A286942 Irregular triangle read by rows: numbers 1 <= k <= (A002110(n) - 1) where gcd(k, A002110(n - 1)) = 1.

1, 2, 1, 3, 5, 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83, 89, 91, 97, 101, 103, 107, 109, 113, 119, 121, 127, 131, 133, 137, 139, 143, 149, 151, 157, 161, 163, 167, 169, 173

Offset: 1

Author

Jamie Morken and Michael De Vlieger, May 16 2017

Keywords

• nonn
• tabf

Comments

From Michael De Vlieger, May 18 2017: (Start)

Row n of a(n) is the list of numbers 1 <= k <= A002110(n) that are coprime to A002110(n-1).

A286941(n) and A279864(n) are subsets of a(n) such that the terms of the rows of each sequence combined and sorted comprise all the terms of a(n).

Row lengths = A005867(n) + A005867(n-1): {2, 3, 10, 56, 528, 6240, 97920, ...}.

1 is coprime to all n thus delimits the rows of a(n).

The smallest prime q in row n of a(n) is gpf(primorial(n)) = A006530(A002110(n)) = prime(n) by definition of primorial.

The smallest composite x in row n of a(n) is q^2 = A001248(n).

The Kummer number A057588(n) = A002110(n) - 1 is the largest term in row n of a(n). (End)

Examples

			The triangle starts:
1, 2;
1, 3, 5;
1, 5, 7, 11, 13, 17, 19, 23, 25, 29
Example1:
To find row n of the irregular triangle A286942, take a running sum for each value in the irregular triangle row n-1 of A286941 with A002110(n-1) b-1 times, where b is the largest prime factor in A002110(n).
For example to find row 3 of A286942: Take a running sum for both 1 and 5 in row n-1 of A286941 with A002110(3-1)=6, 5-1=4 times, where b is the largest prime factor 5 in A002110(3).
Result:
1 5
7 11
13 17
19 23
25 29
Equal to row 3 of A286942: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29.
Example2:
To find row n of the irregular triangle A279864, multiply each value in row n-1 of A286941 with the largest prime factor b in A002110(n).
Example for n=3: b=5.
1*5=5
5*5=25
Example3:
To find row n of the irregular triangle A286941, remove the values that are in row n of the irregular triangle A279864 from the values that are in row n of the irregular triangle A286942.
For n=3.
A286942 row n = 1, 5, 7, 11, 13, 17, 19, 23, 25, 29.
A279864 row n = 5, 25.
Removing values 5, 25 from the values in A286942 row n gives row n of A286941: 1, 7, 11, 13, 17, 19, 23, 29.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..6839 (rows 1 <= n <= 6).
• Eric Weisstein's World of Mathematics, Relatively Prime - _Michael De Vlieger_, May 18 2017
• Seqfan, Formula for the sequence.

Crossrefs

Cf. A002110, A005867, A279864, A286941, A286942.

Programs

Mathematica

Table[Select[Range@ #2, Function[k, CoprimeQ[k, #1]]] & @@ Map[Times @@ # &, {Most@ #, #}] &@ Prime@ Range@ n, {n, 4}] // Flatten (* Michael De Vlieger, May 18 2017 *)

Formula

a(n) = union(A286941(n), A279864(n)) where n consists of all terms in row n of each sequence. - Michael De Vlieger, May 18 2017

Extensions

More terms from Michael De Vlieger, May 18 2017