A080396 -id:A080396 - cp's OEIS frontend

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

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

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

			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

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

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

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

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 *)

A307239 Analog of Pascal's triangle, with A007947 applied to each sum.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 2, 6, 2, 1, 1, 3, 2, 2, 3, 1, 1, 2, 5, 2, 5, 2, 1, 1, 3, 7, 7, 7, 7, 3, 1, 1, 2, 10, 14, 14, 14, 10, 2, 1, 1, 3, 6, 6, 14, 14, 6, 6, 3, 1, 1, 2, 3, 6, 10, 14, 10, 6, 3, 2, 1, 1, 3, 5, 3, 2, 6, 6, 2, 3, 5, 3, 1, 1, 2, 2, 2, 5, 2, 6, 2, 5, 2, 2, 2, 1

Offset: 0

Rémy Sigrist, Apr 01 2019

The parity of the terms in this triangle is the same as in Pascal's triangle (A007318). As a consequence, the number of odd terms in row n is A001316(n).

The distribution of the terms different from 2 in the triangle evokes Sierpinski's triangle; this is also the case for terms that are multiples of 3 (see illustrations in Links section).

			Triangle begins:
  0:                       1
  1:                     1   1
  2:                   1   2   1
  3:                 1   3   3   1
  4:               1   2   6   2   1
  5:             1   3   2   2   3   1
  6:           1   2   5   2   5   2   1
  7:         1   3   7   7   7   7   3   1
  8:       1   2  10  14  14  14  10   2   1
  9:     1   3   6   6  14  14   6   6   3   1
  ...

Rémy Sigrist, Colored representation of the first 1500 rows (where the color is function of log(T(n,k)))
Rémy Sigrist, Representation of the 2's in the first 1500 rows (black pixels correspond to 2's)
Rémy Sigrist, Representation of the multiples of 3 in the first 1500 rows (black pixels correspond to multiples of 3)

Cf. A001316, A007318, A007947, A080396, A307356 (number of 2's in rows).

rad(n) = my (p=factor(n)[,1]~); prod(i=1, #p, p[i])
{ for (r=0, 12, row = vector(r+1, k, if ( k==1||k==r+1, 1, rad(row[k-1]+row[k]))); for (c=1, #row, print1 (row[c] ", "))) }

A098426 a(n)=(1/2)*sum(i=0,n,rad(binomial(n,i))) where rad(k)=A007947(k).

Original entry on oeis.org

1, 2, 4, 6, 16, 27, 64, 66, 94, 287, 1024, 645, 2380, 6723, 12094, 7136, 36772, 38733, 161254, 241083, 571714, 1203173, 4194304, 2005009, 2162496, 8739877, 6699838, 27330423, 112844656, 41322885, 252807388, 201501336, 419398108, 2840231975

Offset: 1

Benoit Cloitre, Oct 26 2004

nonn

For n>1 a(n) odd seems to imply 2*phi(n)

Halved row sums of A080396.

A048686 Number of classes generated by function A007947 when applied to binomial coefficients.

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 4, 4, 4, 6, 6, 6, 7, 8, 8, 9, 8, 10, 8, 10, 11, 12, 12, 12, 13, 12, 12, 15, 15, 16, 14, 14, 14, 15, 16, 19, 18, 18, 18, 21, 19, 22, 21, 22, 23, 24, 24, 24, 22, 24, 25, 27, 23, 25, 24, 25, 29, 29, 28, 31, 31, 32, 30, 28, 29, 33, 31, 32, 34, 36, 35, 37, 36, 35, 36

Offset: 1

Labos Elemer

nonn

			For n=9, A007947({C(9,k)}) = {1,3,6,42,42,42,42,6,3,1} includes 4 distinct values, thus generating 4 classes of k values: {0,9}, {1,8}, {2,7} and {3,4,5,6}. So a(9)=4.

Cf. A007947, A080396.

a[n_] := Length[ Union[ Table[ A007947[ Binomial[ n, k ] ], {k, 0, n} ] ] ]

a(n) = #Set(vector(ceil(n\2)+1, k, factorback(factorint(binomial(n,k-1))[, 1]))); \\ Michel Marcus, May 20 2018

Showing 1-4 of 4 results.

cp's OEIS Frontend

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

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A307239 Analog of Pascal's triangle, with A007947 applied to each sum.

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PARI

A098426 a(n)=(1/2)*sum(i=0,n,rad(binomial(n,i))) where rad(k)=A007947(k).

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A048686 Number of classes generated by function A007947 when applied to binomial coefficients.

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Mathematica

PARI