cp's OEIS Frontend

Also the largest number of the n-th row of A140994.

a(n+1) is the number of n-bit sequences that avoid 1100. - David Callan, Jul 19 2004 [corrected by Kent E. Morrison, Jan 08 2019]. Also the number of n-bit sequences avoiding one of the patterns 1000, 0011, 1110, ... or any binary string of length 4 without overlap at beginning and end. Strings where it is not true are: 1111, 1010, 1001, ... and their bitwise complements. - Alois P. Heinz, Jan 09 2019

Row sums of Riordan array (1/(1-x), x(1+x+x^2)). - Paul Barry, Feb 16 2005

Diagonal sums of Riordan array (1/(1-x)^2, x(1+x)/(1-x)), A104698.

A shifted version of this sequence can be found in Eqs. (4) and (3) on p. 356 of Dunkel (1925) with r = 3. (Equation (3) follows equation (4) in the paper!) The whole paper is a study of the properties of this and other similar sequences indexed by the parameter r. For r = 2, we get a shifted version of A000071. For r = 4, we get a shifted version of A107066. For r = 5, we get a shifted version of A001949. For r = 6, we get a shifted version of A172316. See also the table in A172119. - Petros Hadjicostas, Jun 14 2019

Officially, to match A000073, this should start with a(0)=a(1)=0, a(2)=1. - N. J. A. Sloane, Sep 12 2020

Numbers with tribonacci representation that is a prefix of 100100100100... . - Jeffrey Shallit, Jul 10 2024

From Petros Hadjicostas, Jun 10 2019: (Start)

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)

From Petros Hadjicostas, Feb 09 2021: (Start)

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)

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 10 2019: (Start)

According to the attached picture, the index of asymmetry here is s = 1 and the index of obliqueness (or obliquity) is e = 0.

In the picture, the equation G(n, e*n) = 1 becomes G(n, 0) = 1, while the equations G(n+x+1, n-e*n+e*x-e+1) = 2^x for 0 <= x < s = 1 become G(n+1, n+1) = 1 and G(n+2, n+1) = 2.

Also, in the picture, the recurrence G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{m=1..s+1} G(n+m, k-e*s+m*e-2*e) for k = 1..n+1 becomes G(n+3, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) for k = 1..n+1.

Except for a shifting of the indices by 1, this array is a mirror image of array A140993. We have G(n, k) = A140993(n+1, n-k+1) for 0 <= k <= n. Triangular array A140993 has the same index of asymmetry (i.e., s = 1) but index of obliqueness e = 1.

(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)