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 1-10 of 15 results. Next

A244072 Odd integers n such that for every integer k>0, n*2^k-1 has a divisor in the set {3, 5, 7, 13, 19, 37, 109}.

Original entry on oeis.org

2136283, 2251349, 2924861, 3781541, 8010517, 10645867, 11124703, 16413457, 16593431, 17713229, 21992527, 22146359, 23572417, 23835883, 23909173, 24359437, 24688477, 25124999, 26711801, 26880179, 27094349, 28151593, 30577271, 32126257
Offset: 1

Views

Author

Pierre CAMI, Jun 19 2014

Keywords

Comments

For n > 144 a(n) = a(n-144) + 209191710, the first 144 values are in the table.

Crossrefs

Formula

For n > 144, a(n) = a(n-144) + 209191710.

A244073 Odd integers n such that for every integer k>0, n*2^k-1 has a divisor in the set {3, 5, 7, 13, 19, 73, 109}.

Original entry on oeis.org

1744117, 6975809, 7790113, 11942443, 13006807, 16861093, 16882181, 17207051, 20003369, 20147891, 21013423, 25638127, 42918821, 45113083, 47285977, 48635609, 49884041, 53335151, 53538727, 56592041, 63412693, 63750101, 64062209, 65739209
Offset: 1

Views

Author

Pierre CAMI, Jun 19 2014

Keywords

Comments

For n > 144 a(n) = a(n-144) + 412729590, the first 144 values are in the table.

Crossrefs

Formula

For n > 144, a(n) = a(n-144) + 412729590.

A244074 Odd integers n such that for every integer k>0, n*2^k-1 has a divisor in the set {3, 5, 7, 13, 37, 73, 109}.

Original entry on oeis.org

1830187, 4643293, 17041931, 20787701, 50462309, 52363777, 66659587, 68026001, 71604733, 71817943, 88558303, 91609361, 93193151, 97363751, 118421557, 122606647, 123765359, 124808009, 131118733, 131408411, 134320001, 135411719, 139778591, 142339723
Offset: 1

Views

Author

Pierre CAMI, Jun 19 2014

Keywords

Comments

For n > 144, a(n) = a(n-144) + 803736570, the first 144 values are in the table.

Crossrefs

Formula

For n > 144, a(n) = a(n-144) + 803736570.

A244076 Odd integers n such that for every integer k>0, n*2^k-1 has a divisor in the set {3, 5, 7, 13, 17, 97, 257}.

Original entry on oeis.org

10157893, 71627707, 86571727, 90109601, 98957849, 99023257, 99284501, 114096371, 142399363, 166262293, 207549337, 213185347, 213708611, 215798563, 229298773, 229306949, 242872709, 251719903, 274263943, 276356999, 278326889, 284716807, 289074853, 294317669
Offset: 1

Views

Author

Pierre CAMI, Jun 19 2014

Keywords

Comments

For n > 96 a(n) = a(n-96) + 1156954890, the first 96 values are in the table.

Crossrefs

Formula

For n > 96 a(n) = a(n-96) + 1156954890.

A244561 Odd integers m such that for every integer k > 0, m*2^k+1 has a divisor in the set {3, 5, 7, 13, 17, 241}.

Original entry on oeis.org

271129, 271577, 482719, 575041, 603713, 903983, 965431, 1518781, 1624097, 1639459, 2131043, 2131099, 2541601, 2931767, 2931991, 3083723, 3098059, 3555593, 3608251, 4067003, 4573999, 6134663, 6135559, 6557843, 6676921, 6678713, 6742487, 6799831, 7400371, 7523267, 7523281, 7761437, 7765021, 7892569, 8007257, 8629967, 8840599, 8871323, 9208337, 9454129, 9454157, 9854491, 9854603, 9930469, 9937637, 10192733, 10422109, 10675607
Offset: 1

Views

Author

Pierre CAMI, Jun 30 2014

Keywords

Comments

For n > 48, a(n) = a(n-48) + 11184810; the first 48 values are in the data.
The set {3, 5, 7, 13, 17, 241} is the set of prime divisors of 2^24 - 1. Hence for every p in the set the multiplicative order of 2 modulo p divides 24. Note that twice the product of {3, 5, 7, 13, 17, 241} is 11184810. - Jeppe Stig Nielsen, Mar 10 2019
Subset of provable Sierpiński numbers A076336. - Jeppe Stig Nielsen, Mar 10 2019

Crossrefs

Programs

  • PARI
    D=[3, 5, 7, 13, 17, 241];P=2*lcm(D);M=lcm(apply(d->znorder(Mod(2,d)),D));forstep(k=1,+oo,2,if(k%P==1,print();print());for(n=0,M-1,for(i=1,#D,k*Mod(2,D[i])^n+1==0 && next(2));next(2));print1(k,", ")) \\ Jeppe Stig Nielsen, Mar 10 2019

Formula

For n > 48, a(n) = a(n-48) + 11184810.

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

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+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

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

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

Formula

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

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

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

Formula

For n > 24 a(n) = a(n-24) + 4488211.
Showing 1-10 of 15 results. Next