[QUOTE=science_man_88;230087][CODE](08:36) gp > forstep(k=2,1000,[4,2],if(isprime(k*5^5+1),print(k)))
12[/CODE]

looks like 5 is gone lol[/QUOTE]

Why??? Again, I'm looking for a prime 5*b^b+1, not k*5^5+1!

[QUOTE=kar_bon;230089]Why??? Again, I'm looking for a prime 5*b^b+1, not k*5^5+1![/QUOTE]

sorry a misinterpret so like me.

[CODE](09:42) gp > for(b=2,1000,if((5*b^b+1)%6==1 || (5*b^b+1)%6==5,print1(b",")))
6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96,102,108,114,120,126,132,138,144,1
50,156,162,168,174,180,186,192,198,204,210,216,222,228,234,240,246,252,258,264,2
70,276,282,288,294,300,306,312,318,324,330,336,342,348,354,360,366,372,378,384,3
90,396,402,408,414,420,426,432,438,444,450,456,462,468,474,480,486,492,498,504,5
10,516,522,528,534,540,546,552,558,564,570,576,582,588,594,600,606,612,618,624,6
30,636,642,648,654,660,666,672,678,684,690,696,702,708,714,720,726,732,738,744,7
50,756,762,768,774,780,786,792,798,804,810,816,822,828,834,840,846,852,858,864,8
70,876,882,888,894,900,906,912,918,924,930,936,942,948,954,960,966,972,978,984,9
90,996,
(09:44) gp > for(b=2,1000,if((5*b^b-1)%6==1 || (5*b^b-1)%6==5,print1(b",")))
2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,
58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,90,92,94,96,98,100,102,104,106,1
08,110,112,114,116,118,120,122,124,126,128,130,132,134,136,138,140,142,144,146,1
48,150,152,154,156,158,160,162,164,166,168,170,172,174,176,178,180,182,184,186,1
88,190,192,194,196,198,200,202,204,206,208,210,212,214,216,218,220,222,224,226,2
28,230,232,234,236,238,240,242,244,246,248,250,252,254,256,258,260,262,264,266,2
68,270,272,274,276,278,280,282,284,286,288,290,292,294,296,298,300,302,304,306,3
08,310,312,314,316,318,320,322,324,326,328,330,332,334,336,338,340,342,344,346,3
48,350,352,354,356,358,360,362,364,366,368,370,372,374,376,378,380,382,384,386,3
88,390,392,394,396,398,400,402,404,406,408,410,412,414,416,418,420,422,424,426,4
28,430,432,434,436,438,440,442,444,446,448,450,452,454,456,458,460,462,464,466,4
68,470,472,474,476,478,480,482,484,486,488,490,492,494,496,498,500,502,504,506,5
08,510,512,514,516,518,520,522,524,526,528,530,532,534,536,538,540,542,544,546,5
48,550,552,554,556,558,560,562,564,566,568,570,572,574,576,578,580,582,584,586,5
88,590,592,594,596,598,600,602,604,606,608,610,612,614,616,618,620,622,624,626,6
28,630,632,634,636,638,640,642,644,646,648,650,652,654,656,658,660,662,664,666,6
68,670,672,674,676,678,680,682,684,686,688,690,692,694,696,698,700,702,704,706,7
08,710,712,714,716,718,720,722,724,726,728,730,732,734,736,738,740,742,744,746,7
48,750,752,754,756,758,760,762,764,766,768,770,772,774,776,778,780,782,784,786,7
88,790,792,794,796,798,800,802,804,806,808,810,812,814,816,818,820,822,824,826,8
28,830,832,834,836,838,840,842,844,846,848,850,852,854,856,858,860,862,864,866,8
68,870,872,874,876,878,880,882,884,886,888,890,892,894,896,898,900,902,904,906,9
08,910,912,914,916,918,920,922,924,926,928,930,932,934,936,938,940,942,944,946,9
48,950,952,954,956,958,960,962,964,966,968,970,972,974,976,978,980,982,984,986,9
88,990,992,994,996,998,1000,[/CODE]

looks like for +1 the covering set is 6n and the covering set for -1 is 2n to me.

No. A Covering Set is a limited amount of small prime factors, of which every number of the form (here 5*b^b+1) has a member.

Example:
The Riesel-number 992077*2^n-1 has a Covering Set = [3, 5, 7, 13, 17, 241].

This means, for every n>=1 the number 992077*2^n-1 has a factor out of this set of factors.
Try it.

well then all b values are either have a covering set of [3] or [2] for k = 5

[QUOTE=science_man_88;230096]well then all b values are either have a covering set of [3] or [2] for k = 5[/QUOTE]

Give factorizations for b=1 to 6 for the form 5*b^b+1 please!

[CODE](10:21) gp > for(b=2,20,if((5*b^b+1)%6==1 || (5*b^b+1)%6==5,print1(factor(5*b^>
[263, 1; 887, 1],[127, 1; 1423, 1; 246681361, 1],[449, 1; 577721, 1; 758421438600649, 1],[/CODE]

[CODE]<(5*b^b-1)%6==1 || (5*b^b-1)%6==5,print1(factor(5*b^b-1)",")))
Mat([19, 1]),Mat([1279, 1]),Mat([233279, 1]),[181, 1; 463459, 1],[29, 1; 1724137931, 1],[10831, 1; 4116009809, 1],[13259, 1; 13679, 1; 306335539, 1],[1244696389, 1; 74101380211, 1],[11, 1; 701, 1; 172472381, 1; 147926109109, 1],[31, 1; 169125161290322580645161
29, 1],[/CODE] and heres what I got for the -1 side.

[QUOTE=science_man_88;230101]
b=6 [263, 1; 887, 1]
b=12 [127, 1; 1423, 1; 246681361, 1]
b=18 [449, 1; 577721, 1; 758421438600649, 1][/QUOTE]

As you can see, 2 or 3 are not factors of the numbers 5*b^b+1 here,
so [2] or [3] or [2, 3] is not a full Covering Set!

[QUOTE=kar_bon;230104]As you can see, 2 or 3 are not factors of the numbers 5*b^b+1 here,
so [2] or [3] or [2, 3] is not a full Covering Set![/QUOTE]

I mean't for b not the factorization of of 5*b^b+1

[QUOTE=kar_bon;230083]The case I gave for testing was k=5, so find a prime for 5*b^b+1 for any b>=1![/QUOTE]

Of course since [TEX]\int\frac{db}{log(b^b)}=\log\log b[/TEX], we expect the primes to be very sparse.

There are no examples for b < 3000, so by size alone we'd expect only one below exp(exp(log(log(3000))+1)) ≈ 2.8 billion. With the condition mod 6, it's more like exp(exp(log(log(3000))+2)) ≈ 5 * 10[sup]25[/sup].