A047390 Numbers that are congruent to {0, 3, 5} mod 7.

0, 3, 5, 7, 10, 12, 14, 17, 19, 21, 24, 26, 28, 31, 33, 35, 38, 40, 42, 45, 47, 49, 52, 54, 56, 59, 61, 63, 66, 68, 70, 73, 75, 77, 80, 82, 84, 87, 89, 91, 94, 96, 98, 101, 103, 105, 108, 110, 112, 115, 117, 119, 122, 124, 126, 129, 131, 133, 136, 138, 140, 143

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Also numbers k such that k*(k+2)*(k+4) is divisible by 7. - Bruno Berselli, Dec 28 2017

Links

• Daniel Starodubtsev, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Crossrefs

Cf. A011655, A047391, A049347, A079978.

Programs

Magma

[n: n in [0..122] | n mod 7 in [0, 3, 5]];  // Bruno Berselli, Mar 29 2011

Maple

seq(2*n+floor(n/3)+(n^2 mod 3), n=0..52); # Gary Detlefs, Mar 19 2010

Mathematica

Select[Range[0,150], MemberQ[{0,3,5}, Mod[#,7]]&] (* Harvey P. Dale, Dec 07 2011 *)
CoefficientList[Series[x (3 + 2 x + 2 x^2)/((1 - x)^2 (1 + x + x^2)), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 02 2014 *)

PARI

is(n)=!!setsearch([0,3,5],n%7) \\ Charles R Greathouse IV, Nov 09 2014

PARI

a(n)=(7*n-5)\3 \\ Charles R Greathouse IV, Nov 09 2014

Python

import math
a = lambda n: 2*(n-1)+math.ceil((n-1)/3.0)
for n in range(1,101): print(a(n), end = ", ") # Karl V. Keller, Jr., Nov 01 2014

Formula

a(n) = 2*n + floor(n/3) + (n^2 mod 3), with offset 0, a(0)=0. - Gary Detlefs, Mar 19 2010

From Bruno Berselli, Mar 29 2011: (Start)

G.f.: x^2*(3 + 2*x + 2*x^2)/((1 - x)^2*(1 + x + x^2)).

a(n) = (1/3)*(7*n - 6 - A049347(n-1)) = A047391(n) - A079978(n-1). (End)

a(n) = n + ceiling(4*(n-1)/3) - 1. - Arkadiusz Wesolowski, Sep 18 2012

a(n) = 2*(n-1) + ceiling((n-1)/3). - Karl V. Keller, Jr., Nov 01 2014

From Wesley Ivan Hurt, Jun 10 2016: (Start)

a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.

a(n) = 7*n/3 - 2 - 2*sin(2*n*Pi/3)/(3*sqrt(3)).

a(3*k) = 7*k-2, a(3*k-1) = 7*k-4, a(3*k-2) = 7*k-7. (End)