%I #11 Dec 05 2019 19:13:25
%S 1,3,2,4,3,3,7,6,4,4,6,5,5,5,5,12,11,9,6,6,6,8,7,7,7,7,7,7,15,14,12,
%T 12,8,8,8,8,13,12,12,9,9,9,9,9,9,18,17,15,15,15,10,10,10,10,10
%N A127093 * A000012.
%C Row sums = A001157: (1, 5, 10, 21, 26, 50, ...). Left border = sigma(n), A000203.
%C From _Lechoslaw Ratajczak_, Nov 01 2019: (Start)
%C Let b_n(k) (n = 1,2,3,...) be consecutive finite sequences defined as follows: b_n(k) is the sum of all integers u satisfying the equation: n mod u = k (1 <= u <= n, k = 0,1,2,...,ceiling(n/2)-1). These sequences are consecutive antidiagonals of the triangle (b_n(k) = T(n-k,k+1)).
%C The example for n = 8 (k_max = ceiling(8/2) - 1 = 3):
%C - b_8(0) = T(8-0,0+1) = T(8,1) = 15 = sigma(8) because 8 mod {1,2,4,8} = 0 and 1+2+4+8 = 15;
%C - b_8(1) = T(8-1,1+1) = T(7,2) = 7 = A039653(8-1) because 8 mod 7 = 1;
%C - b_8(2) = T(8-2,2+1) = T(6,3) = 9 because 8 mod {3,6} = 2 and 3+6 = 9;
%C - b_8(3) = T(8-3,3+1) = T(5,4) = 5 because 8 mod 5 = 3.
%C Conjecture: let P(n) be the n-th antidiagonal product (P(n) = Product_{k=0..ceiling(n/2)-1} b_n(k)). Consecutive n satisfying two equations: gcd(P(n),n) = 1 and gcd(P(n+1),n+1) = 2 are consecutive elements of A005383 (primes p such that (p+1)/2 are also primes, save A005383(1) = 3 and A005383(2) = 5). The conjecture is false if for any prime number p belonging to A005383 gcd(P(p),p) = p. The conjecture was checked for 2000 consecutive integers. (End)
%F A127093 * A000012 as infinite lower triangular matrices. Triangle read by rows, partial sums of A127093 terms starting from the right.
%e First few rows of the triangle are:
%e 1;
%e 3, 2;
%e 4, 3, 3;
%e 7, 6, 4, 4;
%e 6, 5, 5, 5, 5;
%e 12, 11, 9, 6, 6, 6;
%e 8, 7, 7, 7, 7, 7, 7;
%e 15, 14, 12, 12, 8, 8, 8, 8;
%e ...
%Y Cf. A127093, A001157, A000203.
%K nonn,tabl
%O 1,2
%A _Gary W. Adamson_, Oct 31 2007