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A176409 Multiples of 3 or 7.

%I A176409 #34 Mar 21 2020 22:30:52
%S A176409 0,3,6,7,9,12,14,15,18,21,24,27,28,30,33,35,36,39,42,45,48,49,51,54,
%T A176409 56,57,60,63,66,69,70,72,75,77,78,81,84,87,90,91,93,96,98,99,102,105,
%U A176409 108,111,112,114,117,119,120,123,126,129,132,133,135,138,140,141,144,147,150,153
%N A176409 Multiples of 3 or 7.
%C A176409 Therefore, this sequence also includes multiples of 21.
%C A176409 First differences: 3, 3, 1, 2, 3, 2, 1, 3, 3, 3, 3, 1, 2, 3, 2, 1, 3, 3, 3, 3, 1, 2, 3, 2, 1, 3, 3, 3, 3, 1, 2, 3, 2, 1, 3, 3, 3 have period {3, 3, 1, 2, 3, 2, 1, 3, 3}.
%C A176409 In general, sequences of numbers divisible by primes p and q will be of the form a(n+p+q-1) = a(n)+p*q. - _Gary Detlefs_, Oct 07 2013
%H A176409 Daniel Starodubtsev, <a href="/A176409/b176409.txt">Table of n, a(n) for n = 1..10000</a>
%H A176409 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,0,1,-1).
%F A176409 a(n+9) = a(n) + 21.
%F A176409 a(n) = 21*floor((n-1)/9) + 2*((n-1) mod 9) + s(((n-1) mod 9)-1) + 1 - floor(((n-2) mod 9)/8), where s(n) = floor(n*sqrt(2)) - 2*floor(n/sqrt(2)). - _Gary Detlefs_, Oct 07 2013
%F A176409 a(n) = 7n/3 + O(1). - _Charles R Greathouse IV_, Feb 13 2017
%F A176409 From _Colin Barker_, Mar 21 2020: (Start)
%F A176409 G.f.: x^2*(3 + 3*x + x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6 + 3*x^7 + 3*x^8) / ((1 - x)^2*(1 + x + x^2)*(1 + x^3 + x^6)).
%F A176409 a(n) = a(n-1) + a(n-9) - a(n-10) for n>10.
%F A176409 (End)
%t A176409 Union@Flatten@{Table[3n, {n, 70}], Table[7n, {n, 30}]}
%t A176409 Select[ Range@ 153, Mod[#, 3] == 0 || Mod[#, 7] == 0 &]
%o A176409 (PARI) concat(0, Vec(x^2*(3 + 3*x + x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6 + 3*x^7 + 3*x^8) / ((1 - x)^2*(1 + x + x^2)*(1 + x^3 + x^6)) + O(x^70))) \\ _Colin Barker_, Mar 21 2020
%Y A176409 Cf. A047229, A065520, A083035.
%K A176409 nonn,easy
%O A176409 1,2
%A A176409 _Zak Seidov_, Dec 07 2010
%E A176409 0 inserted by _Daniel Starodubtsev_, Mar 21 2020