A204087 Reduced Pascal triangle: C_R(n,m) = A003418(n) / max(A003418(m), A003418(n-m)), m=0,...,n.

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, 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, 42, 42, 6, 3, 1, 1, 1, 11, 11, 33, 66, 462, 462, 66, 33, 11, 11, 1

Offset

0,5

Comments

The sixth row is the first one which differs from triangles A080381, A080396.

Links

Alois P. Heinz, Rows n = 0..140, flattened

Example

Triangle begins:
n/m.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1.....2.....1
.3..|..1.....3.....3.....1
.4..|..1.....2.....6.....2.....1
.5..|..1.....5....10....10.....5.....1
.6..|..1.....1.....5....10.....5.....1.....1
.7..|..1.....7.....7....35....35.....7.....7.....1

Maple

g:= proc(n) option remember; `if`(n=0, 1, ilcm(g(n-1), n)) end:
CR:= proc(n, m) option remember; g(n)/max(g(m), g(n-m)) end:
seq (seq (CR(n, m), m=0..n), n=0..11); # Alois P. Heinz, Jan 11 2012

Mathematica

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 *)

Crossrefs

Cf. A007318, A080381, A080396, A186430, A202917, A202941.

Sequence in context: A065133 A343033 A307239 * A080381 A080396 A155582

Adjacent sequences: A204084 A204085 A204086 * A204088 A204089 A204090

Keyword

nonn,tabl

Author

Vladimir Shevelev and Peter J. C. Moses, Jan 10 2012

Status

approved