[QUOTE=CRGreathouse]I must be interpreting you the wrong way, then, since I find 17 million instances below 10[sup]20[/sup]:
[/QUOTE]

No, sm88 was searching from 1 to 10[sup]20[/sup]

Also: I can rig the number to be semiprime: Here's an example:

[code]55762113159666605738616781764882446662982216262889524965004140441546022418664991294967299112901478710275915859674894915215729565500652173245752458035579997432840192000000000000000000000001 = 5846650401949673788584832661 * 9537446114628590934543506256858833034627998057578048449114232547066222757945489740614277575764557950018685132240750168209560234946900002547179019523457586506941[/code]

It's rather cool what you can do with modular exponentiation.

Here's a better example:
[code]289508238708400855287326405336918749133423298889334502223156944093968*150!+1 = 10254288128120939058918287267226262277004901802104088101537671807691 * 1613053686056875801829683444255478970299912076335944456541386114240383845905441917746506601556683535279149882320681534772255292138166095487502436596518188053218548567074676625073642723441629113007713095460989964369203609983928686506798110902140688105800120704029411[/code]

See? I just made a record! :wink: (Imagine someone was in the middle of factoring that one.)

[QUOTE=3.14159;225319]And you would have continued checking for another few thousand years, because it's in the 10^20 range.[/QUOTE]

He would have hit an instance in a few hours, and a prime instance in a few days, if I calculate correctly. Of course they wouldn't be close to 10^20.

[CODE]M(x)= a=0;for(n=0,x-1,a=a+M(n));a=a+x;print(a",")[/CODE]

best i can do in coding my equation.

[QUOTE=CRGreathouse]If you wanted k to be a prime, then[/QUOTE]

Ah: But if you want the cofactor to be prime:

[code]forstep(n=1,a,b,if(isprime((n*22!+1)/b), print(n)))[/code]

Here's an example:
2386628800 * 60! + 1 = 89[sup]4[/sup] * 316518937978282424919538194794587561257209063707380372243938125981964972473329411361

Ex 2:
1974103215468915131950479405951082163*66!+1 = 71[sup]6[/sup] * 89[sup]4[/sup] * 101[sup]8[/sup] *
12347042307394672596544859365776765192013890088140194548358364707496516846616209505096123905841

I know for n>2 MP(n) = 4^k*MP(n-1)+(4^k-1) the hard part is how to define k.

only problem with my codes is deep recursion I think it's called.

I think if i change it to vectors then ask for the sum of the indexes it might work. but it's never a guarantee with me lol.

[QUOTE=science_man_88;225369][CODE]M(x)= a=0;for(n=0,x-1,a=a+M(n));a=a+x;print(a",")[/CODE]

best i can do in coding my equation.[/QUOTE]

Did you mean to return the a, rather than printing it?

[QUOTE=science_man_88;225374]only problem with my codes is deep recursion I think it's called.[/QUOTE]

I'm not sure if the program does what it's intended. If so, then the answer is what you hint at below: memoize the result.

[QUOTE=science_man_88;225375]I think if i change it to vectors then ask for the sum of the indexes it might work. but it's never a guarantee with me lol.[/QUOTE]

[QUOTE=CRGreathouse;225393]Did you mean to return the a, rather than printing it?[/QUOTE]

I've tried both neither give a correct answer for me.

[QUOTE=science_man_88;225395]I've tried both neither give a correct answer for me.[/QUOTE]

I can't correct the program because I don't know what it's trying to do. But certainly the program isn't doing what you intend if it stores the value it returns but doesn't actually return a value.