[QUOTE=GP2;496143][c]worktodo.add[/c] or [c]worktodo.add.txt[/c]

Both of these work on either Windows or Linux.[/QUOTE]

Thank you GP2. Wasn’t GP3 the best formula one PC game from micropose?!
Anyway, I’m done with [14500,15000] and [15500,16000] and the other ranges are still underway.

16500-17000
[CODE]2^16553+1 has a factor: 24255740240695470491 (ECM curve 1, B1=250000, B2=25000000)
2^16553+1 has a factor: 7154053018754849945563697 (ECM curve 97, B1=250000, B2=25000000)
2^16573+1 has a factor: 696649544614857521 (ECM curve 1, B1=250000, B2=25000000)
2^16603+1 has a factor: 15585137074585080458129252635718353 (ECM curve 101, B1=250000, B2=25000000)
2^16607+1 has a factor: 303320328422719887497 (ECM curve 35, B1=250000, B2=25000000)
2^16619+1 has a factor: 1151577374018700125678948761 (ECM curve 72, B1=250000, B2=25000000)
2^16649+1 has a factor: 14738684332216197916018591568291 (ECM curve 328, B1=250000, B2=25000000)
2^16657+1 has a factor: 789629909044670355787468891 (ECM curve 154, B1=250000, B2=25000000)
2^16729+1 has a factor: 27900051005133099320370700483 (ECM curve 279, B1=250000, B2=25000000)
2^16729+1 has a factor: 96650433856277903079913 (ECM curve 12, B1=250000, B2=25000000)
2^16741+1 has a factor: 809804835360656414883168237139 (ECM curve 243, B1=250000, B2=25000000)
2^16763+1 has a factor: 20134727690344808240068001 (ECM curve 136, B1=250000, B2=25000000)
2^16823+1 has a factor: 1679052693166082395351642577 (ECM curve 38, B1=250000, B2=25000000)
2^16829+1 has a factor: 501981017684293624882492979 (ECM curve 66, B1=250000, B2=25000000)
2^16829+1 has a factor: 7030862200846423842337531 (ECM curve 6, B1=250000, B2=25000000)
2^16831+1 has a factor: 52888863505346416714097 (ECM curve 16, B1=250000, B2=25000000)
2^16831+1 has a factor: 6337591265234038183048243 (ECM curve 98, B1=250000, B2=25000000)
2^16871+1 has a factor: 6454068146099797732673515409 (ECM curve 113, B1=250000, B2=25000000)
2^16903+1 has a factor: 640701004988163401 (ECM curve 3, B1=250000, B2=25000000)
[/CODE]

My range is done.


14500-16500

17000-17500
[CODE]2^17021+1 has a factor: 19970118882517760713844757569 (ECM curve 270, B1=250000, B2=25000000)
2^17041+1 has a factor: 46936866960519129408225318409 (ECM curve 130, B1=250000, B2=25000000)
2^17077+1 has a factor: 3553221273890775977131483 (ECM curve 14, B1=250000, B2=25000000)
2^17093+1 has a factor: 6434901203451482177147 (ECM curve 15, B1=250000, B2=25000000)
2^17137+1 has a factor: 74065101935672709731614369 (ECM curve 170, B1=250000, B2=25000000)
2^17159+1 has a factor: 180366271112782247999249803 (ECM curve 9, B1=250000, B2=25000000)
2^17167+1 has a factor: 7199673061767521949212888347 (ECM curve 91, B1=250000, B2=25000000)
2^17239+1 has a factor: 15743980144824856474553 (ECM curve 16, B1=250000, B2=25000000)
2^17239+1 has a factor: 9265020153871021580804299 (ECM curve 125, B1=250000, B2=25000000)
2^17291+1 has a factor: 127151453638503039336813703937 (ECM curve 135, B1=250000, B2=25000000)
2^17291+1 has a factor: 5494662222282781553959883 (ECM curve 44, B1=250000, B2=25000000)
2^17327+1 has a factor: 279194784644349188143579 (ECM curve 44, B1=250000, B2=25000000)
2^17359+1 has a factor: 189729835257594067 (ECM curve 2, B1=250000, B2=25000000)
2^17377+1 has a factor: 1598815760606505821467 (ECM curve 32, B1=250000, B2=25000000)
2^17393+1 has a factor: 310471675824874318553 (ECM curve 1, B1=250000, B2=25000000)
2^17393+1 has a factor: 97019705507132797228107411041 (ECM curve 47, B1=250000, B2=25000000)
2^17401+1 has a factor: 284292996618239242781965355931769 (ECM curve 284, B1=250000, B2=25000000)
2^17419+1 has a factor: 478541207978011275399739 (ECM curve 71, B1=250000, B2=25000000)
2^17477+1 has a factor: 959936791635270283 (ECM curve 1, B1=250000, B2=25000000)
2^17491+1 has a factor: 1127373789196363561966486313683 (ECM curve 67, B1=250000, B2=25000000)
2^17551+1 has a factor: 25735713653693529449 (ECM curve 8, B1=250000, B2=25000000)
[/CODE]

[QUOTE=pinhodecarlos;496520]My range is done.


14500-16500[/QUOTE]

Hmmm... Somehow I missed this message. Do you need another set of tests?

[QUOTE=axn;496649]Hmmm... Somehow I missed this message. Do you need another set of tests?[/QUOTE]

Yes please, same range size.

[QUOTE=pinhodecarlos;496664]Yes please, same range size.[/QUOTE]

Ok, here is the 18000-20000 range.

BTW, did you check the relative performance of P95 vs GMP-ECM on these numbers? In my trials, P95 was way faster so that two core P95 was more throughput than 4 core GMP-ECM.

[QUOTE=axn;496666]Ok, here is the 18000-20000 range.

BTW, did you check the relative performance of P95 vs GMP-ECM on these numbers? In my trials, P95 was way faster so that two core P95 was more throughput than 4 core GMP-ECM.[/QUOTE]

Thank you. Did the trial yesterday and matches your results so I’ll run Prime95.

It appears that the set of all factors of all Mersenne numbers with prime exponent and the set of all factors of all Wagstaff numbers with prime exponent are disjoint. Is there an elementary proof of this?

By contrast, for other pairs of b values in (b[sup]p[/sup] − 1) / (b − 1), there are many factors in common: for b = 2 and 3, 2 and 5, 3 and 5; −2 and 3, −2 and 5... but not 2 and −2.

For example: (3[SUP]5[/SUP] − 1) /2 = [B]11[/B][SUP]2[/SUP] and (5[SUP]5[/SUP] − 1) /4 = [B]11[/B] × 71

Incidentally, that's the only non-squarefree factor of a repunit that I know of. Sure would be nice if we could find one for Mersennes... I keep checking.

For a random pair of repunit bases b[SUB]1[/SUB], b[SUB]2[/SUB] such that b[SUB]1[/SUB] ≠ −b[SUB]2[/SUB], would the set of common factors be finite or infinite?

[QUOTE=GP2;496734]It appears that the set of all factors of all Mersenne numbers with prime exponent and the set of all factors of all Wagstaff numbers with prime exponent are disjoint. Is there an elementary proof of this?
[/QUOTE]

Factors of Mersennes are 8*n+-1 and factors of Wagstaff are 8*m+-3.

[QUOTE=GP2;496734]It appears that the set of all factors of all Mersenne numbers with prime exponent and the set of all factors of all Wagstaff numbers with prime exponent are disjoint. Is there an elementary proof of this?
[/QUOTE]

There is a single exception: 2^2-1 and (2^3+1)/3 are divisible by 3. (and actually they are equal to 3). And there is no more solution:

Assume that 1<d | 2^p-1 and (2^q+1)/3, where p,q are odd primes.
Then obviously p!=q, otherwise d | 2^p-1 and 2^p+1, so d | 2, what is impossible.
We can write:
d | gcd(2^p-1,2^q+1) | gcd(2^p-1,2^(2*q)-1)=2^gcd(p,2*q)-1=2^1-1=1

where used: gcd(p,2*q)=1, because p is odd and p!=q primes.