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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. 15

%I #89 Jan 28 2024 14:21:49

%S 4,12,20,28,33,46,54,63,69,88,168,70,143,161,289,232,567,594,169,376,

%T 399,817,1194,407,609,934,1778,1820,2355,408,975,986,2150,3789,4570,

%U 984,1596,2316,4862,5646,7922,8745,985,2367,2583,9849,10801,16281,16532,4180,5667,17091,23585,30923,32948,2378

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

%C 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).

%C 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).

%C From _Petros Hadjicostas_, Jun 09 2019: (Start)

%C 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.

%C 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.

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

%C 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.

%C 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.

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

%C 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.

%C 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.

%C 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.

%C 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.

%C 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).

%C 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).

%C (End)

%H Petros Hadjicostas, <a href="/A141065/b141065.txt">Table of n, a(n) for n = 1..120</a>

%H Juri-Stepan Gerasimov, <a href="/A140998/a140998.jpg">Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...</a>

%e Pascal-like triangle with y = 1 and z = 0 (i.e., A140998) begins as follows:

%e 1, so no composites.

%e 1 1, so no composites.

%e 1 2 1, so no composites.

%e 1 4 2 1, so a(1) = 4.

%e 1 7 5 2 1, so no composites.

%e 1 12 11 5 2 1, so a(2) = 12.

%e 1 20 23 12 5 2 1, so a(3) = 20.

%e 1 33 46 28 12 5 2 1, so a(4) = 28, a(5) = 33, and a(6) = 46.

%e 1 54 89 63 29 12 5 2 1, so a(7) = 54 and a(8) = 63.

%e 1 88 168 137 69 29 12 5 2 1, so a(9) = 69, a(10) = 88, and a(11) = 168.

%e 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.

%e 1 232 567 594 367 168 70 29 12 5 2 1, so a(16) = 232, a(17) = 567, and a(18) = 594.

%e ... [example edited by _Petros Hadjicostas_, Jun 11 2019]

%p # This is a modification of _R. J. Mathar_'s program for A141031 (for the case y = 4 and z = 0).

%p # Construction of array A140998 (y = 1 and z = 0):

%p 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;

%p # Construction of the current sequence:

%p 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;

%p # Generation of terms of the current sequence:

%p A141065(24);

%p # 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

%Y 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).

%K nonn

%O 1,1

%A _Juri-Stepan Gerasimov_, Jul 14 2008

%E Partially edited by _N. J. A. Sloane_, Jul 18 2008

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Last modified March 29 04:23 EDT 2024. Contains 371264 sequences. (Running on oeis4.)