Problems & Puzzles: Puzzles

 Puzzle 722 2^m-1 consecutive integers having m prime factors In May 2001 Tony Forbes published an interesting article titled "FIFTEEN CONSECUTIVE INTEGERS WITH EXACTLY FOUR PRIME FACTORS".There you may learn that at the much you can get 2^m-1 consecutive integers all of them having m prime factors. Accordingly, for m=4 prime factors, 15  is the largest possible quantity of consecutive integers having four prime factors. Forbes explain in this article his developed strategy to get a specific solution. His result for m=4 starts at the integer 488995430567765317569. As per May 2001 Forbes wrote "we believe to be the only known example of a maximum-length sequence with m = 4." Q1. Can you get another/smaller solution to m=4? In the same article Forbes also wrote: "A search for 31 consecutive integers with the same value of 5 seems to be quite a difficult task, as yet unaccomplished." Q2. Can you try to get now your best solution for m=5 (the closest quantity of consecutive integers to 31, having 5 prime factors)?

Contributions came from Emmanuel Vantieghem, Shyam Sunder Gupta, J. K. Andersen and Vicente Felipe Izquierdo.

***

Emmanuel wrote:

About puzzle 722 : some time ago I submitted the curio http://primes.utm.edu/curios/page.php?short=97524222465
So, I guess it is an answer to the first question of pyzzle 722.

According to A077657 at the OEIS, the curio 97524222465 (which I submitted on 17 oktober 2013) appears to have been known by Martin Fuller in 2006.
Hence, I was not the first to discover that number. I sent a mail to G.L. Honacker to inform him about this and to suggest him to remove my name. In the meantime, I continue my search for the 31-sequence.

***

Shyam wrote:

Regarding Q.1 of Puzzle 722, please refer CYF NO. 8 on my webpage (http://www.shyamsundergupta.com/canyoufind.htm),where
Brian Trial, Ferndale, Michigan U.S.A.has found not one but two solutions in 2002.

97524222465 = 3 * 5 * 42751 * 152081
97524222466 = 2 * 11 * 19 * 233311537
97524222467 = 7 * 29 * 149 * 3224261
97524222468 = 2 * 2 * 3 * 8127018539
97524222469 = 73 * 73 * 251 * 72911
97524222470 = 2 * 5 * 67 * 145558541
97524222471 = 3 * 3 * 13 * 833540363
97524222472 = 2 * 2 * 2 * 12190527809
97524222473 = 17 * 17 * 3499 * 96443
97524222474 = 2 * 3 * 7 * 2322005297
97524222475 = 5 * 5 * 18493 * 210943
97524222476 = 2 * 2 * 23029 * 1058711
97524222477 = 3 * 11 * 18457 * 160117
97524222478 = 2 * 23 * 151 * 14040343
97524222479 = 47 * 181 * 181 * 63337

212220020305 - 212220020319 is the second solution.

***

Andersen wrote:

Q1. Puzzle 428 shows the smallest for m=4 is 97524222465.

Q2. http://oeis.org/A067820 says the smallest case of 14 integers for m=5 starts at 5509463413255, found by Donovan Johnson. It's the best below 10^13.

***

Vicente wrote:

Q1: Si he entendido bien el problema, comencé a buscar números y buscando la secuencia obtenida en OEIS encontré lahttp://oeis.org/A067814
Donde Don Reble dice: "who remarks that the sequence is now complete".
Entiendo con esa afirmación que Don quiere decir que la secuencia está completa, pero falta, evidentemente, el número que encontró Forbes 488995430567765317569, o uno menor.

Q2:Encuentro la secuencia http://oeis.org/A067820. Buscar los siguientes excede la potencia de cálculo de mi máquina.

***

 Records   |  Conjectures  |  Problems  |  Puzzles