look if you multiply 3 consecutive Mersenne primes (MP[SUB]x[/SUB] * MP[SUB]x+1[/SUB] * MP[SUB]x+2[/SUB]) we get a additive root so far for primes it seems this sequence always starts with 3 if we can prove something like always starting with 3,1,4 for primes then we can use this knowledge to predict modulo of the next or further Mersenne primes given the modulo it should be able to make a smaller list of checks to do for a given range until we find the next one then we move on to search the next modulo for a given range as long as we can predict which exponents give a certain modulo we get a small list and then we can limit these to prime exponents to check.

[QUOTE=3.14159;225205]The sum of the digits of pi = 1? So 3.141592653589793238462643383279... = 1? Give us your math proof if this.

For the lulz:

[code]3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420198938095257201065485863278865936153381827968230301952035301852968995773622599413891249721775283479131515574857242454150695950829533116861727855889075098381754637464939319255060400927701671139009848824012858361603563707660104710181942955596198946767837449448255379774726847104047534646208046684259069491293313677028989152104752162056966024058038150193511253382430035587640247496473263914199272604269922796782354781636009341721641219924586315030286182974555706749838505494588586926995690927210797509302955321165344987202755960236480665499119881834797753566369807426542527862551818417574672890977772793800081647060016145249192173217214772350141441973568548161361157352552133475741849468438523323907394143334547762416862518983569485562099219222184272550254256887671790494601653466804988627232791786085784383827967976681454100953883786360950680064225125205117392984896084128488626945604241965285022210661186306744278622039194945047123713786960956364371917287467764657573962413890865832645995813390478027590099465764078951269468398352595709825822620522489407726719478268482601476990902640136394437455305068203496252451749399651431429809190659250937221696461515709858387410597885959772975498930161753928468138268683868942774155991855925245953959431049972524680845987273644695848653836736222626099124608051243884390451244136549762780797715691435997700129616089441694868555848406353422072225828488648158456028506016842739452267467678895252138522549954666727823986456596116354886230577456498035593634568174324112515076069479451096596094025228879710893145669136867228748940560101503308617928680920874760917824938589009714909675985261365549781893129784821682998948722658804857564014270477555132379641451523746234364542858444795265867821051141354735739523113427166102135969536231442952484937187110145765403590279934[/code]

The first 2800 digits of pi. Remember, the sum of those digits is 1! LOL!

You do realize that the only number whose sum of digits is 1 is 1, correct?[/QUOTE]

I mean summation until you get a single digit number they both equal 1.

[QUOTE=science_man_88]I mean summation until you get a single digit number they both equal 1.
[/QUOTE]

Tell me, how can you reduce infinity to 1?

Sum of digits(infinity) = infinity.

How did you avoid the circular loop?

This should be amusing :missingteeth:

[QUOTE=science_man_88;225206]look if you multiply 3 consecutive Mersenne primes (MP[SUB]x[/SUB] * MP[SUB]x+1[/SUB] * MP[SUB]x+2[/SUB]) we get a additive root so far for primes it seems this sequence always starts with 3 if we can prove something like always starting with 3,1,4 for primes then we can use this knowledge to predict modulo of the next or further Mersenne primes given the modulo it should be able to make a smaller list of checks to do for a given range until we find the next one then we move on to search the next modulo for a given range as long as we can predict which exponents give a certain modulo we get a small list and then we can limit these to prime exponents to check.[/QUOTE]

Digital roots of the product of three consecutive Mersenne primes: 3, 1, 4, 4, 4, 4, 1, 4, 7, 7, 7, 4, 4, 1, 1, 1, 4, 4, 7, 7, 1, 1, 1, 7, 4, 4, 4, 4, 1, 4, 7, 7, 7, 7, 1, 1, 7

But what do you mean, "always starts with 3"?

[QUOTE=science_man_88;225207]I mean summation until you get a single digit number they both equal 1.[/QUOTE]

I can't think of any reasonable way to define this. How are you calculating it?

[QUOTE=science_man_88;225194]my idea is this:

[CODE]#M %9[SUB]1[/SUB] %9[SUB]2[/SUB] %9[SUB]3[/SUB]
2 3 1 4 x
3 3 1 4
5 3 [/CODE]

care to use the data in the table to calculate x ?[/QUOTE]

look %9[SUB]3[/SUB] in the 3 row is 4 this is for Mp3*mp4*mp5

if we already know there's a pattern and the first 2 we can predict it for Mp5 and hence Mp4*mp5 which would give us x hence we can use the pattern (if there is one) to predict modulo even at the basic level of a single Mersenne and hence we only have to check possible Mersenne exponents that give a number that can give a number of that modulo.

@ CRG: digit one + digit 2 ....... until the end of both continue until you get one digit that digit is 1 for both in the example you gave.

Ah, I think I have it. Together with my above code take
[code]ProdK(k)=k--;vector(#A43-k,i,prod(j=0,k,drMersenne(A43[i+j]))%9)[/code]
which gives the digital roots of the product of k consecutive Mersenne numbers. So ProdK(1) gives [url=http://oeis.org/classic/A135928]A135928[/url](n), ProdK(2) gives [url=http://oeis.org/classic/A010888]A010888[/url]([url=http://oeis.org/classic/A165223]A165223[/url](n)), etc.

Then you have
[code]>for(k=1,15,print(k"\t"ProdK(k)))
1 [3, 7, 4, 1, 1, 4, 1, 1, 1, 4, 4, 1, 4, 1, 1, 1, 1, 1, 4, 1, 4, 4, 4, 4, 4, 1, 1, 4, 1, 1, 1, 4, 4, 1, 4, 4, 4, 4, 1]
2 [3, 1, 4, 1, 4, 4, 1, 1, 4, 7, 4, 4, 4, 1, 1, 1, 1, 4, 4, 4, 7, 7, 7, 7, 4, 1, 4, 4, 1, 1, 4, 7, 4, 4, 7, 7, 7, 4]
3 [3, 1, 4, 4, 4, 4, 1, 4, 7, 7, 7, 4, 4, 1, 1, 1, 4, 4, 7, 7, 1, 1, 1, 7, 4, 4, 4, 4, 1, 4, 7, 7, 7, 7, 1, 1, 7]
4 [3, 1, 7, 4, 4, 4, 4, 7, 7, 1, 7, 4, 4, 1, 1, 4, 4, 7, 1, 1, 4, 4, 1, 7, 7, 4, 4, 4, 4, 7, 7, 1, 1, 1, 4, 1]
5 [3, 4, 7, 4, 4, 7, 7, 7, 1, 1, 7, 4, 4, 1, 4, 4, 7, 1, 4, 4, 7, 4, 1, 1, 7, 4, 4, 7, 7, 7, 1, 4, 4, 4, 4]
6 [3, 4, 7, 4, 7, 1, 7, 1, 1, 1, 7, 4, 4, 4, 4, 7, 1, 4, 7, 7, 7, 4, 4, 1, 7, 4, 7, 1, 7, 1, 4, 7, 7, 4]
7 [3, 4, 7, 7, 1, 1, 1, 1, 1, 1, 7, 4, 7, 4, 7, 1, 4, 7, 1, 7, 7, 7, 4, 1, 7, 7, 1, 1, 1, 4, 7, 1, 7]
8 [3, 4, 1, 1, 1, 4, 1, 1, 1, 1, 7, 7, 7, 7, 1, 4, 7, 1, 1, 7, 1, 7, 4, 1, 1, 1, 1, 4, 4, 7, 1, 1]
9 [3, 7, 4, 1, 4, 4, 1, 1, 1, 1, 1, 7, 1, 1, 4, 7, 1, 1, 1, 1, 1, 7, 4, 4, 4, 1, 4, 7, 7, 1, 1]
10 [3, 1, 4, 4, 4, 4, 1, 1, 1, 4, 1, 1, 4, 4, 7, 1, 1, 1, 4, 1, 1, 7, 7, 7, 4, 4, 7, 1, 1, 1]
11 [3, 1, 7, 4, 4, 4, 1, 1, 4, 4, 4, 4, 7, 7, 1, 1, 1, 4, 4, 1, 1, 1, 1, 7, 7, 7, 1, 4, 1]
12 [3, 4, 7, 4, 4, 4, 1, 4, 4, 7, 7, 7, 1, 1, 1, 1, 4, 4, 4, 1, 4, 4, 1, 1, 1, 1, 4, 4]
13 [3, 4, 7, 4, 4, 4, 4, 4, 7, 1, 1, 1, 4, 1, 1, 4, 4, 4, 4, 4, 7, 4, 4, 4, 4, 4, 4]
14 [3, 4, 7, 4, 4, 7, 4, 7, 1, 4, 4, 4, 4, 1, 4, 4, 4, 4, 7, 7, 7, 7, 7, 7, 7, 4]
15 [3, 4, 7, 4, 7, 7, 7, 1, 4, 7, 7, 4, 4, 4, 4, 4, 4, 7, 1, 7, 1, 1, 1, 1, 7][/code]

[QUOTE=CRGreathouse]I can't think of any reasonable way to define this. How are you calculating it?
[/QUOTE]


See here:

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[QUOTE=CRGreathouse;225212]Ah, I think I have it. Together with my above code take
[code]ProdK(k)=k--;vector(#A43-k,i,prod(j=0,k,drMersenne(A43[i+j]))%9)[/code]
which gives the digital roots of the product of k consecutive Mersenne numbers. So ProdK(1) gives [url=http://oeis.org/classic/A135928]A135928[/url](n), ProdK(2) gives [url=http://oeis.org/classic/A010888]A010888[/url]([url=http://oeis.org/classic/A165223]A165223[/url](n)), etc.

Then you have[/QUOTE]
well wanted to keep it more specific to Mersenne primes all I can say from this is that the line of 3's going down map on diagonals to the top one.

[QUOTE=science_man_88;225215]well wanted to keep it more specific to Mersenne primes[/QUOTE]

Yes, these are (the digital roots of products of) Mersenne primes.

[QUOTE=science_man_88;225215]all I can say from this is that the line of 3's going down map on diagonals to the top one.[/QUOTE]

I trust you mean that the first column (the first number in each row) is always 3?

Can you prove it?

I think the proof is that the first 2 numbers seem to always give 3:

that means if we only deal with 3 [1,4,7]

once all those combo's appear and give 3 in the next one by my logic all that only use 3,[1,4,7] to start will always give a 3 in the next one hence is there a proof about 1,4,7 being the only ones used in most of these if so we can prove it.