Problems & Puzzles: Puzzles

Puzzle 66.- The SOPF sequences

Patrick De Geest defined the sequence C1, C2, C3, …CL, P, as a succession of L composite numbers and an ending prime, where every number is the sum of the previous one plus the sum of all his prime factors (repetitions included).

Just in order to be precise, let Ci =p1a1.p2a2… , then Ci+1 = Ci + (sum of) (aj . pj).

Example (and the least of this kind of sequences): 4, 8, 14, 23
C1 = 4; C2 = 8 = 4+(2+2); C3 = 14 = 8+(2+2+2); P = 23 = 14+(2+7)

Questions (1, 3 & 5 by C. Rivera; 2 & 4 by Patrick De Geest):

1.  Find three consecutive composites C1, C1+1 and C1+2 such that all they have the same ending prime in their respective sequence, or prove that such triplet cannot exist.

2.   Find the first occurrence of  a chain of 20=>Z=>2 consecutive numbers with the same parity (this chain may be a part of a complete sequence)

Examples

a) 7 consecutive composite even numbers: 60, 72, 84, 98, 114, 138, 166, ….  
b) 5 consecutive composite odd numbers: 213, 287, 335, 407, 455, …

3. Find the least complete sequences formed with Z odd members, for Z = 6, 7, 8, 9 & 10.

Z Sequence Author
3 9, 15, 23 C. Rivera
4 87, 119, 143, 167 C. Rivera
5 69, 95, 119, 143,167 C. Rivera
6 1765 - 2123 - 2327 - 2519 - 2759 - 2879 McCranie / Haga?
7 1845 - 1897 - 2175 - 2217 - 2959 - 3239 - 3359 McCranie / Haga
8 1815 - 1845 - 1897 - 2175 - 2217 - 2959 -
3239 - 3359		 McCranie / Haga
9 10005 - 10065 - 10145 - 12179 - 12839 - 13223 - 15119 - 16295 - 19559 McCranie / Haga
10 24351 - 32471 - 34199 - 37319 - 37943 - 39959 - 41279 - 47183 - 48839 - 55823 McCranie / Haga


4. Find the smallest composite C1 so that 666 iterations of the SOPF procedure are needed to reach the ending prime.

5.- Defining the "Rate of Growth of C1", as RG(C1) = (P/C1)/L, I found the following conspicuous C1:

RG(137222) = (35200439/137222)/99 = 2.59113

Last weekend I asked to certain friends of this site if they could improve this rate. The 9/9/99 Patrick De Geest responded:

RG(13161303) = (31760920439/13161303)/216 = 11.1722

Can you improve the Patrick's result?

Jud McCranie four triplets of consecutive composite numbers-solution for the part 1 of this puzzle: 559-561; 1946-1948; 3155-3157; 17226-17228.

Jud McCranie found some solutions to part 2. of this puzzle:

Z = 9: Starting at 1099
Z=10: Starting at 3476
Z=11: Starting at 8373
Z=12: Starting at 21697
Z=17: Starting at 36333
Z=19: Starting at 294222

Jud McCranie & Enoch Haga independently found solutions to part 3. of this puzzle. See their contributions in the table above.

***

Anton Vrba wrote on June 2011:

Its time to improve Patrick's solution for part 5. A quick random search yielded:

RG(761001172) = (3500188955101/761001172)/278 = 16.5448 
I then improved the result by calculating the possible roots further up the chain
RG(569432586) = (3500188955101/569432586)/279 = 22.0315
 
761001172->951251469->996549168->etc
569432586->664338022->996507035->996549168->etc

***

 

 

 
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