Most rapidly convergent reciprocal prime series equal to 1

Loren Booda

Consider the series with ascending (but not necessarily sequential) primes p_n,

1/p₁+1/p₂+1/p₃+ . . . +1/p_N=1, as N approaches infinity.

Determine the p_n that most rapidly converge (minimize the terms in) this series. That set of primes I call the "Booda set."

CRGreathouse

The greedy algorithm will give you this sequence: 1/2 + 1/3 + 1/7 + 1/43 + 1/1811 + ....

I would call it 'Brown's sequence' or A075442, though.

CRGreathouse

Here are the first 14 terms of the sequence, which is quadratically convergent. This goes well beyond the numbers listed on Sloane's page. I've used Primo to certify the larger primes.
2
3
7
43
1811
654149
27082315109
153694141992520880899
337110658273917297268061074384231117039
84241975970641143191937729259599673223984401210591284715138038691334074 74043 20300753813848234767829276831523557673198257859510554046165697181301918 53238327044022755919864137905072640628543408883454336841275913139590457 97597991 20323705381471272842875007371173702206412895749160166236301878018992708 09662722013418929902317831751006111097116615853489126142076385434061799 78226940671084971215364367952969493044517652693946209790164697536189882 63465387278662806174549394372153368168260151461401541201311531962164939 4434192763213 12748246592672078196456890919024474284594743657470862905083260928719689 34441678929237475467817711388277347282149565899239406969419583821733090 03674136734939635115911950185278856723905533317225869173143727757500864 10818240231722576087740591033469073562795684524988391126969122472879001 41740937470738379901364986139220871590989651563229484468197967188842956 15061983569948724603504968839507379509261264747361678387764680551635257 88015393743382726349905518583986563440741065636223906511352254334452001 58465460170931637161622534065436173292137151681083456917963425025261497 067420708715953110886963 46749025165138838243664698962569790435030599967917325862637151546898709 63133266450292892088862390672424409867999645482221757721207196479746435 72000690459305213094725046619048203695087101866824416353314603738482920 34362405115080548749894965507880598314418658406437798694372172445745413 37850767835040012514512703738156921978701630977322730122546270523284807 59551281140216011454365371209771328854238199676846300717946758502707784 40166182810374411060170777998358547717747161860864444904090117995057017 24896401840681020362481900499087346271773033220422019991901141612377682 70000006231670384958459369063924655713055231870250283372131061561915672 66324728855828646197600610596627624025414262521467265794280272778423896 99957696510073479039406587560346782163529705803311869179162174002249378 89838002862341181454273784302528524084411974610556336380952781108810301 86294080044293011383477965234844092767904799477182329506286670646528359 36768508718306501132895941531885319833404442841994117641768056545938093 52944475750786022266578711531348405180506728690735099462124157884883340 91805130678536494514780382632655422655987121849477225834844630314832270 13131692074997274649313665355869250350888941

I'm working on 15 and 16.

Hurkyl

Is there any particular reason to be interested in it?


By the way, is there a proof that the greedy algorithm converges the fastest?

CRGreathouse

None that I can think of, except that the question was asked and that it's a Sloane sequence.

Any set of primes whose reciprocals sum to 1 must be infinite, so no sequence will 'finish' and thus be faster in that sense.

I suppose it's conceivable that a sequence could have a few partial sums smaller than the greedy sequence* and then outpace it thereafter. I suppose to put that on a rigorous foundation we'd need to have a good definition of "fastest".

* Any subsequence of the primes must have a partial sum less than the same partial sum of this sequence, if the sum of the reciprocals add to 1.

Loren Booda

Hurkyl

I was interested in the rapidity with which the series converged.

I also wondered if it had any relation to the Riemann hypothesis.

It seemed a problem simple to state with an elegant algorithm.

I attempted to find my place in history.

CRGreathouse

Quadratically.

The RH could be used to put tight bounds on exactly how fast the sequence converges, but that's all that comes to mind at the moment.

I rather agree.

Hey, why not? I do have two things to say about that:
* Don't name things after yourself, let others do that. This is just the way it's done in math.
* You'll actually have to prove interesting things about (whatever) to have it named after you, in general. Even if you discover it, if someone else proves the cool/useful/etc. things about it, the object may be named after them instead of you.

Loren Booda

Well received, CRGreathouse!

CRGreathouse

Thanks.

If you want a good place to get something named after you, try http://www.primepuzzles.net/ - the site often names things after people who solve the problems. Of course you won't see you name in a textbook or anything crazy like that, but it can still be pretty cool.