A134804 Remainder of triangular number A000217(n) modulo 9.

0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6, 3, 1, 0, 0, 1, 3, 6, 1, 6

Offset

0,3

Comments

Periodic with period 9 since A000217(n+9) = A000217(n)+9(n+5) .

From Jacobsthal numbers A001045, A156060 = 0,1,1,3,5,2,3,7,4,0,8, = b(n). a(n)=A156060(n)*A156060(n+1) mod 9. Same transform (a(n)*a(n+1) mod 9 or b(n)*b(n+1) mod 9) in A157742, A158012, A158068, A158090. - Paul Curtz, Mar 25 2009

Links

Table of n, a(n) for n=0..104.

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 1).

Formula

a(n) = A010878(A000217(n)) = A010878(A055263(n)) = a(n-9).

O.g.f.: (-2x+2)/[3(x^2+x+1)]+(-3+3x^5)/(x^6+x^3+1)-7/[3(x-1)].

Mathematica

LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 1, 3, 6, 1, 6, 3, 1, 0}, 105] (* Ray Chandler, Aug 26 2015 *)

Crossrefs

Sequence in context: A307281 A089078 A355072 * A145389 A055263 A004157

Adjacent sequences: A134801 A134802 A134803 * A134805 A134806 A134807

Keyword

easy,nonn

Author

R. J. Mathar, Jan 28 2008

Status

approved