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 A008364 11-rough numbers: not divisible by 2, 3, 5 or 7. 37
 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 (list; graph; refs; listen; history; text; internal format)
 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 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) 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 1s and 169s 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 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 REFERENCES Diatomic sequence of 4th prime: A. de Polignac (1849), J. Dechamps J. (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,  Springer-Verlag, 1980. LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 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). FORMULA Starting with a(49) = 211, a(n) = a(n-48) + 210. - Zak Seidov, Apr 11 2011 a(n) = a(n-1) + a(n-48) - a(n-49). - 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) = 35n/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, 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 !! (n-1) a008364_list = 1 : filter ((> 7) . a020639) [1..] -- Reinhard Zumkeller, Mar 26 2012 CROSSREFS First differences give A049296. Cf. A002110, A048597. 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 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 EXTENSIONS New name from Charles R Greathouse IV, Dec 21 2011 based on comment from Michael B. Porter, Oct 10 2009 STATUS approved

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Last modified May 26 09:37 EDT 2020. Contains 334620 sequences. (Running on oeis4.)