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A208644 Consider any chain of consecutive primes which divides n; take all longest such chains; maximize the product of the primes in the chain. 0
1, 2, 3, 2, 5, 6, 7, 2, 3, 5, 11, 6, 13, 7, 15, 2, 17, 6, 19, 5, 7, 11, 23, 6, 5, 13, 3, 7, 29, 30, 31, 2, 11, 17, 35, 6, 37, 19, 13, 5, 41, 6, 43, 11, 15, 23, 47, 6, 7, 5, 17, 13, 53, 6, 11, 7, 19, 29, 59, 30, 61, 31, 7, 2, 13, 6, 67, 17, 23, 35, 71, 6, 73 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

LINKS

Table of n, a(n) for n=1..73.

EXAMPLE

No primes divide 1, so the empty product 1 = a(1).

4290 = 2 * 3 * 5 * 11 * 13 which is divisible by three consecutive primes 2, 3, and 5 so a(4290) = 2 * 3 * 5 = 30. (11 * 13 is larger but a shorter chain.)

PROG

(PARI) a(n)=if(n<4, n, my(f=factor(n)[, 1], runLen=1, runProd=f[1], rLen=1, rProd=f[1]); for(i=2, #f, if(nextprime(f[i-1]+1)==f[i], runProd*=f[i]; runLen++, if(runLen>=rLen, rLen=runLen; rProd=runProd); runLen=1; runProd=f[i])); if(runLen<rLen, rProd, runProd)) \\ Charles R Greathouse IV, Mar 02 2012

CROSSREFS

Sequence in context: A056552 A049273 A053590 * A162323 A165743 A086297

Adjacent sequences:  A208641 A208642 A208643 * A208645 A208646 A208647

KEYWORD

nonn

AUTHOR

Charles R Greathouse IV, Mar 02 2012

STATUS

approved

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Last modified July 11 01:51 EDT 2020. Contains 335600 sequences. (Running on oeis4.)