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%I #42 Jun 17 2019 03:20:38
%S 2,5,7,11,23,29,89,137,311,367,1021,3217,5441,2377,12619,65761,5741,
%T 144593,13859,78511,1462397,33461,469957,2552939,11096497,5930669,
%U 6343133,26512597,470831,127626137,372222703,15955507,538270693,531077333,11401285549,38613943,15433507333,92554537183,113828092793
%N List of different primes in Pascal-like triangles with index of asymmetry y = 1 and index of obliquity 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 In each row of A140998, the primes not appearing in earlier rows are collected, sorted, and added to the sequence. [_R. J. Mathar_, Apr 28 2010]
%C From Petros Hadjicostas, Jun 10 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 Hence, 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 (paraphrasing R. J. Mathar above!).
%C (End)
%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 prime.
%e 1 1, so no primes.
%e 1 2 1, so a(1) = 2.
%e 1 4 2 1, so no new primes.
%e 1 7 5 2 1, so a(2) = 5 and a(3) = 7.
%e 1 12 11 5 2 1, so a(4) = 11.
%e 1 20 23 12 5 2 1, so a(5) = 23.
%e 1 33 46 28 12 5 2 1, so no new primes.
%e 1 54 89 63 29 12 5 2 1, so a(6) = 29 and a(7) = 89.
%e 1 88 168 137 69 29 12 5 2 1, so a(8) = 137.
%e 1 143 311 289 161 70 29 12 5 2 1, so a(9) = 311.
%e 1 232 567 594 367 168 70 29 12 5 2 1, so a(10) = 367.
%e ...
%e [edited by _Petros Hadjicostas_, Jun 11 2019]
%p # This is a modification _R. J. Mathar_'s program from A141031 (for the case y = 4 and z = 0).
%p # Construct 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 # Construct the current sequence:
%p A141064 := 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 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;
%p # Generate terms of the current sequence:
%p A141064(38);
%p # If one wants to get the primes sorted, then replace RETURN(a) in the Maple 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
%Y Cf. A140993, A140994, A140995, A140996, A140997, A140998, A141021, A141022, A141031, A141066, A141067.
%K nonn
%O 1,1
%A _Juri-Stepan Gerasimov_, Jul 14 2008
%E Partially edited by _N. J. A. Sloane_, Jul 18 2008
%E More terms from _R. J. Mathar_, Apr 28 2010
%E More terms from _Petros Hadjicostas_, Jun 11 2019
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