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Revision History for A342876 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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Numbers that can be written as a product of k consecutive composite numbers and also of k+1 consecutive composite numbers, for some k>1.
(history; published version)

#8 by N. J. A. Sloane at Sun Mar 28 18:39:31 EDT 2021

STATUS

editing

approved

#7 by N. J. A. Sloane at Sun Mar 28 18:39:20 EDT 2021

COMMENTS

Similar to but different from A175340 which has an additional condition that says that factors cannot be used twice. So 1814400 is here via 8*9*10*12*14*15 = 15*16*18*20*21 but not there as 15 is used twice.

STATUS

proposed

editing

Discussion

Sun Mar 28

18:39

N. J. A. Sloane: small edit

#6 by David A. Corneth at Sun Mar 28 17:25:53 EDT 2021

STATUS

editing

proposed

#5 by David A. Corneth at Sun Mar 28 15:06:42 EDT 2021

KEYWORD

nonn,more,changed

STATUS

proposed

editing

Discussion

Sun Mar 28

15:09

David A. Corneth: Michel, done, thanks

15:10

David A. Corneth: Hugo, that seems likely. Also all terms are 31-smooth. I don't think all terms will be 31-smooth. But I don't see a term divisible by 103 any time soon.

16:12

Michael S. Branicky: David, won't you get something divisible by any number n using your c1 * c2 * product(S) construction?

16:30

Michael S. Branicky: Is this worth a comment? "All primes p appear as a factor of some term, choosing c1 = 4 and c2 = 2*p in the construction above."

17:20

David A. Corneth: yes we can find a multiple of any n at some point. It just seems that they get pretty big. Just guessing but even at 1.1*10^14 we haven't found a multiple of 37 so 103 (for example) seems far away.

17:22

David A. Corneth: Not sure your choices for c1 and c2 work as they most be consecutive composites. c1 = 4p and c2 = the next composite after c1 might work.

17:26

Michael S. Branicky: Got it.  Thank you.

#4 by David A. Corneth at Sun Mar 28 12:28:12 EDT 2021

STATUS

editing

proposed

Discussion

Sun Mar 28

12:44

Michel Marcus: needs keyword more

13:16

Hugo Pfoertner: All terms divisible by 120?

#3 by David A. Corneth at Sun Mar 28 12:27:16 EDT 2021

COMMENTS

Similar but different from A175340 which has an additional condition that says that factors cannot be used twice. So 1814400 is here via 8*9*10*12*14*15 = 15*16*18*20*21 but not there as 15 is used twice.

This sequence is infinite. Proof:

Choose two consecutive composites c1 and c2 such that c1 < c2. Let S be the composites strictly between c2 and c1 * c2. Let product(S) be the product of terms in S. Then c1 * c2 * product(S) is a term.

CROSSREFS

Cf. A175340.

#2 by David A. Corneth at Sun Mar 28 12:21:28 EDT 2021

NAME

allocated Numbers that can be written as a product of k consecutive composite numbers and also of k+1 consecutive composite numbers, for David Asome k>1. Corneth

DATA

1680, 4320, 120960, 166320, 175560, 215760, 725760, 1814400, 1080647568000, 104613949440000, 115880067072000

OFFSET

1,1

KEYWORD

allocated

nonn

AUTHOR

David A. Corneth, Mar 28 2021

STATUS

approved

editing

#1 by David A. Corneth at Sun Mar 28 12:20:14 EDT 2021

NAME

allocated for David A. Corneth

KEYWORD

allocated

STATUS

approved