A075368 -id:A075368 - cp's OEIS frontend

A075365 Smallest k such that (n+1)(n+2)...(n+k) is divisible by the product of all the primes up to n.

0, 2, 3, 2, 5, 4, 7, 6, 5, 5, 11, 10, 13, 12, 11, 10, 17, 16, 19, 18, 17, 16, 23, 22, 21, 20, 19, 18, 29, 28, 31, 30, 29, 28, 27, 26, 37, 36, 35, 34, 41, 40, 43, 42, 41, 40, 47, 46, 45, 44, 43, 42, 53, 52, 51, 50, 49, 48, 59, 58, 61, 60, 59, 58, 57, 56, 67, 66, 65, 64, 71, 70

Offset: 1

Amarnath Murthy, Sep 20 2002

nonn

			a(6) = 4 as (6+1)*(6+2)*(6+3)*(6+4) is divisible by 2*3*5 but (6+1)*(6+2)*(6+3) is not.

T. D. Noe, Table of n, a(n) for n=1..1000

Cf. A075366, A075367, A075368.

a[n_] := Module[{div, k, pr}, div=Times@@Prime/@Range[PrimePi[n]]; For[k=0; pr=1, True, k++; pr*=n+k, If[Mod[pr, div]==0, Return[k]]]]

If p <= n < q, where p and q are consecutive primes, then a(n) = 2p-n, unless n=10. Sketch of proof: a(n) >= 2p-n, to make (n+1)...(n+a(n)) divisible by p. If r is a prime less than p and r does not divide (n+1)...(2p), then r > 2p-n and 2r <= n, so 4p < 3n < 3q. But q/p is known to be < 4/3 for all primes p >= 11.

Edited by Dean Hickerson, Oct 28 2002

A075366 Smallest product (n+1)(n+2)...(n+k) that is divisible by the product of all the primes up to n.

Original entry on oeis.org

1, 12, 120, 30, 30240, 5040, 17297280, 2162160, 240240, 360360, 28158588057600, 2346549004800, 64764752532480000, 4626053752320000, 308403583488000, 19275223968000, 830034394580628357120000, 46113021921146019840000

Offset: 1

Amarnath Murthy, Sep 20 2002

Reinhard Zumkeller, Table of n, a(n) for n = 1..300

Cf. A075365, A075367, A075368.

Cf. A000040.

a075366 n = a075366_list !! (n-1)
a075366_list = 1 : f 2 1 a000040_list where
   f x pp ps'@(p:ps)
     | p <= x    = f x (p * pp) ps
     | otherwise = g $ dropWhile (< pp) $ scanl1 (*) [x+1, x+2 ..]
     where g (z:zs) | mod z pp == 0 = z : f (x + 1) pp ps'
                    | otherwise     = g zs
-- Reinhard Zumkeller, May 18 2015

a75365[n_] := Module[{div, k, pr}, div=Times@@Prime/@Range[PrimePi[n]]; For[k=0; pr=1, True, k++; pr*=n+k, If[Mod[pr, div]==0, Return[k]]]]; a[n_] := Times@@Range[n+1, n+a75365[n]]

If p <= n < q, where p and q are consecutive primes, then a(n) = (2p)!/n!, unless n=10.

Edited by Dean Hickerson, Oct 28 2002

A075367 Smallest value of lcm(n+1, n+2, ..., n+k) (for k >= 0) that is divisible by the product of all the primes up to n.

Original entry on oeis.org

1, 12, 60, 30, 2520, 2520, 360360, 180180, 60060, 60060, 232792560, 232792560, 26771144400, 26771144400, 26771144400, 13385572200, 144403552893600, 144403552893600, 5342931457063200, 5342931457063200, 5342931457063200

Offset: 1

Amarnath Murthy, Sep 20 2002

T. D. Noe, Table of n, a(n) for n = 1..200

Cf. A075365, A075366, A075368.

a75365[n_] := Module[{div, k, pr}, div=Times@@Prime/@Range[PrimePi[n]]; For[k=0; pr=1, True, k++; pr*=n+k, If[Mod[pr, div]==0, Return[k]]]]; a[1]=1; a[n_] := LCM@@Range[n+1, n+a75365[n]]

a(n) = lcm(n+1, ..., n + A075365(n)).

Edited by Dean Hickerson, Oct 28 2002

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A075365 Smallest k such that (n+1)(n+2)...(n+k) is divisible by the product of all the primes up to n.

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A075366 Smallest product (n+1)(n+2)...(n+k) that is divisible by the product of all the primes up to n.

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A075367 Smallest value of lcm(n+1, n+2, ..., n+k) (for k >= 0) that is divisible by the product of all the primes up to n.

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Author

Keywords

Links

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Mathematica

Formula

Extensions