Problems & Puzzles: Puzzles

Puzzle 450. K extended factors for sum of two consecutive primes. J. M. Bergot sent the following puzzle: It is noticed that for two consecutive primes p,q that 113+127=240 has six extended factors, 2*2*2*2*3*5. Q. Find sums of consecutive primes having more than six extended factors.

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Contributions came from J. K. Andersen, Farideh Firoozbakht, Jan van Delden, Anton Vrba, J.C. Rosa & F. Schneider.

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J.K. Andersen wrote:

The largest known twin primes are 2003663613*2^195000+/-1, found in 2007 by
Eric Vautier, Dmitri Gribenko and Patrick W. McKibbon in the projects Twin
Prime Search and PrimeGrid, using the programs NewPGen and LLR.
The sum of the two primes is 2003663613*2^195001.
2003663613 = 3*7*487*195919 so the sum has 195005 prime factors.

The html title of the page is still "Puzzle 449. Count N composites after p".
Latest News says "Week 14-29, June 2008" instead of 14-20.

Puzzle 223 has solutions where the sum is a power of 2.
I found the consecutive probable primes 2^9229+/-2211 summing to 2^9230.

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Farideh wrote:

113+127=240=2*2*2*2*3*5 has six extended factors.
436207613+436207619=872415232=2^26*13. 1879048183+1879048201=3758096384=2^29*7, so it has 30

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Jan van Delden wrote:

Some large answers are already known if p,q of certain special forms are considered.

For instance if p,q are twins, say p=6n-1, q=6n+1, then p+q=12n and the number of factors depend on the factorization of n.

If we take p=k*2^n+/-1 (special twins), then the number of extended factors can easily be calculated using the factorization of k.
The prime pages of Chris Caldwell gives 2003663613 2^195000 +/-1 as largest example known today (not necessarily the greatest number of extended factors), giving a total of 195005 extended factors.

Of course one could also use other special twin forms like k*b^n+/-1, with b even.

Are there (large) solutions of the forms: n!+/-1 or n#+/-1?

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Anton wrote:

2^14 * p(200)# ± 1741 are consecutive 517 digit primes
their sum equals 2^15 * p(200)#
and has 215 extended factors.

2^66 *p(303)# ± p(595) are consecutive 863 digit primes
and their sum has 370 extended factors

But, todays record must be the sum of the
largest known Twin Primes 2003663613 * 2^195000 ± 1
which has 195005 extended factors.
 

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J. C. Rosa wrote:

About the puzzle 450 here are some results :

P Q S=P+Q K factors
71 73 144=2*2*2*2*3*3 6
61 67 128=2^7 7
191 193 384 =2*2*2*2*2*2*2*3 8
953 967 1920=(2^7)*3*5 9
1151 1153 2304=(2^8)*(3^2) 10
3833 3847 7680=(2^9)*3*5 11
7159 7177 14336=(2^11)*7 12
4093 4099 8192=2^13 13
30713 30727 61440=(2^12)*3*5 14

P Q S=P+Q K different factors
15013 15017 30030=2*3*5*7*11*13 6
285283 285287 570570=2*3*5*7*11*13*19 7
10140583 10140587 20281170=2*3*5*7*13*17*19*23 8
140645501 140645509 281291010=2*3*5*7*11*13*17*19*29 9
4127218087 4127218103 8254436190=2*3*5*7*11*13*17*19*23*37 10


P Q R S=P+Q+R K factors
3797 3803 3821 11421=(3^5)*47 6
7351 7369 7393 22113=(3^5)*7*13 7
18701 18713 18719 56133=(3^6)*7*11 8
123191 123203 123209 369603=(3^7)*(13^2) 9
198997 199021 199033 597051=(3^8)*7*13 10
1178767 1178803 1178809 3536379=(3^8)*(7^2)*11 11
1240021 1240027 1240039 3720087=(3^11)*7 12


P Q R S=P+Q+R K different factors
299903 299909 299933 899745=3*5*7*11*19*41 6
7234517 7234541 7234547 21703605=3*5*7*11*19*23*43 7
158055193 158055199 158055223 474165615=3*5*7*11*17*19*31*41 8

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F. Schneider wrote:

There are many easy-to-find answers. The example is not even minimal
: 61 + 67 = 128 = 2^7 is a lower and "better" case than the example.

I stopped at 500K. There are two 17 cases

360439 = 360457 = 720896 = 2^16 * 11
368633 + 368647 = 737280 = 2^14 * 3^2 * 5

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