Problems & Puzzles: Puzzles

Puzzle 560.- Same quantity of prime factors

JM Bergot sent the following puzzle:

See this: Adding two consecutive prime factors you get:

Three sums, Three factors
3+5 = 8  = 2*2*2
5+7 = 12 = 2*2*3
7+11 = 18 = 2*3*3
11+13 = 24 = 2*2*2*3 (fail)

Three sums, Four factors
41+43 = 84 = 2*2*3*7
43+47 = 90 = 2*3*3*5
47+53 = 100 = 2*2*5*5
53+59 = 112 = 2*2*2*2*7 (fail)

Q. Can there be more than three of these sums having on the right the same quantity of factors?

Contributions came from: Torbjörn Alm, Dan Dima, Jan van Delden, Jeff Heleen, Seiji Tomita, Fred Schalekamp, JC Rosa & Emmanuel Vantieghem, Farid Lian.

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Torbjörn Alm wrote:

Sum count: 4  Factors: 4
571+577 = 1148 = 2*2*7*41
577+587 = 1164 = 2*2*3*97
587+593 = 1180 = 2*2*5*59
593+599 = 1192 = 2*2*2*149

Sum count: 5  Factors: 4
11593+11597 = 23190 = 2*3*5*773
11597+11617 = 23214 = 2*3*53*73
11617+11621 = 23238 = 2*3*3*1291
11621+11633 = 23254 = 2*7*11*151
11633+11657 = 23290 = 2*5*17*137

Sum count: 6  Factors: 4
9811+9817 = 19628 = 2*2*7*701
9817+9829 = 19646 = 2*11*19*47
9829+9833 = 19662 = 2*3*29*113
9833+9839 = 19672 = 2*2*2*2459
9839+9851 = 19690 = 2*5*11*179
9851+9857 = 19708 = 2*2*13*379

Sum count: 7  Factors: 5
498361+498367 = 996728 = 2*2*2*23*5417
498367+498391 = 996758 = 2*7*7*7*1453
498391+498397 = 996788 = 2*2*13*29*661
498397+498401 = 996798 = 2*3*11*11*1373
498401+498403 = 996804 = 2*2*3*3*27689
498403+498409 = 996812 = 2*2*17*107*137
498409+498439 = 996848 = 2*2*2*2*62303

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Dan Dima wrote:

Seven sums, Four factors
9803+9811 = 19614 = 2*3*7*467  
9811+9817 = 19628 = 2*2*7*701  
9817+9829 = 19646 = 2*11*19*47
9829+9833 = 19662 = 2*3*29*113
9833+9839 = 19672 = 2*2*2*2459
9839+9851 = 19690 = 2*5*11*179
9851+9857 = 19708 = 2*2*13*379
9857+9859 = 19716 = 2*2*3*31*53  (fail)

 

I see no reason to get a bound for the number of sums having the same number of factors (as it seems there is no bound for the number of factors of such a sum).

Twelve sums, Six factors
1630324763+1630324777 = 3260649540 = 2*2*3*5*4297*12647
1630324777+1630324793 = 3260649570 = 2*3*5*53*977*2099
1630324793+1630324799 = 3260649592 = 2*2*2*61*631*10589
1630324799+1630324807 = 3260649606 = 2*3*17*151*269*787
1630324807+1630324823 = 3260649630 = 2*3*5*7*7*2218129
1630324823+1630324831 = 3260649654 = 2*3*3*3*3*20127467
1630324831+1630324853 = 3260649684 = 2*2*3*29*1217*7699
1630324853+1630324903 = 3260649756 = 2*2*3*7*13*2985943
1630324903+1630324921 = 3260649824 = 2*2*2*2*2*101895307
1630324921+1630324951 = 3260649872 = 2*2*2*2*67*3041651
1630324951+1630324991 = 3260649942 = 2*3*3*11*379*43451
1630324991+1630325029 = 3260650020 = 2*2*3*5*83*654749
1630325029+1630325033 = 3260650062 = 2*3*23*1039*22741  (fail)

Here is the smallest first prime in a series that produces at least k such sums:
 k            p-prime
 3                   3
 4                571
 5              9803
 6              9803
 7              9803
 8          959659
 9       20692817
10     188088421
11   1173258563
12   1630324763
and the smallest first prime in a series that produces exactly k such sums 
(only for k=7 sums the sequence is not increasing and for k=5,6 the sequences differ):
 k            p-prime
 3                   3
 4                571
 5            11593
 6            34651
 7              9803
 8          959659
 9       20692817
10     188088421
11   1173258563
12   1630324763

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Jan van Delden wrote:

#sums    #factors    first prime

1             1            2

2             3           19

3             3            3

4             4            571

5             4            11593

6             4            34561

7             4            9803

8             5            959659

9             4            20692817

10           5            188088421

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Jeff Heleen wrote:

For puzzle 560 I have found the following:

      No. of Sums = 3
No. of   start           end
factors  prime           prime
 3       3               11 
 4       41              53 
 5       337             353 
 6       1019            1033 
 7       13553           13591 
 8       113933          113963 
 9       816961          817027 
 10      9311503         9311569 

      No. of Sums = 4
No. of   start           end
factors  prime           prime
 3       1511            1549 
 4       571             599 
 5       1637            1669 
 6       19681           19709 
 7       154591          154643 
 8       7260511         7260553
 9       12906937        12907007 
 10      96697177        96697273 

      No. of Sums = 5
No. of   start           end
factors  prime           prime
 3       134741          134837 
 4       11593           11657
 5       22147           22189 
 6       508513          508567 
 7       251761          251833 
 8       5306219         5306311 
 9       182247743       182247827 
 10      none less than 2^32

      No. of Sums = 6
No. of   start           end
factors  prime           prime
 3       1081441         1081631 
 4       34651           34703
 5       135841          135911 
 6       636241          636287 
 7       21721717        21721787 
 8       656158313       656158397 
 9       none less than 2^32
 10      none less than 2^32

      No. of Sums = 7
No. of   start           end
factors  prime           prime
 3       46341601        46341791 
 4       9803            9857 
 5       498361          498439 
 6       6350353         6350471 
 7       78271847        78271979 
 8       3681644779      3681644921 
 9       none less than 2^32
 10      none less than 2^32

      No. of Sums = 8
No. of   start           end
factors  prime           prime
 3       none less than 2^32
 4       4702349         4702433 
 5       959659          959773 
 6       42079441        42079549
 7       305563331       305563417 
 8       none less than 2^32
 9       none less than 2^32
 10      none less than 2^32

      No. of Sums = 9
No. of   start           end
factors  prime           prime
 3       none less than 2^32 
 4       20692817        20693039 
 5       47117647        47117857 
 6       24870017        24870161 
 7       3922410389      3922410617 
 8       none less than 2^32
 9       none less than 2^32
 10      none less than 2^32

      No. of Sums = 10
No. of   start           end
factors  prime           prime
 3       none less than 2^32
 4       286363937       286364077 
 5       188088421       188088617 
 6       218545751       218545919 
 7       none less than 2^32
 8       none less than 2^32
 9       none less than 2^32
 10      none less than 2^32

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Seiji Tomita wrote:

Smallest twelve sums that having same quantity of prime factors.

 1630324763+1630324777 = (2)^2*(3)*(5)*(12647)*(4297)
 1630324777+1630324793 = (2)*(3)*(5)*(53)*(977)*(2099)
 1630324793+1630324799 = (2)^3*(61)*(631)*(10589)
 1630324799+1630324807 = (2)*(3)*(17)*(151)*(269)*(787)
 1630324807+1630324823 = (2)*(3)*(5)*(7)^2*(2218129)
 1630324823+1630324831 = (2)*(3)^4*(20127467)
 1630324831+1630324853 = (2)^2*(3)*(29)*(1217)*(7699)
 1630324853+1630324903 = (2)^2*(3)*(7)*(13)*(2985943)
 1630324903+1630324921 = (2)^5*(101895307)
 1630324921+1630324951 = (2)^4*(67)*(3041651)
 1630324951+1630324991 = (2)*(3)^2*(11)*(379)*(43451)
 1630324991+1630325029 = (2)^2*(3)*(5)*(83)*(654749)

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Fred Schalekamp wrote:

Four sums, Nine factors
12906937 +   12906953 =   25813890
12906953 +   12906967 =   25813920
12906967 +   12906977 =   25813944
12906977 +   12907007 =   25813984

Five sums, Eight factors
 5306219 +    5306221 =   10612440
 5306221 +    5306267 =   10612488
 5306267 +    5306293 =   10612560
 5306293 +    5306309 =   10612602
 5306309 +    5306311 =   10612620

Six sums, Seven factors
21721717 +   21721723 =   43443440
21721723 +   21721727 =   43443450
21721727 +   21721741 =   43443468
21721741 +   21721747 =   43443488
21721747 +   21721781 =   43443528
21721781 +   21721787 =   43443568

Seven sums, Seven factors
78271847 +   78271889 =  156543736
78271889 +   78271903 =  156543792
78271903 +   78271913 =  156543816
78271913 +   78271943 =  156543856
78271943 +   78271951 =  156543894
78271951 +   78271969 =  156543920
78271969 +   78271979 =  156543948

Eight sums, Six  factors
42079441 +   42079451 =   84158892
42079451 +   42079459 =   84158910
42079459 +   42079481 =   84158940
42079481 +   42079483 =   84158964
42079483 +   42079501 =   84158984
42079501 +   42079507 =   84159008
42079507 +   42079511 =   84159018
42079511 +   42079549 =   84159060

Nine sums, Six factors
24870017 +   24870059 =   49740076
24870059 +   24870073 =   49740132
24870073 +   24870077 =   49740150
24870077 +   24870103 =   49740180
24870103 +   24870113 =   49740216
24870113 +   24870119 =   49740232
24870119 +   24870137 =   49740256
24870137 +   24870143 =   49740280
24870143 +   24870161 =   49740304

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JC Rosa wrote:

I have found solutions for 4,5,6,7 and 8 sums having the same

quantity of factors. Here is my best result :

 

959659+959677=1919336=2*2*2*29*8273

959677+959681=1919358=2*3*3*7*15233

959681+959689=1919370=2*3*5*137*467

959689+959719=1919408=2*2*2*2*119963

959719+959723=1919442=2*3*7*23*1987

959723+959737=1919460=2*2*3*5*31991

959737+959759=1919496=2*2*2*3*79979

959759+959773=1919532=2*2*3*19*8419

 

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Emmanuel Vantieghem wrote:

Let  F(t) = the number of primefactors of  t, counted with multiplicity  (usually called ‘the big omega’ of  t).

For several values of  k,  I found sequences of  k  consecutive primes  p(m+1), p(m+2), ... , p(m+k)  such that the  k-1 numbers F(p(m+j)+p(m+j+1))  ( j = 1, 2, ... , k-1) were all the same.

The smallest value for  p(m+1)  was equal to

3 when  k = 3

571  when  k = 4

9803  when  k < 8

959659  when  k = 8

20692817  when  k = 9

188088421  when  k = 10

1173258563  when  k = 11

1630324763  when  k = 12

When  k = 13, the smallest  p(m+1)  must be bigger than 2*10^9.

 

I think there is a good reason why there exist an  m  for every value of  k.  Indeed, if  t  is smaller than  2^N, then  F(t) < N.  So, for suitable  N, the  k-1 -tuple

                        (  F(p(m+1)+p(m+2)), F(p(m+2)+p(m+3)), ... , F(p(m+k-1)+p(m+k))  )          (*)

will look like a  k-1 –tuple of randomly chosen numbers from the interval  [2,N-1].  The number of such  k-1 –tuples is  (N-2)^(k-1),  whence the probability that all the elements are the same is  1 / (N-1)^(k-2).  This is very small, but enough for ‘Murphy’s law’.  Besides, the number of  k-1 –tuples of the form  (*)  is as great as there are prime numbers < 2^N  and that increases the probability.

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Farid wrote:

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