A141073 -id:A141073 - cp's OEIS frontend

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).

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

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

A141070 Number of primes in rows of Pascal-like triangles with index of asymmetry y = 3 and index of obliquity z = 0 or z = 1.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 3, 3, 3, 3, 3, 5, 4, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 5, 3, 4, 3, 3, 3, 5, 3, 4, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 4, 3, 4, 4, 3, 3, 3, 3, 5, 4, 4, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 5, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 5, 3, 3, 3

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 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 two triangular arrays A140995 and A140996, which are described above, are mirror images of each other. Thus, we get the same sequence no matter which one we use.
Even though the numbering of the rows of both triangular arrays A140995 and A140996 starts at n = 0, the author of this sequence set up the offset at n = 1; that is, a(n) = number of primes in row n - 1 for A140995 (or for A140996) for n >= 1.
Finally, we mention that in the attached picture about Stepan's triangles, 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).
(End)

			Pascal-like triangle with y = 3 and z = 0 (i.e, A140996) begins as follows:
  1, so a(1) = 0.
  1 1, so a(2) = 0.
  1 2 1, so prime 2 and a(3) = 1.
  1 4 2 1, so prime 2 and a(4) = 1.
  1 8 4 2 1, so prime 2 and a(5) = 1.
  1 16 8 4 2 1, so prime 2 and a(6) = 1.
  1 31 17 8 4 2 1, so primes 2, 17, 31 and a(7) = 3.
  1 60 35 17 8 4 2 1, so primes 2, 17 and a(8) = 2.
  1 116 72 35 17 8 4 2 1, so primes 2, 17 and a(9) = 2.
  1 224 148 72 35 17 8 4 2 1, so primes 2, 17 and a(10) = 2.
  1 432 303 149 72 35 17 8 4 2 1, so primes 2, 17, 149 and a(11) = 3.
  ...

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

Cf. A140993, A140994, A140995, A140996, A140997, A140998, A141065, A141066, A141067, A141068, A141069, A141072, A141073.

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]]];
A141070 = {}; For[n = 0, n <= nlim, n++, c = 0;
 For[k = 0, k <= n, k++, If[PrimeQ[G[n, k]], c++]];
 AppendTo[A141070, c]];
A141070 (* Robert Price, Jul 03 2019 *)

Partially edited by N. J. A. Sloane, Jul 18 2008
More terms and comments edited by Petros Hadjicostas, Jun 13 2019
a(52)-a(100) from Robert Price, Jul 03 2019

A141072 Sum of diagonal numbers in a Pascal-like triangle with index of asymmetry y = 3 and index of obliquity z = 0 (going upwards, left to right).

Original entry on oeis.org

1, 1, 2, 3, 6, 11, 22, 42, 83, 162, 319, 626, 1231, 2419, 4756, 9349, 18380, 36133, 71036, 139652, 274549, 539748, 1061117, 2086100, 4101165, 8062677, 15850806, 31161863, 61262610, 120439119, 236777074, 465491470, 915132135, 1799102406, 3536942203, 6953445286

Offset: 1

Juri-Stepan Gerasimov, Jul 16 2008

nonn

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 for k = 4..(n+4). (This is array A140995.)
From Petros Hadjicostas, Jun 13 2019: (Start)
In the example below the author uses array A140996 to create this sequence. If we use array A140995, which is the mirror image of A140996, and follow the same process, we get a different sequence: 1, 1, 2, 3, 4, 7, 8, 15, 16, 31, 33, 63, 68, 127, 140, 255, 288, 512, 592, ...
Even though array A140996 starts at row n = 0, the offset of the current sequence was set at n = 1. In other words, a(n) = Sum_{ceiling((n-1)/2) <= i <= n-1} G(i, n-1-i) = G(n-1, 0) + G(n-2, 1) + ... + G(ceiling((n-1)/2), floor((n-1)/2)) for n >= 1, where G(n, k) = A140996(n, k).
To get the g.f. of this sequence, we take the bivariate g.f. of sequence A140996, and set x = y. We multiply the result by x because the offset here was set at n = 1.
Finally, we mention that in the attached photograph about Stepan's triangle, the index of asymmetry is denoted by s (rather than y) and the index of obliqueness is denoted by e (rather than z). For the Pascal triangle, s = y = 0.
(End)

			Pascal-like triangle with y = 3 and z = 0 (i.e, A140996) begins as follows:
  1, so a(1) = 1.
  1   1, so a(2) = 1.
  1   2   1, so a(3) = 1 + 1 = 2.
  1   4   2   1, so a(4) = 1 + 2 = 3.
  1   8   4   2  1, so a(5) = 1 + 4 + 1 = 6.
  1  16   8   4  2  1, so a(6) = 1 + 8 + 2 = 11.
  1  31  17   8  4  2  1, so a(7) = 1 + 16 + 4 + 1 = 22.
  1  60  35  17  8  4  2 1, so a(8) = 1 + 31 + 8 + 2 = 42.
  1 116  72  35 17  8  4 2 1, so a(9) = 1 + 60 + 17 + 4 + 1 = 83.
  1 224 148  72 35 17  8 4 2 1, so a(10) = 1 +  116 + 35 + 8 + 2 = 162.
  1 432 303 149 72 35 17 8 4 2 1, so a(11) = 1 + 224 + 72 + 17 + 4 + 1 = 319.
... [edited by _Petros Hadjicostas_, Jun 13 2019]

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

Cf. A140993, A140994, A140995, A140996, A140997, A140998, A141065, A141066, A141067, A141068, A141069, A141073.

From Petros Hadjicostas, Jun 13 2019: (Start)
a(n) = Sum_{ceiling((n-1)/2) <= i <= n-1} G(i, n-1-i) for n >= 1, where G(n, k) = A140996(n, k) for 0 <= k <= n.
G.f.: x*(1 - x^2 - x^3 - x^4 - x^5)/((1 - x)*(1 + x)*(1 - x - x^2 - x^3 - x^4 - x^5)) = x*(1 - x^2 - x^3 - x^4 - x^5)/(1 - x - 2*x^2 + x^6 + x^7).
Recurrence: a(n) = -(3 + (-1)^n)/2 + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) for n >= 6 with a(1) = a(2) =  1, a(3) = 2, a(4) = 3, and a(5) = 6.
(End)

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

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

A129847 a(n) = number of set partitions of {1, 2, ..., n} whose blocks consist only of elements that differ by two or less (that is, have only the forms {i}, {i,i+1}, {i,i+2}, or {i,i+1,i+2}).

Original entry on oeis.org

1, 1, 2, 5, 10, 20, 42, 87, 179, 370, 765, 1580, 3264, 6744, 13933, 28785, 59470, 122865, 253838, 524428, 1083466, 2238435, 4624595, 9554390, 19739321, 40781336, 84254032, 174068400, 359624425, 742982225, 1534997482, 3171296957, 6551883314, 13536157460

Offset: 0

Rhodes Peele (rpeele(AT)mail.aum.edu), May 22 2007

(1) Bryce Duncan and I found this sequence while studying a graph invariant we call the Bell number. For a simple graph G = (V,E), we define B(G) to be the number of partitions P of V in which each block of P is an independent set of G. The sequence considered here results from choosing V = {1, 2, ..., n} and E = {(i,j) : |i - j| > 2}. (The classical Bell numbers B(n) come from the edgeless graph on n vertices.)
(2) The constant r in the formula is the dominant root of the characteristic equation of a linear homogeneous recurrence relation that also defines {a(n)}. (This recurrence relation, along with initial conditions, appears in the Mathematica program given below. The formula for a(n) is analogous to one version of the Binet formula for the Fibonacci numbers, namely F(n) = the integer nearest to (1/sqrt(5)) p^n where p is the golden mean. The shifted Fibonacci numbers F(n+1) are also graphical Bell numbers: replace the condition |i - j| > 2 with |i - j| > 1 to obtain them.
(3) Bell numbers for simple graphs satisfy the deletion-contraction identity B(G) = B(G\e) - B(G/e), but not the product identity B(G union H) = B(G)B(H) for disjoint graphs G and H. However, we do have B(B join H) = B(G)B(H) for the join of graphs G and H. The join graph G join H results from adding to their disjoint union, all possible edges that join a vertex of G to a vertex of H.
Diagonal sums of triangle A158687. - Paul Barry, Mar 24 2009
a(n) is the number of compositions (ordered partitions) of n into parts 1, 2, 3, and 4 where there are two kinds of part 3. - Joerg Arndt, Sep 16 2016
a(n) is the number of ways to tile a skew double-strip of n cells using squares, "double", and "triangular triple" tiles. Here is the skew double-strip corresponding to n=12, with 12 cells:
   _ ___ _ ___ _ ___
  |   |   |   |   |   |   |
 |__|___|_|___| |___|
|   |   |   |   |   |   |
|_|___|_|___|_|___|,
and here are the three possible "double" tiles and two possible "triangular triple" tiles:
   _    _                     _      _____
  |   |  |   |                   |   |    |       |
 |  |  |  |     _____     |   |   |     |
|   |      |   |   |       |   |       |    |   |
|_|,     |_|,  |_____|,  |_____|,   |_|
As an example, here is one of the b(12) = 3264 ways to tile the skew double-strip of 12 cells:
   _ ___ _____ _______
  |   |   |       |   |   |
 |__|_  |    |   | _|
|       |   |   |       |
|_____|___|_|___ _|. - Greg Dresden and Ruotong Li, Jun 12 2024

			a(4) = 10, with set partitions {{1},{2},{3}, {4}}; {{1,2}, {3}, {4}}; {{1,3}, {2}, {4}}; {{1}, {2,3}, {4}}; {{1}, {2,4}, {3}}; {{1}, {2}, {3,4}}; {{1,2,3}, {4}}; {{1}, {2,3,4}}; {{1,2}, {3,4}} and {{1,3}, {2,4}}.

Herbert S. Wilf, Generatingfunctiononogy (2nd Edition), Academic Press 1990, pp. 1 - 10 and pp. 20 - 23.
Arthur T. Benjamin and Jennifer J. Quinn, Proofs that Really Count, Dolciani Mathematical Expositions (MAA), (2003), p. 1. (A relevant combinatorial interpretation of Fibonacci numbers.)

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Kassie Archer, Ethan Borsh, Jensen Bridges, Christina Graves, and Millie Jeske, Cyclic permutations avoiding patterns in both one-line and cycle forms, arXiv:2312.05145 [math.CO], 2023. See p. 2.
B. Duncan and R. Peele, Bell and Stirling numbers for graphs, JIS 12 (2009), Article 09.7.1.
W. Kuszmaul, Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations, arXiv preprint arXiv:1509.08216 [cs.DM], 2015.
W. Kuszmaul, Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations, Mathematics of Computation 87 (2018), 987-1011.

Cf. A000110, A000045, A141073, A158687, A276893.

Column k=3 of A276719.

a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <1|2|1|1>>^n)[4, 4]:
seq(a(n), n=0..35);  # Alois P. Heinz, Sep 15 2016

a[1] = 1; a[2] = 2; a[3] = 5; a[4] = 10
a[n_] := a[n] = a[n-1] + a[n-2] + 2 a[n-3] + a[n-4]
Table[a[n],{n,1,30}]

a(n) = the integer nearest to s r^n, where r = 2.0659948920 ... and s = 0.53979687305... . (More information in comment (2).)
a(n) = Sum_{k=0..floor(n/2)} Sum_{j=0..n-2k} C(n-j-k,k)*C(2k,j). - Paul Barry, Mar 24 2009
G.f.: 1/(1 - x - x^2 - 2*x^3 - x^4). - Colin Barker, May 02 2012
a(n) = a(n-1) + a(n-2) + 2*(n-3) + a(n-4) with a(0) = a(1) = 1, a(2) = 2, a(3) = 5. - Taras Goy, Aug 04 2017
a(2*n) = a(n)^2 - a(n-1)^2 + a(n-2)^2 + 2*a(n-1)*(a(n+1)-a(n)). - Greg Dresden, Jul 03 2024

a(0)=1 prepended by Alois P. Heinz, Sep 15 2016

A309462 Limiting row sequence for Pascal-like triangle A140995 (Stepan's triangle with index of asymmetry s = 3).

Original entry on oeis.org

1, 2, 4, 8, 17, 35, 72, 149, 308, 636, 1314, 2715, 5609, 11588, 23941, 49462, 102188, 211120, 436173, 901131, 1861732, 3846329, 7946496, 16417420, 33918306, 70075047, 144774689, 299103768, 617946857, 1276675050, 2637604132, 5449276664, 11258177753, 23259337731

Offset: 0

Petros Hadjicostas, Aug 03 2019

In the attached photograph, we see that the index of asymmetry is denoted by s and the index of obliqueness by e. 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 arrays A141020 (with e = 0) and A141021 (with e = 1). In some of these arrays, the indices n and k are shifted.
For the triangular array G(n, k) = A140995(n, k), 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).
With G(n, k) = A140995(n, k), the current sequence (a(k): k >= 0) is defined by a(k) = lim_{n -> infinity} G(n, k) for k >= 0. Then a(k) = a(k-4) + 2*a(k-3) + a(k-2) + a(k-1) for k >= 4 with a(x) = 2^x for x = 0..3.

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, A141065, A141066, A141067, A141068, A141069, A141070, A141072, A141073, A308808.

a(0) = 1, a(1) = 2, a(2) = 4, a(3) = 8, and a(k) = a(k-1) + a(k-2) + 2*a(k-3) + a(k-4) for k >= 4.

G.f.: (x^2 + x + 1)/(-x^4 - 2*x^3 - x^2 - x + 1).

cp's OEIS Frontend

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

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

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A141070 Number of primes in rows of Pascal-like triangles with index of asymmetry y = 3 and index of obliquity z = 0 or z = 1.

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A141072 Sum of diagonal numbers in a Pascal-like triangle with index of asymmetry y = 3 and index of obliquity z = 0 (going upwards, left to right).

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A129847 a(n) = number of set partitions of {1, 2, ..., n} whose blocks consist only of elements that differ by two or less (that is, have only the forms {i}, {i,i+1}, {i,i+2}, or {i,i+1,i+2}).