A208644 Consider any chain of consecutive primes which divides n; take all longest such chains; maximize the product of the primes in the chain.

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

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