirror images of each other.

As suggested by R. J. Mathar for sequence A141064, in each row of A140998, the composites not appearing in earlier rows are collected, sorted, and added to the sequence.

Obviously, instead of working with A140998, we may work with A140993: in each row of A140993, the primes not appearing in earlier rows may be collected, sorted, and added to the sequence.

Finally, we explain the meaning of the double recurrence in the attached photograph (about Stepan's triangles and Pascal's triangles).

The creator of the stone slab uses the notation G_n^k to denote either one of the two double arrays G(n, k) described above.

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 = z equals 0 or 1.

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 = y + 1 (= 2 here) to n + s + 1 = n + y + 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 = y. 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).

Apparently, for convenience, the author of the current sequence has shifted the indices of the recurrences that appear on the stone slab (see at the beginning of the comments).

(End)

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)

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)