# Most rapidly convergent reciprocal prime series equal to 1

1. Feb 3, 2007

### Loren Booda

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

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

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

2. Feb 4, 2007

### 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.

3. Feb 4, 2007

### 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
8424197597064114319193772925959967322398440121059128471513803869133407474043 203007538138482347678292768315235576731982578595105540461656971813019185323832704402275591986413790507264062854340888345433684127591313959045797597991 203237053814712728428750073711737022064128957491601662363018780189927080966272201341892990231783175100611109711661585348912614207638543406179978226940671084971215364367952969493044517652693946209790164697536189882634653872786628061745493943721533681682601514614015412013115319621649394434192763213 1274824659267207819645689091902447428459474365747086290508326092871968934441678929237475467817711388277347282149565899239406969419583821733090036741367349396351159119501852788567239055333172258691731437277575008641081824023172257608774059103346907356279568452498839112696912247287900141740937470738379901364986139220871590989651563229484468197967188842956150619835699487246035049688395073795092612647473616783877646805516352578801539374338272634990551858398656344074106563622390651135225433445200158465460170931637161622534065436173292137151681083456917963425025261497067420708715953110886963 4674902516513883824366469896256979043503059996791732586263715154689870963133266450292892088862390672424409867999645482221757721207196479746435720006904593052130947250466190482036950871018668244163533146037384829203436240511508054874989496550788059831441865840643779869437217244574541337850767835040012514512703738156921978701630977322730122546270523284807595512811402160114543653712097713288542381996768463007179467585027077844016618281037441106017077799835854771774716186086444490409011799505701724896401840681020362481900499087346271773033220422019991901141612377682700000062316703849584593690639246557130552318702502833721310615619156726632472885582864619760061059662762402541426252146726579428027277842389699957696510073479039406587560346782163529705803311869179162174002249378898380028623411814542737843025285240844119746105563363809527811088103018629408004429301138347796523484409276790479947718232950628667064652835936768508718306501132895941531885319833404442841994117641768056545938093529444757507860222665787115313484051805067286907350994621241578848833409180513067853649451478038263265542265598712184947722583484463031483227013131692074997274649313665355869250350888941

I'm working on 15 and 16.

4. Feb 4, 2007

### Hurkyl

Staff Emeritus
Is there any particular reason to be interested in it?

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

5. Feb 4, 2007

### 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.

6. Feb 4, 2007

### 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.

7. Feb 4, 2007

### CRGreathouse

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.

8. Feb 5, 2007