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%I #22 Jul 04 2017 09:27:22
%S 1,1,1,2,1,1,3,2,1,1,5,3,2,1,1,6,4,2,1,1,1,8,5,3,2,1,1,1,9,6,3,2,1,1,
%T 1,1,11,7,4,3,2,1,1,1,1,14,9,5,4,2,2,1,1,1,1,15,10,6,4,2,2,1,1,1,1,1,
%U 18,12,7,5,3,2,2,1,1,1,1,1,20,13,8,5,3,3,2,2,1,1,1,1,1
%N Triangle T(n,k) read by rows: T(n,k) = floor(prime(n)/prime(k)), n >= k >= 1.
%C Alternate name: triangle of quotients of prime(n)/prime(k), n >= k >= 1.
%e Triangle starts:
%e n/k 1 2 3 4 5 6 7 8 9 10 11 12
%e 1 1
%e 2 1 1
%e 3 2 1 1
%e 4 3 2 1 1
%e 5 5 3 2 1 1
%e 6 6 4 2 1 1 1
%e 7 8 5 3 2 1 1 1
%e 8 9 6 3 2 1 1 1 1
%e 9 11 7 4 3 2 1 1 1 1
%e 10 14 9 5 4 2 2 1 1 1 1
%e 11 15 10 6 4 2 2 1 1 1 1 1
%e 12 18 12 7 5 3 2 2 1 1 1 1 1
%e T(11,3) = 6 because prime(11) = 31 and prime(3) = 5, and floor(31/5) = 6.
%o (PARI) T(n,k) = prime(n)\prime(k);
%o tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n, k), ", ")); print()); \\ _Michel Marcus_, Jun 06 2017
%Y Cf. A000040 (primes), A130290 (1st column), A144769 (2nd column), A116572 (3rd column).
%K nonn,tabl
%O 1,4
%A _Bob Selcoe_, Jun 02 2017
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