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

***

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

***

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:

 last prime # (k) first prime last prime # of sums # factors 3 3 5 1 3 4 3 7 2 3 5 3 11 3 3 109 571 599 4 4 1,214 9,803 9,839 5 4 1,215 9,803 9,851 6 4 1,216 9,803 9,857 7 4 75,601 959,659 959,773 8 5 1,311,759 20,692,817 20,693,039 9 4 10,455,221 188,088,421 188,088,617 10 5 59,175,242 1,173,258,563 1,173,258,763 11 6 80,881,875 1,630,324,763 1,630,325,029 12 6 1,116,236,010 25,580,875,711 25,580,875,937 13 5 1,116,236,011 25,580,875,711 25,580,875,939 14 5

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