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 1-6 of 6 results.

A243974 Integers n not of form 3m+1 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

10176, 17601, 19361, 25827, 27147, 27686, 35916, 36048, 45462, 47213, 48036, 49248, 54638, 62864, 64184, 64899, 72953, 73085, 82499, 85073, 86285, 93435, 101760, 101936
Offset: 1

Views

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+1 then n*10^k-1 is always divisible by 3. - Jens Kruse Andersen, Jul 09 2014

Examples

			10176*10^k-1 is divisible by 11 for k of form 6m, 6m+2, 6m+4, by 7 for k of form 6m+1, by 37 for 6m+3 (and also 6m), and by 13 for 6m+5. This covers all k. {7, 11, 13, 37} is called a covering set. - _Jens Kruse Andersen_, Jul 09 2014

Crossrefs

Cf. A076337, A243969, A243974, A244070, A244071, A244072, A244073, A244074, A244076.

Formula

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

Extensions

Definition corrected by Jens Kruse Andersen, Jul 09 2014

A244211 Integers n such that for every integer k>0, n*6^k-1 has a divisor in the set { 7, 13, 31, 37, 43 }.

Original entry on oeis.org

133946, 213410, 299144, 33845, 367256, 803676, 1214450, 1250446, 1280460, 1704478, 1780150, 1792762, 1794864, 2003070, 2004962, 2203536, 2798489, 3014465, 3027709, 3041998, 3053350, 3194549, 3326301, 4244794
Offset: 1

Views

Author

Pierre CAMI, Jun 23 2014

Keywords

Comments

For n > 24 a(n) = a(n-24) + 4488211, the first 24 values are in the data.
When the number a(n) has 1 or 6 as the last digit, the number a(n)*6^k-1 is always divisible by 5 and always has another divisor in the set { 7, 13, 31, 37, 97 } for every k.

Crossrefs

Cf. A076337, A243969, A244070, A244071, A244072, A244073, A244074, A244076.

Formula

For n > 24, a(n) = a(n-24) + 4488211.

A244545 Integers n such that for every integer k>0, n*6^k+1 has a divisor in the set { 7, 13, 31, 37, 43 }.

Original entry on oeis.org

243417, 1161910, 1293662, 1434861, 1446213, 1460502, 1473746, 1689722, 2284675, 2483249, 2485141, 2693347, 2695449, 2708061, 2783733, 3207751, 3237765, 3273761, 3684535, 4120955, 4154366, 4189067, 4274801, 4354265
Offset: 1

Views

Author

Pierre CAMI, Jun 29 2014

Keywords

Comments

For n > 24, a(n) = a(n-24) + 4488211, the first 24 values are in the data.
When the number a(n) has 4 or 9 as the last digit, the number a(n)*6^k-1 is always divisible by 5 and always has a divisor in the set { 7, 13, 31, 37, 97 } for every k.

Crossrefs

Cf. A076337, A243969, A244070, A244071, A244072, A244073, A244074, A244076, A244211.

Formula

For n > 24 a(n) = a(n-24) + 4488211.

A244351 Integers n such that for every integer k>0, n*6^k-1 has a divisor in the set { 7, 13, 31, 37, 97 }.

Original entry on oeis.org

84687, 429127, 508122, 1273238, 1570311, 1656045, 2574762, 2847748, 3048732, 3345805, 3849481, 5076399, 5324003, 5338292, 5908351, 6961919, 7639428, 8167823, 8508662, 8994775, 9078721, 9421866, 9936270, 9950261
Offset: 1

Views

Author

Pierre CAMI, Jun 26 2014

Keywords

Comments

For n > 24 a(n) = a(n-24) + 10124569, the first 24 values are in the data.
When the number a(n) has 1 or 6 as the last digit the number a(n)*6^k-1 is always divisible by 5 and have always a divisor in the set { 7, 13, 31, 37, 97 } for every k.

Crossrefs

Cf. A076337, A243969, A244070, A244071, A244072, A244073, A244074, A244076, A244211.

Formula

For n>24 a(n) = a(n-24) + 10124569.

A244549 Integers m such that for every integer k>0, m*6^k+1 has a divisor in the set { 7, 13, 31, 37, 97 }.

Original entry on oeis.org

174308, 188299, 702703, 1045848, 1129794, 1615907, 1956746, 2485141, 3162650, 4216218, 4786277, 4800566, 5048170, 6275088, 6778764, 7075837, 7276821, 7549807, 8468524, 8554258, 8851331, 9616447, 9695442, 10039882
Offset: 1

Views

Author

Pierre CAMI, Jun 29 2014

Keywords

Comments

For n > 24 a(n) = a(n-24) + 10124569, the first 24 values are in the data.
When the number a(n) has 4 or 9 as the last digit the number a(n)*6^k-1 is always divisible by 5 and have always a divisor in the set { 7, 13, 31, 37, 97 } for every k.

Crossrefs

Cf. A076337, A243969, A244070, A244071, A244072, A244073, A244074, A244076, A244211, A244545.

Formula

For n > 24 a(n) = a(n-24) + 10124569.

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

Cf. A243969, A243974, A244348.

Formula

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

Extensions

a(9)-a(27) from Jason Yuen, Nov 10 2024
Showing 1-6 of 6 results.

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