A047350 Numbers that are congruent to {1, 2, 4} mod 7.

1, 2, 4, 8, 9, 11, 15, 16, 18, 22, 23, 25, 29, 30, 32, 36, 37, 39, 43, 44, 46, 50, 51, 53, 57, 58, 60, 64, 65, 67, 71, 72, 74, 78, 79, 81, 85, 86, 88, 92, 93, 95, 99, 100, 102, 106, 107, 109, 113, 114, 116, 120, 121, 123, 127, 128, 130, 134, 135, 137, 141

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

a(n+1) = a(n) + (a(n) mod 7). - Ben Paul Thurston, Jan 09 2008

Also defined by: a(1)=1, and a(n) = smallest number larger than a(n-1) such that a(n)^3 - a(n-1)^3 is divisible by 7. - Zak Seidov, Apr 21 2009

Union of A047353 and A017029. - R. J. Mathar, Apr 28 2009

Indices of the even numbers in the Padovan sequence. - Francesco Daddi, Jul 31 2011

Euler's problem (see Link lines, English translation by David Zao): Finding the values of a so that the form a^3-1 is divisible by 7. The three residuals that remain after the division of any square by 7 are 1, 2 and 4. Hence the values are 7n+1, 7n+2, 7n+4. - Bruno Berselli, Oct 24 2012

Links

• Leonhard Euler, The Euler Archive, Theoremata circa divisores numerorum (E134), Novi Commentarii academiae scientiarum imperialis Petropolitanae, Volume 1 (1750), p. 40 (Theorem II, example 2).
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Crossrefs

Cf. A000931, A017029, A047353, A134720.

Programs

Magma

[n : n in [0..150] | n mod 7 in [1, 2, 4]]; // Wesley Ivan Hurt, Jun 13 2016

Maple

A047350:=n->(21*n-21-6*cos(2*n*Pi/3)+4*sqrt(3)*sin(2*n*Pi/3))/9: seq(A047350(n), n=1..100); # Wesley Ivan Hurt, Jun 13 2016

Mathematica

Select[Range[0, 150], MemberQ[{1, 2, 4}, Mod[#, 7]] &] (* Wesley Ivan Hurt, Jun 13 2016 *)

PARI

a(n)=n\3*7+[-3,1,2][n%3+1] \\ Charles R Greathouse IV, Jul 31 2011

Formula

From R. J. Mathar, Apr 28 2009: (Start)

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

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

a(n) = a(n-3) + 7 for n > 3. (End)

From Wesley Ivan Hurt, Jun 13 2016: (Start)

a(n) = (21*n - 21 - 6*cos(2*n*Pi/3) + 4*sqrt(3)*sin(2*n*Pi/3))/9.

a(3k) = 7k-3, a(3k-1) = 7k-5, a(3k-2) = 7k-6. (End)

a(n) = 4*n - 3 - 2*floor(n/3) - 3*floor((n+1)/3). - Ridouane Oudra, Nov 23 2022