this is probably already done but I think:

[TEX]M(x)= \sum_{n=1}^{x-1} + 5[/TEX]

not sure though.

[B]Edit:[/B] [TEX]M(x)= \sum_{n=1}^{x-1} M(n)+ 5[/TEX] or [TEX]M(x)= \sum_{n=0}^{x-1} M(n)+ 5[/TEX]

another edit x>3

I have learn to think over it now lol technically replace 5 with [TEX]\sum _{p=0}^{2}M(p) [/TEX]all add (x-4)

[QUOTE=3.14159;225306]20004 * 901! + 1 is prime! (≈2280-2285 digits)[/QUOTE]

2278 digits!

I should say all that thinking came up to:

for x>3,[TEX]M(x)=\sum_{n=0}^{x-1}M(n)[/TEX] all plus x.

[QUOTE=Karsten]2278 digits!
[/QUOTE]

You still haven't found me a solution to n * 71[sup]6[/sup] = 1 mod 1124000727777607680000.

[QUOTE=3.14159;225314]You still haven't found me a solution to:

[tex]$71^6 \equiv 1(mod 22!)$[/tex][/QUOTE]

outside of proving it i have no idea lol

71^6 = 128100283921
22! = 1124000727777607680000 ?

so if pari did what I think you are wrong.

now i saw the n and altered my pari code so far i've checked over 1 million. going toward 1 billion.

I gave up.

[QUOTE=science_m]outside of proving it i have no idea lol

71^6 = 128100283921
22! = 1124000727777607680000 ?

so if pari did what I think you are wrong.[/QUOTE]

Nope. n * 71[sup]6[/sup] = 1 mod 1124000727777607680000.

[QUOTE=3.14159;225317]Nope. n * 71[sup]6[/sup] = 1 mod 1124000727777607680000.[/QUOTE]

I bow out I checked again to n=100 million and found nothing.

[QUOTE=science_man_88]I bow out I checked again to n=100 million and found nothing.
[/QUOTE]

And you would have continued checking for another few thousand years, because it's in the 10^20 range.

So Pi have you ever heard of:
for x>3[TEX]M(x)=\sum_{n=0}^{x-1}M(n) + x[/TEX]

?

if so who came up with it.

I realized this works for all of them above M(0) = 0

[QUOTE=3.14159;225302]Find a number of the form (k * 1124000727777607680000 + 1) which is divisible by 71[sup]6[/sup].[/QUOTE]

[code]> Mod(-1,71^6)/1124000727777607680000
%1 = Mod(36882197250, 128100283921)[/code]

so 41455616551037280586874880000001, 185440428906759455741024993280001, ...

If you wanted k to be a prime, then
[code]forstep(n=36882197250,1e15,128100283921,if(isprime(n),return(n)))[/code]
takes 0ms to come up with the smallest example, 1702185888223.

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

I must be interpreting you the wrong way, then, since I find 17 million instances below 10[SUP]20[/SUP]:
[code]s=0;forstep(n=36882197250,1e20,128100283921,if(ispseudoprime(n),s++));s
time = 24mn, 50,380 ms.
%1 = 17584374[/code]