login
A194731
Digital roots of the nonzero octagonal numbers.
0
1, 8, 3, 4, 2, 6, 7, 5, 9, 1, 8, 3, 4, 2, 6, 7, 5, 9, 1, 8, 3, 4, 2, 6, 7, 5, 9, 1, 8, 3, 4, 2, 6, 7, 5, 9, 1, 8, 3, 4, 2, 6, 7, 5, 9, 1, 8, 3, 4, 2, 6, 7, 5, 9, 1, 8, 3, 4, 2, 6, 7, 5, 9, 1, 8, 3, 4, 2, 6, 7, 5, 9, 1, 8, 3, 4, 2, 6, 7, 5, 9, 1, 8, 3, 4, 2
OFFSET
1,2
COMMENTS
This is a periodic sequence with period 9 and cycle 1, 8, 3, 4, 2, 6, 7, 5, 9 - which are also the coefficients of x in the numerator of the generating function.
FORMULA
a(n) = a(n-9).
a(n) = 45 -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) = 2(cos(14*n*Pi/9) +cos(10*n*Pi/9) +cos(2*n*Pi/3) +cos(2*n*Pi/9) +cos(Pi/3)) +(n +8*n^3 +n^4 +8*n^5 +7*n^6 +8*n^7 +4*n^8) mod 9.
G.f.: x(1 +8*x +3*x^2 +4*x^3 +2*x^4 +6*x^5 +7*x^6 +5*x^7 +9*x^8)/((1-x)*(1 +x +x^2)*(1 +x^3 +x^6)).
a(n) = A010888(A000567(n)). - Michel Marcus, Aug 10 2015
EXAMPLE
The sixth nonzero octagonal number is A000567(6)=96. As 9+6=15 and 1+5=6, this has digital root 6 and so a(6)=6.
MATHEMATICA
DigitalRoot[n_]:=FixedPoint[Plus@@IntegerDigits[#]&, n]; Table[DigitalRoot[n*(3*n - 2)], {n, 100}]
CROSSREFS
Sequence in context: A117889 A021927 A145594 * A021849 A338462 A201754
KEYWORD
nonn,easy,base
AUTHOR
Ant King, Sep 02 2011
STATUS
approved