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)?

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

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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.

Later he added

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.

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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.

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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.

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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.

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