A007775 Numbers not divisible by 2, 3 or 5.

1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83, 89, 91, 97, 101, 103, 107, 109, 113, 119, 121, 127, 131, 133, 137, 139, 143, 149, 151, 157, 161, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 203, 209

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Also numbers n such that the sum of the 4th powers of the first n positive integers is divisible by n, or A000538(n) = n*(n+1)(2*n+1)(3*n^2+3*n-1)/30 is divisible by n. - Alexander Adamchuk, Jan 04 2007

Also the 7-rough numbers: positive integers that have no prime factors less than 7. - Michael B. Porter, Oct 09 2009

a(n) mod 3 has period 8, repeating [1,1,2,1,2,1,2,2] = (n mod 2) + floor(((n-1) mod 8)/7) - floor(((n-2) mod 8)/7) + 1. floor(a(n)/3) is the set of numbers k such that k is congruent to {0,2,3,4,5,6,7,9} mod 10 = floor((5*n-2)/4)-floor((n mod 8)/6). - Gary Detlefs, Jan 08 2012

Numbers k such that C(k+3,3)==1 (mod k) and C(k+5,5)==1 (mod k). - Gary Detlefs, Sep 15 2013

a(n) mod 30 has period 8 repeating [1, 7, 11, 13, 17, 19, 23, 29]. The mean of these 8 numbers is 120/8 = 15. (a(n)-15) mod 30 has period 8 repeating [-14, -8, -4, -2, 2, 4, 8, 14]. One half of the absolute value produces the symmetric sequence [7, 4, 2, 1, 1, 2, 4, 7] = A061501(((n-1) + 16) mod 8). - Gary Detlefs, Sep 24 2013

a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)*(n + 3*m)(n + 4*m)/120 is integral. Cf. A007310. - Peter Bala, Nov 13 2015

The asymptotic density of this sequence is 4/15. - Amiram Eldar, Sep 30 2020

If a(n) + a(n+1) = 0 (mod 30), then a(n-j) + a(n+j+1) = a(n) + a(n+1) for each j in [1, n-1]. - Alexandre Herrera, Jun 27 2023

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..8000
• Peter Bala, A note on A007775.
• Abel Jansma, E_8 Symmetry Structures in the Ising model, Master's thesis, University of Amsterdam, 2018.
• Eric Weisstein's World of Mathematics, Rough Number.
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,1,-1).
• Index entries for sequences related to smooth numbers

Crossrefs

Cf. A000538, A054403, A145011 (first differences).

For k-rough numbers with other values of k, see A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063.

Complement is A080671.

For digital root of Fibonacci numbers indexed by this sequence, see A227896.

Programs

Haskell

a007775 n = a007775_list !! (n-1)
a007775_list = 1 : filter ((> 5) . a020639) [1..]
-- Reinhard Zumkeller, Jan 06 2013

Magma

I:=[1, 7, 11, 13, 17, 19, 23, 29, 31]; [n le 9 select I[n] else Self(n-1) +Self(n-8) - Self(n-9): n in [1..80]]; // G. C. Greubel, Oct 22 2018

Maple

for i from 1 to 500 do if gcd(i,30) = 1 then print(i); fi; od;
for k from 1 to 300 do if ((k^2 mod 48=1) or (k^2 mod 48=25)) and ((k^2 mod 120=1) or (k^2 mod 120=49)) then print(k) fi od. # Gary Detlefs, Dec 30 2011

Mathematica

Select[ Range[ 300 ], GCD[ #1, 30 ]==1& ]
Select[Range[250], Mod[#, 2]>0&&Mod[#, 3]>0&&Mod[#, 5]>0&] (* Vincenzo Librandi, Feb 08 2014 *)
a[ n_] := Quotient[ n, 8, 1] 30 + {1, 7, 11, 13, 17, 19, 23, 29}[[Mod[n, 8, 1]]]; (* Michael Somos, Jun 02 2014 *)
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 1, -1}, {1, 7, 11, 13, 17, 19, 23, 29, 31}, 100] (* Mikk Heidemaa, Dec 07 2017 *)
Cases[Range@1000, x_ /; NoneTrue[Array[Prime, 3], Divisible[x, #] &]] (* Mikk Heidemaa, Dec 07 2017 *)
CoefficientList[ Series[(x^8 + 6x^7 + 4x^6 + 2x^5 + 4x^4 + 2x^3 + 4x^2 + 6x + 1)/((x - 1)^2 (x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)), {x, 0, 55}], x] (* Robert G. Wilson v, Dec 07 2017 *)

PARI

isA007775(n) = gcd(n,30)==1 \\ Michael B. Porter, Oct 09 2009

PARI

{a(n) = n\8 * 30 + [ -1, 1, 7, 11, 13, 17, 19, 23][n%8 + 1]} /* Michael Somos, Feb 05 2011 */

PARI

{a(n) = n\8 * 6 + 9 + 3 * (n+1)\2 * 2 - max(5, (n-2)%8) * 2} /* Michael Somos, Jun 02 2014 */

PARI

Vec(x*(1+6*x+4*x^2+2*x^3+4*x^4+2*x^5+4*x^6+6*x^7+x^8)/((1+x)*(x^2+1)*(x^4+1)*( x-1)^2) + O(x^100)) \\ Altug Alkan, Nov 16 2015

Python

def A007775(n): return ((m:=n-1)<<2|1)-(m>>2&-2)+(2,0,-2,0)[m-1>>1&3] # Chai Wah Wu, Feb 02 2025

Sage

a = lambda n: ((((n-1)<< 2)-((n-1)>>2))|1) + ((((n-1)<<1)-((n-1)>> 1)) & 2)
print([a(n) for n in (1..56)]) # after Andrew Lelechenko, Peter Luschny, Jul 08 2017

Formula

A141256(a(n)) = n+1. - Reinhard Zumkeller, Jun 17 2008

From R. J. Mathar, Feb 27 2009: (Start)

a(n+8) = a(n) + 30.

a(n) = a(n-1) + a(n-8) - a(n-9).

G.f.: x*(1 + 6*x + 4*x^2 + 2*x^3 + 4*x^4 + 2*x^5 + 4*x^6 + 6*x^7 + x^8)/((1 + x)*(x^2 + 1)*(x^4 + 1)*(x-1)^2). (End)

a(n) = 4*n - 3 - 2*floor((n-1)/8) + (1 + (-1)^floor((n-2)/2))*(-1)^floor((n-2)/4), n >= 1. - Timothy Hopper, Mar 14 2010

a(1 - n) = -a(n). - Michael Somos, Feb 05 2011

Numbers k such that ((k^2 mod 48=1) or (k^2 mod 48=25)) and ((k^2 mod 120=1) or (k^2 mod 120=49)). - Gary Detlefs, Dec 30 2011

Numbers k such that k^2 mod 30 is 1 or 19. - Gary Detlefs, Dec 31 2011

a(n) = 3*(floor((5*n-2)/4) - floor((n mod 8)/6)) + (n mod 2) + floor(((n-1) mod 8)/7) - floor(((n-2) mod 8)/7) + 1. - Gary Detlefs, Jan 08 2012

a(n) = 4*n - 3 + 2*(floor((n+6)/8) - floor((n+4)/8) - floor((n+2)/8) + floor(n/8) - floor((n-1)/8)), n >= 1. From the o.g.f. given above by R. J. Mathar (with the denominator written as (1-x^8)*(1-x)), and a two-step reduction of the floor functions. Compare with Hopper's and Detlefs's formulas above. - Wolfdieter Lang, Jan 26 2012

a(n) = (6*f(n) - 3 + (-1)^f(n))/2, where f(n)= n + floor(n/4)+ floor(((n+4) mod 8)/6). - Gary Detlefs, Sep 15 2013

a(n) = 30*floor((n-1)/8) + 15 + 2*f((n-1) mod 8 + 16)*(-1)^floor(((n+3) mod 8)/4), where f(n) = (n*(n+1)/2+1) mod 10. - Gary Detlefs, Sep 24 2013

a(n) = 3*n + 6*floor(n/8) + (n mod 2) - 2*floor(((n-2) mod 8)/6) - 2*floor(((n-2) mod 8)/7) + 1. - Gary Detlefs, Jun 01 2014

a(n+1) = ((n << 2 - n >> 2) || 1) + ((n << 1 - n >> 1) && 2), where << and >> are bitwise left and right shifts, || and && are bitwise "or" and "and". - Andrew Lelechenko, Jul 08 2017

a(n) = 2*n + 2*floor(1/2 + (7*n)/8) + 2*(91 mod (2 - ((3*n)/4 + n^2/4) mod 2)) - 3 (n > 0). - Mikk Heidemaa, Dec 06 2017

Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(23 + sqrt(5) - sqrt(6*(5 + sqrt(5))))*Pi/15. - Amiram Eldar, Dec 13 2021