cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A243969 Integers n not of form 3m+2 such that for any integer k > 0, n*10^k+1 has a divisor in the set { 7, 11, 13, 37 }.

Original entry on oeis.org

9175, 9351, 17676, 24826, 26038, 28612, 38026, 38158, 46212, 46927, 48247, 56473, 61863, 63075, 63898, 65649, 75063, 75195, 83425, 83964, 85284, 91750, 93510, 100935
Offset: 1

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Author

Pierre CAMI, Jun 16 2014

Keywords

Comments

For n>24 a(n) = a(n-24) + 111111, the first 24 values are in the data.
If n is of form 3m+2 then n*10^k+1 is always divisible by 3. The sequence is a base 10 variant of provable Sierpiński numbers (A076336). It is currently unknown whether 7666*10^k+1 is always composite but based on heuristics it probably has large undiscovered primes. 7666 is the only remaining base 10 Sierpiński candidate below 9175. - Jens Kruse Andersen, Jul 09 2014

Examples

			9175*10^k+1 is divisible by 11 for k of form 6m+1, 6m+3, 6m+5, by 37 for k of form 6m (and also 6m+3), by 13 for 6m+2, and by 7 for 6m+4. This covers all k. {7, 11, 13, 37} is called a covering set. - _Jens Kruse Andersen_, Jul 09 2014
		

Crossrefs

Formula

For n>24 a(n) = a(n-24) + 111111.

Extensions

Definition corrected by Jens Kruse Andersen, Jul 09 2014

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

Original entry on oeis.org

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

Views

Author

Pierre CAMI, Jun 28 2014

Keywords

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 }.
		

Crossrefs

Formula

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

Extensions

More terms from Giovanni Resta, Nov 23 2019

A244598 Integers n such that for every k > 0, n*10^k-1 has a divisor in the set { 11, 73, 101, 137 }.

Original entry on oeis.org

152206, 1522060, 4109489, 4459665, 6001522, 7761557, 9489041, 10948904, 11263317, 12633171, 15220600, 15570776, 17112633, 18872668, 20600152, 22060015, 22374428, 23744282, 26331711, 26681887, 28223744, 29983779, 31711263, 33171126, 33485539, 34855393, 37442822
Offset: 1

Views

Author

Pierre CAMI, Jul 01 2014

Keywords

Comments

For n > 8, a(n) = a(n-8) + 11111111, the first 8 values are given in the data.
If n is of the form 3*m+1 then n*10^k-1 is always divisible by 3 but also has a divisor in the set { 11, 73, 101, 137 }.

Examples

			Consider n = 152206.
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 73.
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, n*10^(8*j+4)-1 is divisible by 137.
This covers all k, so the covering set is { 11, 73, 101, 137 }.
		

Crossrefs

Formula

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

Extensions

a(9)-a(27) from Jason Yuen, Nov 10 2024
Showing 1-3 of 3 results.