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A134804 Remainder of triangular number A000217(n) modulo 9. 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, 3, 1, 0, 0, 1, 3, 6, 1, 6 (list; graph; refs; listen; history; text; internal format)
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)].

a(n) = (1/108)*{7*(n mod 9)+19*[(n+1) mod 9]+31*[(n+2) mod 9]+43*[(n+3) mod 9]-53*[(n+4) mod 9]+67*[(n+5) mod 9]-29*[(n+6) mod 9]-17*[(n+7) mod 9]-5*[(n+8) mod 9]}, with n>=0. - Paolo P. Lava, Jan 30 2008

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: A175032 A078768 A089078 * A145389 A055263 A004157

Adjacent sequences:  A134801 A134802 A134803 * A134805 A134806 A134807

KEYWORD

easy,nonn

AUTHOR

R. J. Mathar, Jan 28 2008

STATUS

approved

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Last modified October 17 09:58 EDT 2018. Contains 316276 sequences. (Running on oeis4.)