A347563 Binomial complement triangle, T(n,k) = LCM(1,...,n)/binomial(n,k) for 0 <= k <= n, a(0) = T(0,0) = 0, read by rows.

0, 1, 1, 2, 1, 2, 6, 2, 2, 6, 12, 3, 2, 3, 12, 60, 12, 6, 6, 12, 60, 60, 10, 4, 3, 4, 10, 60, 420, 60, 20, 12, 12, 20, 60, 420, 840, 105, 30, 15, 12, 15, 30, 105, 840, 2520, 280, 70, 30, 20, 20, 30, 70, 280, 2520

Offset: 0

Author

Gary Waters, Sep 06 2021

Keywords

• nonn
• tabl
• easy

Comments

The one's complement of each carry value, in base prime p, defined in Lucas's Theorem. Also works using Erdős's method (see formula below).

At row n of the triangle, the values are symmetrical with the largest values occurring at T(n,0) = T(n,n) = LCM(1,...,n). The smallest value(s) occur at k = n/2 when n is even, and at k = floor(n/2) and k = floor(n/2)+1 when n is odd. T(n,k) = T(n,n-k).

Conjecture: For all n, T(n,0) mod A213999(n-1,n-1) = 0, and T(n,k+1) mod A213999(n,k) = 0 for 0 <= k <= n-1 (computed and verified for rows = 0..2000).

Examples

			T(7,3) = 12. Triangle T(n,k) begins:
     0;
     1,   1;
     2,   1,  2;
     6,   2,  2,  6;
    12,   3,  2,  3, 12;
    60,  12,  6,  6, 12, 60;
    60,  10,  4,  3,  4, 10, 60;
   420,  60, 20, 12, 12, 20, 60, 420;
   840, 105, 30, 15, 12, 15, 30, 105, 840;
  2520, 280, 70, 30, 20, 20, 30,  70, 280, 2520;
		

Crossrefs

Cf. A003418, A007318, A213999.

Programs

Mathematica

Flatten[Table[(LCM@@Range(1,n))/Binomial[n, k], {n, 0, 11}, {k, 0, n}]]

PARI

row(n) = vector(n+1, k, k--; lcm([1..n])/binomial(n,k)); \\ Michel Marcus, Sep 13 2021

Formula

T(n,k) = Product_{p<=n} p^u_p, where u_p = i_max - Sum_{i=1..i_max} v_p(i) = Sum_{i=1..i_max} NOT(v_p(i)), with v_p(i) = floor(n/p^i) - floor(k/p^i) - floor((n-k)/p^i) = {0, 1} and i_max = floor(log(n)/log(p)) (using Erdős's method).