A075367 -id:A075367 - 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

A075368 Smallest integer value of lcm(n+1, n+2, ..., n+k) (for k >= 0) divided by the product of all the primes up to n.

Original entry on oeis.org

1, 6, 10, 5, 84, 84, 1716, 858, 286, 286, 100776, 100776, 891480, 891480, 891480, 445740, 282861360, 282861360, 550835280, 550835280, 550835280, 550835280, 42222721680, 42222721680, 8444544336, 8444544336, 2814848112, 2814848112

Offset: 1

Amarnath Murthy, Sep 20 2002

			a(3) = 10 as the product of primes <= (n = 3) is 6 and the smallest integer of the form lcm(3+1, 3+2, ..., 3+k) = lcm(4, 5, 6) = 60 giving a(3) = 60/6 = 10. - _David A. Corneth_, Dec 05 2023

David A. Corneth, Table of n, a(n) for n = 1..2236

Cf. A075365, A075366, A075367.

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]]/Times@@Prime/@Range[PrimePi[n]]

a(n) = {if(n==1, return(1));
	my(pp = vecprod(primes(primepi(n))), l = n+1);
	for(k = n+2, 2*n,
		l = lcm(l, k);
		if(l%pp == 0,
			return(l\pp)
		)
	)
} \\ David A. Corneth, Dec 05 2023

a(n) = A075367(n)/A034386(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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A075368 Smallest integer value of lcm(n+1, n+2, ..., n+k) (for k >= 0) divided by the product of all the primes up to n.

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