A164799 -id:A164799 - cp's OEIS frontend

A164798 a(n) = the smallest integer >= n such that a(n)!/(n-1)! is divisible by every prime from 2 to the largest prime divisor of a(n)!/(n-1)!. (a(1)=1.)

1, 2, 4, 4, 6, 6, 10, 8, 10, 14, 15, 12, 22, 15, 16, 16, 26, 18, 34, 21, 38, 38, 38, 24, 46, 46, 46, 46, 46, 30, 58, 32, 62, 62, 36, 36, 62, 74, 74, 74, 74, 82, 82, 86, 86, 86, 86, 48, 94, 94, 94, 94, 94, 54, 106, 106, 106, 106, 106, 60, 118, 122, 66, 64, 122, 122, 122, 134

Offset: 1

Leroy Quet, Aug 26 2009

nonn

a(n) = A164799(n) + n -1.

			Consider the products of consecutive integers, m!/9!, m >= 10. First, 10 is divisible by 2 and 5, but there is a prime gap since 3 is missing from the factorization. 10*11 is divisible by 2, 5, and 11, but 3 and 7 are missing. 10*11*12 is divisible by 2, 3, 5, and 11, but 7 is missing. 10*11*12*13 is divisible by all primes up to 13, except 7. But 10*11*12*13*14 is indeed divisible by every prime from 2 to 13. So a(10) = 14.

Cf. A164799

Terms beyond a(13) from R. J. Mathar, Feb 27 2010

A164858 a(n) = largest prime dividing A164857(n). a(1)=1.

Original entry on oeis.org

1, 2, 3, 2, 5, 3, 7, 2, 5, 13, 13, 3, 19, 7, 5, 2, 23, 3, 31, 7, 37, 37, 37, 3, 43, 43, 43, 43, 43, 5, 53, 2, 61, 61, 7, 3, 61, 73, 73, 73, 73, 79, 79, 83, 83, 83, 83, 3, 89, 89, 89, 89, 89, 3, 103, 103, 103, 103, 103, 5, 113, 113, 13, 2, 113, 113, 113, 131, 131, 131, 131, 3, 139

Offset: 1

Leroy Quet, Aug 28 2009

nonn

A164857(n) is divisible by every prime from 2 to A164858(n).

Cf. A055932, A164798, A164799, A164857.

Extended by Ray Chandler, Mar 14 2010

A164857 a(n) = A164798(n)/(n-1)!.

Original entry on oeis.org

1, 2, 12, 4, 30, 6, 5040, 8, 90, 240240, 360360, 12, 2346549004800, 210, 240, 16, 19275223968000, 18, 46113021921146019840000, 420, 214978908196382744494080000, 10237090866494416404480000

Offset: 1

Leroy Quet, Aug 28 2009

nonn

Each term a(n) is divisible by every prime from 2 to the largest prime dividing a(n). I.e., all terms are in sequence A055932.

Cf. A055932, A164798, A164799, A164858.

Extended by Ray Chandler, Mar 14 2010

A325480 a(n) is the largest integer m such that the product of n consecutive integers starting at m is divisible by at most n primes.

Original entry on oeis.org

16, 24, 24, 45, 48, 49, 120, 120, 125, 189, 240, 240, 350, 350, 350, 350, 374, 494, 494, 714, 714, 714, 714, 825, 832, 1078, 1078, 1078, 1078, 1425, 1440, 1440, 1856, 2175, 2175, 2175, 2175, 2175, 2175, 2175, 2870, 2870, 2870, 2871, 2880, 2880, 2880, 3219

Offset: 3

Onno M. Cain, Sep 06 2019

nonn

Each term is only conjectured and has been verified up to 10^6.

Note a(2) is undefined if there are infinitely many Mersenne primes.

			For example, a(3) = 16 because 16 * 17 * 18 = 2^5 * 3^2 * 17 admits only three prime divisors (2, 3, and 17) and appears to be the largest product of three consecutive integers with the property.

Cf. A002182, A045619, A163264, A164799, A239035, A244656, A006549.

for r in range(3, 100):
  history = []
  M = 0
  for n in range(1, 100000):
    primes = {p for p, _ in factor(n)}
    history.append(primes)
    history = history[-r:]
    total = set()
    for s in history: total |= s
    # Skip if too many primes.
    if len(total) > r: continue
    if n > M: M = n
  print(r, M-r+1)

cp's OEIS Frontend

A164798 a(n) = the smallest integer >= n such that a(n)!/(n-1)! is divisible by every prime from 2 to the largest prime divisor of a(n)!/(n-1)!. (a(1)=1.)

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A164858 a(n) = largest prime dividing A164857(n). a(1)=1.

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A164857 a(n) = A164798(n)/(n-1)!.

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Extensions

A325480 a(n) is the largest integer m such that the product of n consecutive integers starting at m is divisible by at most n primes.

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Comments

Examples

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SageMath