A008366 - OEIS (original) (raw)

A008366

Smallest prime factor is >= 17.

14

1, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283

COMMENTS

Also the 17-rough numbers: positive integers that have no prime factors less than 17. - Michael B. Porter, Oct 10 2009

a(n) - (1001/192) n is periodic with period 5760. - Robert Israel, Mar 18 2016

The product of two 17-rough numbers is a 17-rough number and the prime factors of a 17-rough number are 17-rough numbers.

Let k equal either 13, 14, 15 or 16. Then the product of k numbers n*(n + a)*(n + 2*a)*...*(n + (k-1)*a) in arithmetical progression is divisible by k! for all integer n if and only if a is a 17-rough number.

The sequence terms satisfy the congruence x^60 = 1 (mod 30030), where 30030 = 2*3*5*7*11*13. (End)

The asymptotic density of this sequence is 192/1001. - Amiram Eldar, Sep 30 2020

FORMULA

Numbers n > 1 such that ((Sum_{k=1..n} k^10) mod n = 0) and ((Sum_{k=1..n} k^12) mod n = 0) (conjecture). - Gary Detlefs, Dec 27 2011

a(n) = a(n-1) + a(n-5760) - a(n-5761). - Vaclav Kotesovec, Mar 18 2016

G.f: x*P(x)/(1 - x - x^5760 + x^5761) where P(x) is a polynomial of degree 5760. - Benedict W. J. Irwin, Mar 23 2016

a(n) = (1001/192)*n + O(1), where the O(1) term is bounded by +/- 19. - Charles R Greathouse IV, Oct 13 2022

MAPLE

for i from 1 to 500 do if gcd(i, 30030) = 1 then print(i); fi; od;

MATHEMATICA

Select[ Range[ 300 ], GCD[ #1, 30030 ]==1& ]

Join[{1}, Select[Range[300], FactorInteger[#][[1, 1]]>=17&]] (* Harvey P. Dale, Mar 28 2020 *)