A194732 - OEIS

%I #17 Aug 11 2015 09:17:38

%S 1,8,1,0,5,6,3,6,5,0,1,8,1,0,5,6,3,6,5,0,1,8,1,0,5,6,3,6,5,0,1,8,1,0,

%T 5,6,3,6,5,0,1,8,1,0,5,6,3,6,5,0,1,8,1,0,5,6,3,6,5,0,1,8,1,0,5,6,3,6,

%U 5,0,1,8,1,0,5,6,3,6,5,0,1,8,1,0,5,6

%N Units' digits of the nonzero octagonal numbers.

%C This is a periodic sequence with period 10 and cycle 1, 8, 1, 0, 5, 6, 3, 6, 5, 0.

%F a(n) = a(n-10).

%F a(n) = 35 -a(n-1) -a(n-2) -a(n-3) -a(n-4) -a(n-5) -a(n-6) -a(n-7) -a(n-8) -a(n-9).

%F a(n) = (n*(3*n-2)) mod 10.

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

%F a(n) = A010879(A000567(n)). - _Michel Marcus_, Aug 10 2015

%e The seventh nonzero octagonal number is A000567(7)=133, which has units' digit 3. Hence a(7)=3.

%t Table[Mod[n (3 n - 2), 10], {n, 100}]

%Y Cf. A000567, A010879.

%K nonn,easy,base

%O 1,2

%A _Ant King_, Sep 02 2011