A244348 Integers n such that for every integer k>0, n*10^k+1 has a divisor in the set { 11, 73, 101, 137 }.

162207, 1622070, 3349554, 5109589, 6651446, 7001622, 9589051, 10958905, 11273318, 12733181, 14460665, 16220700, 17762557, 18112733, 20700162, 22070016, 22384429, 23844292, 25571776, 27331811, 28873668, 29223844, 31811273, 33181127, 33495540, 34955403, 36682887

Offset: 1

Author

Pierre CAMI, Jun 28 2014

Keywords

• nonn
• easy

Comments

For n > 8, a(n) = a(n-8) + 11111111, the first 8 values are in the data.

If n is of the form 3*m+2, n*10^k+1 is always divisible by 3 but also has a divisor in the set { 11, 73, 101, 137 }.

Examples

			Consider n = 162207.
If k is of the form 2*j+1, n*10^(2*j+1)+1 is divisible by 11.
If k is of the form 8*j, n*10^(8*j)+1 is divisible by 137.
If k is of the form 4*j+2, n*10^(4*j+2)+1 is divisible by 101.
If k is of the form 8*j+4 then n*10^(8*j+4)+1 is divisible by 73.
This covers all k, so the covering set is { 11, 73, 101, 137 }.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000 (a(1) .. a(27) from _Giovanni Resta_)

Crossrefs

Cf. A243974, A244070.

Formula

For n > 8, a(n) = a(n-8) + 11111111.

Extensions

More terms from Giovanni Resta, Nov 23 2019