%I #33 Oct 09 2021 16:07:38

%S 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,

%T 420,60,20,12,12,20,60,420,840,105,30,15,12,15,30,105,840,2520,280,70,

%U 30,20,20,30,70,280,2520

%N 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.

%C 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).

%C 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).

%C 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).

%F 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).

%e T(7,3) = 12. Triangle T(n,k) begins:

%e 0;

%e 1, 1;

%e 2, 1, 2;

%e 6, 2, 2, 6;

%e 12, 3, 2, 3, 12;

%e 60, 12, 6, 6, 12, 60;

%e 60, 10, 4, 3, 4, 10, 60;

%e 420, 60, 20, 12, 12, 20, 60, 420;

%e 840, 105, 30, 15, 12, 15, 30, 105, 840;

%e 2520, 280, 70, 30, 20, 20, 30, 70, 280, 2520;

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

%o (PARI) row(n) = vector(n+1, k, k--; lcm([1..n])/binomial(n,k)); \\ _Michel Marcus_, Sep 13 2021

%Y Cf. A003418, A007318, A213999.

%K nonn,tabl,easy

%O 0,4

%A _Gary Waters_, Sep 06 2021