cp's OEIS Frontend

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.

%I A141070 #38 Jul 03 2019 23:14:58
%S A141070 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,
%T A141070 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,
%U A141070 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
%N 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.
%C A141070 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.)
%C A141070 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.)
%C A141070 From _Petros Hadjicostas_, Jun 13 2019: (Start)
%C A141070 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.
%C A141070 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.
%C A141070 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).
%C A141070 (End)
%H A141070 Juri-Stepan Gerasimov, <a href="/A140998/a140998.jpg">Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...</a>
%e A141070 Pascal-like triangle with y = 3 and z = 0 (i.e, A140996) begins as follows:
%e A141070   1, so a(1) = 0.
%e A141070   1 1, so a(2) = 0.
%e A141070   1 2 1, so prime 2 and a(3) = 1.
%e A141070   1 4 2 1, so prime 2 and a(4) = 1.
%e A141070   1 8 4 2 1, so prime 2 and a(5) = 1.
%e A141070   1 16 8 4 2 1, so prime 2 and a(6) = 1.
%e A141070   1 31 17 8 4 2 1, so primes 2, 17, 31 and a(7) = 3.
%e A141070   1 60 35 17 8 4 2 1, so primes 2, 17 and a(8) = 2.
%e A141070   1 116 72 35 17 8 4 2 1, so primes 2, 17 and a(9) = 2.
%e A141070   1 224 148 72 35 17 8 4 2 1, so primes 2, 17 and a(10) = 2.
%e A141070   1 432 303 149 72 35 17 8 4 2 1, so primes 2, 17, 149 and a(11) = 3.
%e A141070   ...
%t A141070 nlim = 100;
%t A141070 For[n = 0, n <= nlim, n++, G[n, 0] = 1];
%t A141070 For[n = 1, n <= nlim, n++, G[n, n] = 1];
%t A141070 For[n = 2, n <= nlim, n++, G[n, n-1] = 2];
%t A141070 For[n = 3, n <= nlim, n++, G[n, n-2] = 4];
%t A141070 For[n = 4, n <= nlim, n++, G[n, n-3] = 8];
%t A141070 For[n = 5, n <= nlim, n++, For[k = 1, k < n-3, k++,
%t A141070    G[n, k] = G[n-4, k-1] + G[n-4, k] + G[n-3, k] + G[n-2, k] +
%t A141070      G[n-1, k]]];
%t A141070 A141070 = {}; For[n = 0, n <= nlim, n++, c = 0;
%t A141070  For[k = 0, k <= n, k++, If[PrimeQ[G[n, k]], c++]];
%t A141070  AppendTo[A141070, c]];
%t A141070 A141070 (* _Robert Price_, Jul 03 2019 *)
%Y A141070 Cf. A140993, A140994, A140995, A140996, A140997, A140998, A141065, A141066, A141067, A141068, A141069, A141072, A141073.
%K A141070 nonn,more
%O A141070 1,7
%A A141070 _Juri-Stepan Gerasimov_, Jul 16 2008
%E A141070 Partially edited by _N. J. A. Sloane_, Jul 18 2008
%E A141070 More terms and comments edited by _Petros Hadjicostas_, Jun 13 2019
%E A141070 a(52)-a(100) from _Robert Price_, Jul 03 2019