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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.
1
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
0,4
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).
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).
EXAMPLE
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;
MATHEMATICA
Flatten[Table[(LCM@@Range(1, n))/Binomial[n, k], {n, 0, 11}, {k, 0, n}]]
PROG
(PARI) row(n) = vector(n+1, k, k--; lcm([1..n])/binomial(n, k)); \\ Michel Marcus, Sep 13 2021
CROSSREFS
KEYWORD
nonn,tabl,easy
AUTHOR
Gary Waters, Sep 06 2021
STATUS
approved