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A252941 Irregular triangle T(n,k) read by rows: T(1,1) = 1, otherwise row n lists the prime factors of A098550(n), with duplicates omitted.

%I A252941 #36 Mar 30 2015 21:52:34
%S A252941 1,2,3,2,3,2,3,5,2,7,5,2,3,5,2,3,5,7,2,7,2,5,3,7,2,5,3,2,11,3,13,11,
%T A252941 13,3,11,2,13,3,5,2,7,3,13,2,17,2,3,5,17,2,3,5,11,2,17,5,13,2,3,7,13,
%U A252941 2,3,5,7,2,19,3,7,19,2,3,7,5,19,2,11
%N A252941 Irregular triangle T(n,k) read by rows: T(1,1) = 1, otherwise row n lists the prime factors of A098550(n), with duplicates omitted.
%C A252941 Row n contains the distinct prime factors of A098550(n), in increasing order. For example, when n=13, A098550(13) = 35 and T(13,k) = [5,7].
%C A252941 Because A098550 is a permutation of the natural numbers, this sequence is infinite and contains every prime infinitely often.
%C A252941 Primes appear in order; that is, first appearance of prime(j) occurs prior to first appearance of prime(j+1).
%C A252941 T(n,1) = A251101(n), which are the smallest prime factors of A098550(n), n>1.
%C A252941 For n>1, let each coefficient in T(n,1) be prime(i). The ratio that each coefficient appears in T(j,1) {j=1..n} approaches A038110(i)/A038111(i) as j increases.  For example, prime(4) = 7: as j increases, the ratio that 7 appears in T(j,1) approaches 4/105, because A038110(4)/A038111(4) = 4/105.
%H A252941 David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, <a href="http://arxiv.org/abs/1501.01669">The Yellowstone Permutation</a>, arXiv preprint arXiv:1501.01669, 2015.
%e A252941 Triangle begins T(1,1):
%e A252941 1
%e A252941 2
%e A252941 3
%e A252941 2
%e A252941 3
%e A252941 2
%e A252941 3 5
%e A252941 2 7
%e A252941 5
%e A252941 2 3
%e A252941 5
%e A252941 2 3
%e A252941 5 7
%e A252941 2
%e A252941 7
%e A252941 2 5
%e A252941 3 7
%e A252941 2 5
%e A252941 3
%e A252941 2 11
%e A252941 e.g., n=13: A098550(13) = 35; T(13,k) = 5,7.
%Y A252941 Cf. A098550.
%K A252941 nonn,tabf
%O A252941 1,2
%A A252941 _Bob Selcoe_, Mar 22 2015