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%I #22 Mar 12 2015 05:20:10
%S 1,1,1,1,2,1,1,3,3,1,1,2,6,2,1,1,5,10,10,5,1,1,1,5,10,5,1,1,1,7,7,35,
%T 35,7,7,1,1,2,14,14,70,14,14,2,1,1,3,6,42,42,42,42,6,3,1,1,1,3,6,42,
%U 42,42,6,3,1,1,1,11,11,33,66,462,462,66,33,11,11,1
%N Reduced Pascal triangle: C_R(n,m) = A003418(n) / max(A003418(m), A003418(n-m)), m=0,...,n.
%C The sixth row is the first one which differs from triangles A080381, A080396.
%H Alois P. Heinz, <a href="/A204087/b204087.txt">Rows n = 0..140, flattened</a>
%e Triangle begins:
%e n/m.|..0.....1.....2.....3.....4.....5.....6.....7
%e ==================================================
%e .0..|..1
%e .1..|..1.....1
%e .2..|..1.....2.....1
%e .3..|..1.....3.....3.....1
%e .4..|..1.....2.....6.....2.....1
%e .5..|..1.....5....10....10.....5.....1
%e .6..|..1.....1.....5....10.....5.....1.....1
%e .7..|..1.....7.....7....35....35.....7.....7.....1
%p g:= proc(n) option remember; `if`(n=0, 1, ilcm(g(n-1), n)) end:
%p CR:= proc(n, m) option remember; g(n)/max(g(m), g(n-m)) end:
%p seq (seq (CR(n,m), m=0..n), n=0..11); # _Alois P. Heinz_, Jan 11 2012
%t g[n_] := g[n] = If[n == 0, 1, LCM[g[n-1], n]]; CR[n_, m_] := CR[n, m] = g[n]/Max[ g[m], g[n-m]]; Table[Table[CR[n, m], {m, 0, n}], {n, 0, 11}] // Flatten (* _Jean-François Alcover_, Mar 12 2015, after _Alois P. Heinz_ *)
%Y Cf. A007318, A080381, A080396, A186430, A202917, A202941.
%K nonn,tabl
%O 0,5
%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Jan 10 2012
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