[seqfan] Re: Programs in A280992

jonscho at hiwaay.net jonscho at hiwaay.net
Sun Dec 31 22:15:03 CET 2017


I find that 1, 3, 6, 15, 105, 210, and 255255 are the only terms of  
A280992 through at least 10^32, and I think it shouldn't be too  
difficult to raise that bound to at least 10^100, and perhaps far  
above that.  I think it's nearly certain that 255255 is the final term  
of the sequence.

Unfortunately, I don't have time to pursue it farther right now, but  
-- in case any of it's of interest -- I'll offer the following  
thoughts (with apologies for any of them that are so obviously true  
that they only serve to illustrate my ignorance of some relevant areas  
of mathematics):

- Obviously, since the only triangular number that is prime is 3,  
every term other than 3 and the empty product (1) is the product of at  
least two consecutive primes.

- If the k-th triangular number is the product of two consecutive  
primes p and q, then

      k*(k+1)/2 = p*q
      k*(k+1) = 2*p*q

from which it's not hard to show that k must be q or 2p and that the  
only possible values of k are 3 (which yields 3*4/2 = 6), 5 (which  
yields 5*6/2 = 15), and (if they exist) larger values such that q =  
2*p +- 1, which would require (since p and q are consecutive) that  
there's an interval of at least [p+1, 2*p-2] that is completely devoid  
of any primes.  Is there such a thing, for p > 3?  I would assume that  
the nonexistence of such an interval is either something that was  
already proved long ago or, if not, then it's known that no such p  
exists up to some huge value.  (It didn't take long for me to prove  
that any such p exceeds 10^250, so the corresponding triangular number  
would exceed 10^500.)

- A triangular number less than, say, 10^100 can't have more than 53  
consecutive prime factors (since 2*3*5*...*prime(54) > 10^100), and,  
after generating an array containing all the primes from 2 through a  
few hundred million, it's easy to test the triangularity (if that's a  
word!) of each product P of j _consecutive_ primes such that P <  
10^100 for j = 53, 52, 51, ..., etc., down to some lower bound; e.g.,  
I get that there are only 7 such products of 50 consecutive primes, 47  
such products of 40 consecutive primes, 309 of 30, 9583 of 20, 325154  
of 15, ....  (Of course, to test whether a product P is triangular,  
simply test whether 8*T+1 is a square.)

- The considerations above make it easy to rule out the existence of  
any terms > 255255 up to, say, 10^100 unless they're products of j  
consecutive primes where 3 <= j <= 11.  But (and here's the part I'd  
look into further, if I had time, and I'll offer additional apologies  
now in case any of what follows is obviously flawed!) if a triangular  
number T(k) = k*(k+1)/2 is the product of j consecutive primes, then  
we've got k*(k+1) = 2 * p_1 * p_2 * ... * p_j (with p_1 < p_2 < ... <  
p_j), i.e., j+1 factors (including the factor 2), which can be  
partitioned into two subsets whose products are very nearly equal:   
one is k, the other is k+1.  *If* p_1 gets large enough, then  
partitioning the j+1 factors of k*(k+1) into two such subsets will  
require certain results, I think:

-- if j is odd, then each subset will have to have exactly (j+1)/2 of  
the j+1 prime factors of k*(k+1), yet p_1 * p_2 * ... * p_{(j+1)/2}  
must not exceed 2 * p_{(j+3)/2} * p_{(j+5)/2} * ... * p_j, which  
implies that primes are unusually rare between p_{(j+1)/2} and p_j;

-- if j is even, then one subset will have to have j/2 of the prime  
factors of k*(k+1), and the other will get the other j/2+1, yet 2 *  
p_1 * p_2 * ... * p_{j/2} must not exceed p_{j/2+1} * p_{j/2+2} * ...  
* p_j, which implies that primes are unusually rare between p_{j/2}  
and p_j.

I think the above ideas (if my poor way of stating them hasn't made  
them incomprehensible) provide a basis for concluding, without too  
much computational difficulty, that at least one of the following is  

1.  There are no terms > 255255 through at least 10^100 (and probably  
none up through an easily-verified limit well beyond that).

2.  Jon's an idiot.  ?:-/

In any case, I'd like to take this opportunity to wish a very happy  
new year to all!  :-)

-- Jon

Quoting rgwv at rgwv.com:

> Seqfans, Here is what I have for A280992:
> fQ[n_] := Block[{fi = Sort@ Join[ FactorInteger[ n/If[ OddQ@ n, 1,  
> 2]], FactorInteger[(n +1)/If[ OddQ@ n, 2, 1]]]}, Times @@ (Last@# &  
> /@ fi) == 1 && NextPrime[fi[[1, 1]], Length at fi -1] == fi[[-1, 1]]];
> k = 3; lst = {1, 2}; While[k < 100000001, If[fQ@ k, AppendTo[lst,  
> k]; Print at k]; k++]; lst(lst +1)/2
> Instead of factoring the n-th triangular number, I factor n and n+1  
> separately and join the results. The other speed up was not  
> computing PrimePi of all the factors. Instead if k was the number of  
> primes, I looked for the (k-1)-th prime after the first prime and  
> was that the last prime? Between the two, my program is about 10  
> times quicker that the one presented.
> As a result I have been able to verify that the terms {1, 3, 6, 15,  
> 105, 210, 255255} are the only terms up to the 10^8-th triangular  
> number. This took approximately 1:28. Bob.
> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of  
> Felix Fröhlich
> Sent: Saturday, 30 December, 2017 5:06 AM
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Programs in A280992
> Dear SeqFans
> I believe that several of the programs in A280992 need to be fixed.  
> Is that correct?
> Best regards
> Felix
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> Seqfan Mailing list - http://list.seqfan.eu/
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