

A008364


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


41



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

First 48 terms give reduced residue system for 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
From Peter Bala, May 03 2018: (Start)
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
From Ruediger Jehn, Nov 05 2020: (Start)
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

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Alphonse de Polignac, Recherches Nouvelles sur les Nombres Premiers, in Comptes Rendus de l'Académie des Sciences, October 15 1849. See also Rectification.
A. de Polignac, Six propositions arithmologiques déduites du crible d'Ératosthène, Nouvelles annales de mathématiques : journal des candidats aux écoles polytechnique et normale, Série 1, Tome 8 (1849), pp. 423429.
Eric Weisstein's World of Mathematics, Rough Number
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).
Index entries for sequences related to smooth numbers


FORMULA

Starting with a(49) = 211, a(n) = a(n48) + 210.  Zak Seidov, Apr 11 2011
a(n) = a(n1) + a(n48)  a(n49).  Charles R Greathouse IV, Dec 21 2011
a020639(a(n)) > 7.  Reinhard Zumkeller, Mar 26 2012
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
a(n) = 35*n/8 + O(1).  Charles R Greathouse IV, Sep 14 2015


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 *)


PROG

(PARI) isA008364(n) = gcd(n, 210)==1 \\ Michael B. Porter, Oct 10 2009
(Haskell)
a008364 n = a008364_list !! (n1)
a008364_list = 1 : filter ((> 7) . a020639) [1..]
 Reinhard Zumkeller, Mar 26 2012


CROSSREFS

First differences give A049296. Cf. A002110, A048597.
For krough numbers with other values of k, see A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063.  Michael B. Porter, Oct 10 2009
Cf. A092695, A210679, A080672 (complement).
Sequence in context: A056758 A322273 A096489 * A140461 A120533 A226108
Adjacent sequences: A008361 A008362 A008363 * A008365 A008366 A008367


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


EXTENSIONS

New name from Charles R Greathouse IV, Dec 21 2011 based on comment from Michael B. Porter, Oct 10 2009


STATUS

approved



