A141065 List of different composite numbers in Pascal-like triangles with index of asymmetry y = 1 and index of obliqueness z = 0 or z = 1.

4, 12, 20, 28, 33, 46, 54, 63, 69, 88, 168, 70, 143, 161, 289, 232, 567, 594, 169, 376, 399, 817, 1194, 407, 609, 934, 1778, 1820, 2355, 408, 975, 986, 2150, 3789, 4570, 984, 1596, 2316, 4862, 5646, 7922, 8745, 985, 2367, 2583, 9849, 10801, 16281, 16532, 4180, 5667, 17091, 23585, 30923, 32948, 2378

Offset: 1

Author

Juri-Stepan Gerasimov, Jul 14 2008

Keywords

• nonn

Comments

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)

Examples

			Pascal-like triangle with y = 1 and z = 0 (i.e., A140998) 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 7 5 2 1, so no composites.
  1 12 11 5 2 1, so a(2) = 12.
  1 20 23 12 5 2 1, so a(3) = 20.
  1 33 46 28 12 5 2 1, so a(4) = 28, a(5) = 33, and a(6) = 46.
  1 54 89 63 29 12 5 2 1, so a(7) = 54 and a(8) = 63.
  1 88 168 137 69 29 12 5 2 1, so a(9) = 69, a(10) = 88, and a(11) = 168.
  1 143 311 289 161 70 29 12 5 2 1, so a(12) = 70, a(13) = 143, a(14) = 161, and a(15) = 289.
  1 232 567 594 367 168 70 29 12 5 2 1, so a(16) = 232, a(17) = 567, and a(18) = 594.
  ... [example edited by _Petros Hadjicostas_, Jun 11 2019]
		

Links

• Petros Hadjicostas, Table of n, a(n) for n = 1..120
• Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Crossrefs

Cf. A140993 (mirror image of A140998 with y = 1 and z = 1), A140994 (triangle when y = 2 and z = 1), A140995 (triangle when y = 3 and z = 1), A140996 (triangle when y = 3 and z = 0), A140997 (triangle when y = 2 and z = 0), A140998 (has the above triangle with y = 1 and z = 0), A141020, A141021, A141064 (has primes for y = 1), A141066 (has composites when y = 2), A141067 (has primes when y = 2), A141068 (has primes when y = 3), A141069 (has composites when y = 3).

Programs

Maple

# This is a modification of R. J. Mathar's program for A141031 (for the case y = 4 and z = 0).
# Construction of array A140998 (y = 1 and z = 0):
A140998 := 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; else procname(n - 1, k) + procname(n - 2, k) + procname(n - 2, k - 1); end if; end proc;
# Construction of the current sequence:
A141065 := 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 := A140998(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 terms of the current sequence:
A141065(24);
# If one wishes to sort composites, then one may replace RETURN(a) in the above Maple code 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

Extensions

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