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A008364 11-rough numbers: not divisible by 2, 3, 5 or 7.

%I A008364 #147 May 07 2025 17:20:01
%S A008364 1,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,
%T A008364 103,107,109,113,121,127,131,137,139,143,149,151,157,163,167,169,173,
%U A008364 179,181,187,191,193,197,199,209,211,221,223,227,229,233,239,241,247
%N A008364 11-rough numbers: not divisible by 2, 3, 5 or 7.
%C A008364 The first A005867(4) = 48 terms give the reduced residue system for the 4th primorial number 210 = A002110(4).
%C A008364 This sequence is closed under multiplication: any product of terms is also a term. - _Labos Elemer_, Feb 26 2003
%C A008364 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
%C A008364 From _Peter Bala_, May 03 2018: (Start)
%C A008364 The above conjecture is true. Let m be even and let the m-th 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 11-rough numbers. (End)
%C A008364 Conjecture: these are numbers n such that (n^6 mod 210 = 1) or (n^6 mod 210 = 169). - _Gary Detlefs_, Dec 30 2011
%C A008364 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
%C A008364 It is well-known 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
%C A008364 From _Ruediger Jehn_, Nov 05 2020: (Start)
%C A008364 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.
%C A008364 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.
%C A008364 (End)
%D A008364 Diatomic sequence of 4th prime: A. de Polignac (1849), J. Dechamps (1907).
%D A008364 Dickson L. E., History of the Theory of Numbers, Vol. 1, p. 439, Chelsea, 1952.
%D A008364 K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory,  Springer-Verlag, 1980.
%H A008364 Reinhard Zumkeller, <a href="/A008364/b008364.txt">Table of n, a(n) for n = 1..10000</a>
%H A008364 Alphonse de Polignac, <a href="https://gallica.bnf.fr/ark:/12148/bpt6k2986m/f401.item">Recherches Nouvelles sur les Nombres Premiers</a>, in Comptes Rendus de l'Académie des Sciences, October 15 1849. See also <a href="https://gallica.bnf.fr/ark:/12148/bpt6k2986m/f742.item">Rectification</a>.
%H A008364 Alphonse de Polignac, <a href="http://www.numdam.org/item/NAM_1849_1_8__423_1/">Six propositions arithmologiques déduites du crible d'Ératosthène</a>, Nouvelles annales de mathématiques : journal des candidats aux écoles polytechnique et normale, Série 1, Tome 8 (1849), pp. 423-429.
%H A008364 Han-Lin Li, Shu-Cherng Fang, and Way Kuo, <a href="https://doi.org/10.4236/apm.2024.145023">The Periodic Table of Primes</a>, Advances in Pure Mathematics, Volume 14, Issue 5, May 2024.
%H A008364 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RoughNumber.html">Rough Number</a>
%H A008364 <a href="/index/Rec#order_49">Index entries for linear recurrences with constant coefficients</a>, 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).
%H A008364 <a href="/index/Sk#smooth">Index entries for sequences related to smooth numbers</a>
%F A008364 Starting with a(49) = 211, a(n) = a(n-48) + 210. - _Zak Seidov_, Apr 11 2011
%F A008364 a(n) = a(n-1) + a(n-48) - a(n-49). - _Charles R Greathouse IV_, Dec 21 2011
%F A008364 A020639(a(n)) > 7. - _Reinhard Zumkeller_, Mar 26 2012
%F A008364 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
%F A008364 a(n) = 35*n/8 + O(1). - _Charles R Greathouse IV_, Sep 14 2015
%F A008364 A007775 INTERSECT A206547. - _R. J. Mathar_, Apr 10 2024
%p A008364 for i from 1 to 500 do if gcd(i,210) = 1 then print(i); fi; od;
%p A008364 t1:=[]; for i from 1 to 1000 do if gcd(i,210) = 1 then t1:=[op(t1),i]; fi; od: t1;
%p A008364 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
%t A008364 Select[ Range[ 300 ], GCD[ #1, 210 ] == 1 & ]
%t A008364 Select[Range[250], Mod[#, 2]>0 && Mod[#, 3]>0 && Mod[#, 5]>0 && Mod[#, 7]>0 &] (* _Vincenzo Librandi_, Nov 16 2015 *)
%t A008364 Cases[Range@1000, x_ /; NoneTrue[Array[Prime, 4], Divisible[x, #] &]] (* _Mikk Heidemaa_, Dec 07 2017 *)
%t A008364 Select[Range[250],Union[Divisible[#,{2,3,5,7}]]=={False}&] (* _Harvey P. Dale_, Sep 24 2021 *)
%o A008364 (PARI) isA008364(n) = gcd(n,210)==1 \\ _Michael B. Porter_, Oct 10 2009
%o A008364 (Haskell)
%o A008364 a008364 n = a008364_list !! (n-1)
%o A008364 a008364_list = 1 : filter ((> 7) . a020639) [1..]
%o A008364 -- _Reinhard Zumkeller_, Mar 26 2012
%Y A008364 First differences give A049296. Cf. A002110, A048597.
%Y A008364 For k-rough numbers with other values of k, see A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063. - _Michael B. Porter_, Oct 10 2009
%Y A008364 Cf. A005867, A092695, A210679, A080672 (complement).
%K A008364 nonn,easy
%O A008364 1,2
%A A008364 _N. J. A. Sloane_
%E A008364 New name from _Charles R Greathouse IV_, Dec 21 2011 based on comment from _Michael B. Porter_, Oct 10 2009