A134559 - OEIS

A134559

A127093 * A000012.

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, 12, 8, 8, 8, 8, 13, 12, 12, 9, 9, 9, 9, 9, 9, 18, 17, 15, 15, 15, 10, 10, 10, 10, 10

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OFFSET

1,2

COMMENTS

Row sums = A001157: (1, 5, 10, 21, 26, 50, ...). Left border = sigma(n), A000203.

From Lechoslaw Ratajczak, Nov 01 2019: (Start)

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)).

The example for n = 8 (k_max = ceiling(8/2) - 1 = 3):

- 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;

- b_8(1) = T(8-1,1+1) = T(7,2) = 7 = A039653(8-1) because 8 mod 7 = 1;

- b_8(2) = T(8-2,2+1) = T(6,3) = 9 because 8 mod {3,6} = 2 and 3+6 = 9;

- b_8(3) = T(8-3,3+1) = T(5,4) = 5 because 8 mod 5 = 3.

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)

LINKS

Table of n, a(n) for n=1..55.

FORMULA

A127093 * A000012 as infinite lower triangular matrices. Triangle read by rows, partial sums of A127093 terms starting from the right.

EXAMPLE

First few rows of the triangle are:

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;