OFFSET

1,1

COMMENTS

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)

LINKS

Petros Hadjicostas, Table of n, a(n) for n = 1..64

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

EXAMPLE

Pascal-like triangle with y = 2 and z = 0 (i.e., A140997) begins as follows:

1, so no primes.

1 1, so no primes.

1 2 1, then a(1) = 2.

1 4 2 1, so no new primes.

1 8 4 2 1, so no new primes.

1 15 9 4 2 1, so no new primes.

1 28 19 9 4 2 1, so a(2) = 19.

1 52 40 19 9 4 2 1, so no new primes.

1 96 83 41 19 9 4 2 1, so a(3) = 41 and a(4) = 83.

1 177 170 88 41 19 9 4 2 1, so no new primes.

1 326 345 188 88 41 19 9 4 2 1, so no new primes.

1 600 694 400 189 88 41 19 9 4 2 1, so no new primes.

... [edited by Petros Hadjicostas, Jun 12 2019]

Terms a(5) = 3719 and a(6) = 5431 appear in row k = 14, while terms a(7) = 1873 and a(8) = 3989 appear in row k = 15.

MAPLE

# 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:

A141067 := 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 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:

A141067(50);

# If one wishes to get the primes sorted, then he or she should replace RETURN(a) in the above Maple code with RETURN(sort(a)). In such a case, however, the sequence is not uniquely defined because it depends on the maximum n. - Petros Hadjicostas, Jun 15 2019

CROSSREFS

KEYWORD

nonn

AUTHOR

Juri-Stepan Gerasimov, Jul 14 2008

EXTENSIONS

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

STATUS

approved