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%I #21 Aug 09 2015 02:10:39
%S 1,3,5,1,7,3,9,5,1,11,7,3,13,9,5,1,15,11,7,3,17,13,9,5,1,19,15,11,7,3,
%T 21,17,13,9,5,1,23,19,15,11,7,3,25,21,17,13,9,5,1,27,23,19,15,11,7,3,
%U 29,25,21,17,13,9,5,1,31,27,23,19,15,11,7,3,33
%N Triangle read by rows: T(n,k) = 2*n-4*k+5 (n >= 0, 1 <= k <= 1+floor(n/2)).
%C All terms are odd, decreasing across rows. Row sums = A000217, the triangular numbers.
%C From _Johannes W. Meijer_, Sep 08 2013: (Start)
%C Triangle read by rows formed from the antidiagonals of triangle A099375.
%C The alternating row sums equal A098181(n). (End)
%H Nathaniel Johnston, <a href="/A152204/b152204.txt">Rows n = 0..200 of irregular triangle, flattened</a>
%F By columns, odd terms in every column, n-th column starts at row (2*n).
%F From _Johannes W. Meijer_, Sep 08 2013: (Start)
%F T(n, k) = A099375(n-k+1, k-1), n >= 0 and 1 <= k <= 1+floor(n/2)).
%F T(n, k) = A158405(n+1, n-2*k+2). (End)
%e First few rows of the triangle =
%e 1
%e 3
%e 5 1
%e 7 3
%e 9 5 1
%e 11 7 3
%e 13 9 5 1
%e 15 11 7 3
%e 17 13 9 5 1
%e 19 15 11 7 3
%e 21 17 13 9 5 1
%e ...
%p T := proc(n,k) return 2*n-4*k+5: end: seq(seq(T(n,k), k=1..1+floor(n/2)), n=0..20); # _Nathaniel Johnston_, May 01 2011
%Y Cf. A000217.
%K nonn,tabf,easy
%O 0,2
%A _Gary W. Adamson_, Nov 29 2008
%E Edited by _N. J. A. Sloane_, Sep 25 2010, following a suggestion from Emeric Deutsch
%E Offset corrected by _Johannes W. Meijer_, Sep 07 2013
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