

A008364


11rough numbers: not divisible by 2, 3, 5 or 7.


44



1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 209, 211, 221, 223, 227, 229, 233, 239, 241, 247
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OFFSET

1,2


COMMENTS

The first A005867(4) = 48 terms give the reduced residue system for the 4th primorial number 210 = A002110(4).
This sequence is closed under multiplication: any product of terms is also a term.  Labos Elemer, Feb 26 2003
Conjecture: these are numbers n such that (Sum_{k=1..n} k^4) mod n = 0 and (Sum_{k=1..n} k^6) mod n = 0.  Gary Detlefs, Dec 20 2011
The above conjecture is true. Let m be even and let the mth Bernoulli number be written in reduced form as Bernoulli(m) = N(m)/D(m). Apply Ireland and Rosen, Proposition 15.2.2, to show the congruence D(m)*( Sum_{k = 1..n} k^m )/n = N(m) (mod n) holds for all n >= 1. It follows easily from this congruence that ( Sum_{k = 1..n} k^m )/n is integral iff n is coprime to D(m). Now Bernoulli(4) = 1/(2*3*5) and Bernoulli(6) = 1/(2*3*7) so the numbers n such that both (Sum_{k=1..n} k^4) mod n = 0 and (Sum_{k=1..n} k^6) mod n = 0 are exactly those numbers coprime to the primes 2, 3, 5 and 7, that is, the 11rough numbers. (End)
Conjecture: these are numbers n such that (n^6 mod 210 = 1) or (n^6 mod 210 = 169).  Gary Detlefs, Dec 30 2011
The second Detlefs conjecture above is true and extremely easy to verify with some basic properties of congruences: take the terms of this sequence up to 209 and compute their sixth powers modulo 210: there should only be 1's and 169's there. Then take the complement of this sequence up to 210, where you will see no instances of 1 or 169.  Alonso del Arte, Jan 12 2014
It is wellknown that the product of 7 consecutive integers is divisible by 7!. Conjecture: This sequence is exactly the set of positive values of r such that ( Product_{k = 0..6} n + k*r )/7! is an integer for all n.  Peter Bala, Nov 14 2015
This conjecture is true. The first part of the proof deals with numbers not in A008364, i.e., numbers which are divisible by p (p either 2, 3, 5, 7). Let r = p*s and n = 1, then (Product_{k = 0..6} n + k*r) is not divisible by p, because none of the factors 1 + k*p*s are divisible by p. Hence dividing the product by 7! does not return an integer.
The second part deals with numbers in A008364. If r and q are coprime, then for any i < q there exists k < q with (k*r mod q) = i. From this, it also follows that for any n there exists k < q with ((n + k*r) mod q) = 0. But this means that Product_{k = 0..6} n + k*r is divisible by all numbers from 2 to 7 because there is always a factor that is divisible. We still have to show that the product is also divisible by 2 times 3 times 4 times 6. If the k_1 with ((n + k_1*r) mod 4) = 0 is even, then (n mod 2) = ((n + 2*r) mod 2) = ((n + 4*r) mod 2) = ((n + 6*r) mod 2) = 0. If this k_1 is odd, then ((n + r) mod 2) = ((n + 3*r) mod 2) = ((n + 5*r) mod 2) = 0. In both cases there are at least 2 other factors divisible by 2. If the k_2 with ((n + k_2*r) mod 6) = 0 is smaller than 4, then ((n + (k_2 + 3)*r) mod 3) = 0. Otherwise, ((n + (k_2  3)*r) mod 3) = 0. In both cases there is at least 1 other factor divisible by 3. And therefore Product_{k = 0..6} n + k*r is divisible by 7! for any n.
(End)


REFERENCES

Diatomic sequence of 4th prime: A. de Polignac (1849), J. Dechamps (1907).
Dickson L. E., History of the Theory of Numbers, Vol. 1, p. 439, Chelsea, 1952.
K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, SpringerVerlag, 1980.


LINKS

Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1).


FORMULA

Starting with a(49) = 211, a(n) = a(n48) + 210.  Zak Seidov, Apr 11 2011
G.f.: x*(x^48 + 10*x^47 + 2*x^46 + 4*x^45 + 2*x^44 + 4*x^43 + 6*x^42 + 2*x^41 + 6*x^40 + 4*x^39 + 2*x^38 + 4*x^37 + 6*x^36 + 6*x^35 + 2*x^34 + 6*x^33 + 4*x^32 + 2*x^31 + 6*x^30 + 4*x^29 + 6*x^28 + 8*x^27 + 4*x^26 + 2*x^25 + 4*x^24 + 2*x^23 + 4*x^22 + 8*x^21 + 6*x^20 + 4*x^19 + 6*x^18 + 2*x^17 + 4*x^16 + 6*x^15 + 2*x^14 + 6*x^13 + 6*x^12 + 4*x^11 + 2*x^10 + 4*x^9 + 6*x^8 + 2*x^7 + 6*x^6 + 4*x^5 + 2*x^4 + 4*x^3 + 2*x^2 + 10*x + 1) / (x^49  x^48  x + 1).  Colin Barker, Sep 27 2013


MAPLE

for i from 1 to 500 do if gcd(i, 210) = 1 then print(i); fi; od;
t1:=[]; for i from 1 to 1000 do if gcd(i, 210) = 1 then t1:=[op(t1), i]; fi; od: t1;
S:= (j, n)> sum(k^j, k=1..n): for n from 1 to 247 do if (S(4, n) mod n = 0) and (S(6, n) mod n = 0) then print(n) fi od; # Gary Detlefs, Dec 20 2011


MATHEMATICA

Select[ Range[ 300 ], GCD[ #1, 210 ] == 1 & ]
Select[Range[250], Mod[#, 2]>0 && Mod[#, 3]>0 && Mod[#, 5]>0 && Mod[#, 7]>0 &] (* Vincenzo Librandi, Nov 16 2015 *)
Cases[Range@1000, x_ /; NoneTrue[Array[Prime, 4], Divisible[x, #] &]] (* Mikk Heidemaa, Dec 07 2017 *)
Select[Range[250], Union[Divisible[#, {2, 3, 5, 7}]]=={False}&] (* Harvey P. Dale, Sep 24 2021 *)


PROG

(Haskell)
a008364 n = a008364_list !! (n1)
a008364_list = 1 : filter ((> 7) . a020639) [1..]


CROSSREFS

For krough numbers with other values of k, see A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063.  Michael B. Porter, Oct 10 2009


KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



