A033940 - OEIS

%I #60 Sep 08 2022 08:44:51

%S 1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,

%T 4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,

%U 2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6

%N a(n) = 10^n mod 7.

%C This sequence can be employed in a test for divisibility by seven. Given the decimal expansion of some natural number, it is easily shown that the following sum has the same remainder under division by seven as the original number and that this sum is strictly smaller than the original number: Successively take the digits of the number in reverse order and multiply each of them by the respective term of the sequence A033940, then sum the products. By repeating this process, since the sums decrease in size, one ends up with seven if and only if the initial number is divisible by seven. Example: 43638 is divisible by seven since 8*1 + 3*3 + 6*2 + 3*6 + 4*4 = 63 and 3*1 + 6*3 = 21 and 1*1 + 2*3 = 7. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 16 2007

%C Representation of (3^n) in the circle with seven equidistant points, (10^n) mod 7=(3^n) mod 7. - _Eric Desbiaux_, Feb 15 2009

%C Representation of multiples of 3 in the circle (with seven equidistant points), see the Chryzodes links. - _Eric Desbiaux_, Feb 14 2009

%C Equivalently 3^n mod 7. - _Zerinvary Lajos_, Nov 24 2009

%C Continued fraction expansion of (269+11*sqrt(1086))/490. Decimal expansion of 1195/9009. - _Klaus Brockhaus_, May 24 2010

%C Period 6: Repeat [1, 3, 2, 6, 4, 5]. - _Wesley Ivan Hurt_, Jul 06 2014

%H Vincenzo Librandi, <a href="/A033940/b033940.txt">Table of n, a(n) for n = 0..1000</a>

%H Jean-Paul Sonntag, <a href="http://www.chryzode.org/english/ligne.htm">Chryzodes "3in7"</a>

%H Jean-Paul Sonntag, <a href="http://www.chryzode.org/">Chryzodes</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,-1,1).

%F a(n) = 10^n mod 7 = 3^n mod 7.

%F a(n) = a(n-1) - a(n-3) + a(n-4) for n>3; a(n) = a(n-6) for n>5; G.f.: (1+2*x-x^2+5*x^3)/((1-x)*(1+x)*(1-x+x^2)); a(n) = 7/2 -7*(-1)^n/6 -4*A010892(n)/3-A010892(n-1)/3. - _R. J. Mathar_, Feb 13 2009

%F a(n) = (21 - 7*cos(n*Pi) - 8*cos(n*Pi/3) - 4*sqrt(3)*sin(n*Pi/3))/6. - _Wesley Ivan Hurt_, Jun 23 2016

%p A033940:=n->3^n mod 7: seq(A033940(n), n=0..100); # _Wesley Ivan Hurt_, Jul 05 2014

%t Table[PowerMod[10, n, 7], {n, 0, 200}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 10 2011 *)

%t Mod[3^Range[0, 100], 7] (* _Wesley Ivan Hurt_, Jul 06 2014 *)

%o (Sage) [power_mod(10,n, 7)for n in range(0,106)] # _Zerinvary Lajos_, Nov 24 2009

%o (Sage) [power_mod(3, n, 7)for n in range(0,106)] # _Zerinvary Lajos_, Nov 24 2009

%o (Magma) [Modexp(10, n, 7): n in [0..100]]; // _Vincenzo Librandi_, Feb 05 2011

%o (PARI) a(n)=3^n%7 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A010892, A178247.

%K nonn,easy

%O 0,2

%A _Jeff Burch_