1,1

After 0 it cycles again from 166 (a(165)=0 so there are 165 (A*B*C) terms).

This is another variation on A256496, where a(n) = B*C*(n mod A) + A*C*(n mod B) + A*B*(n mod C), modified to take the values A=3, B=5, C=11 and still maintain the equivalence a(n) mod ABC = n mod ABC.

Here modification is required (to maintain that equivalence) so that 'BC' + 'AC' + 'AB' = ABC + 1 where 'BC', 'AC' and 'AB' are the coefficients. Therefore, a(n)= B*C*(n mod A) + 2A*C*(n mod B) + 3A*B*(n mod C) so that 5*11 + 2*3*11 + 3*3*5 = 3*5*11 = 55 + 66 + 45 = 166.

This is an example with 2 modifications.

Ray Chandler, Table of n, a(n) for n = 1..1000 (first 165 terms from Bruno Berselli, full cycle)

Index entries for linear recurrences with constant coefficients, signature (-2, -3, -3, -3, -2, -1, 0, 0, 0, 0, 1, 2, 3, 3, 3, 2, 1).

G.f.: -x*(824*x^15 +2306*x^14 +4280*x^13 +5921*x^12 +7229*x^11 +7710*x^10 +7530*x^9 +6855*x^8 +6180*x^7 +5505*x^6 +4830*x^5 +3826*x^4 +2659*x^3 +1495*x^2 +664*x +166) / ((x -1)*(x^2 +x +1)*(x^4 +x^3 +x^2 +x +1)*(x^10 +x^9 +x^8 +x^7 +x^6 +x^5 +x^4 +x^3 +x^2 +x +1)). - Colin Barker, Apr 14 2015

(Magma) A:=3; B:=5; C:=11; [B*C*(n mod A)+2*A*C*(n mod B)+3*A*B*(n mod C): n in [1..165]]; // Bruno Berselli, Apr 14 2015

Cf. A255818 for an example with 1 modification and A256668 for 3 modifications.

Sequence in context: A261288 A038007 A020361 * A250735 A105987 A051963

Adjacent sequences: A256640 A256641 A256642 * A256644 A256645 A256646

nonn,easy

Aaron Kastel, Apr 07 2015

Definition corrected by Bruno Berselli, Apr 14 2015

approved