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%I #58 Dec 12 2023 07:39:14
%S 0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,
%T 0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,
%U 1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1,1,0,2,0,1,1
%N Period 6: repeat [0, 1, 1, 1, 0, 2].
%C The number of divisors d of n of the form d=2 or 3. - _Vladimir Shevelev_, May 21 2010
%C a(n) = s(n+6), where s(k) is the number of partitions of k into distinct parts such that max(p) = 2 + min(p) for k >= 1, and (s(0)..s(6)) = (0,0,0,0,1,0,2). - _Clark Kimberling_, Apr 15 2014
%C Number of r X s integer-sided rectangles such that r < s, r + s = 2n, r | s and (s - r)/2 | s. - _Wesley Ivan Hurt_, Apr 24 2020
%C Number of positive integer solutions, (r,s,t), of the equation r^2 + t*s^2 = (n + 6)^2, where r + s = n + 6 and t < r <= s. For example, when n=6 we have the two solutions (4,8,2) and (6,6,3) since 4^2 + 2*8^2 = 12^2 and 6^2 + 3*6^2 = 12^2. - _Wesley Ivan Hurt_, Oct 04 2020
%H Vladimir Shevelev, <a href="http://arXiv.org/abs/0903.1743">A recursion for divisor function over divisors belonging to a prescribed finite sequence of positive integers and a solution of the Lahiri problem for divisor function sigma_x(n)</a>, arXiv:0903.1743 [math.NT], 2009. [From _Vladimir Shevelev_, May 21 2010]
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-1,0,1,1).
%F a(n) = A115357(n-2) for n>1. - _R. J. Mathar_, Dec 09 2009
%F a(2) = 1, a(3) = 1, a(5) = 0, otherwise a(n) = a(n-2) + a(n-3) - a(n-5), where we put a(n) = 0, if n<0. - _Vladimir Shevelev_, May 21 2010
%F a(n) = floor(((n+1) mod 6)/3) + 2*floor(((n+5) mod 6)/5). - _Gary Detlefs_, Feb 15 2014
%F From _Wesley Ivan Hurt_, Aug 27 2014: (Start)
%F G.f.: (2+2*x+x^2)/(1+x-x^3-x^4).
%F a(n) + a(n-1) = a(n-3) + a(n-4) for n>4.
%F a(n) = (1 + floor((n-3)^2/2)) mod 3. (End)
%F a(n) = (5 + 3*cos(n*Pi) + 4*cos(2*n*Pi/3))/6. - _Wesley Ivan Hurt_, Jun 19 2016
%F From _Amiram Eldar_, Sep 16 2023: (Start)
%F Additive with a(p^e) = 1 if p <= 3, and 0 otherwise.
%F a(n) = A059841(n) + A079978(n).
%F a(n) = A001221(A089128(n)).
%F a(n) = A001221(A065331(n)). (End)
%p A171182:=n->(1+floor((n-3)^2/2)) mod 3: seq(A171182(n), n=1..100); # _Wesley Ivan Hurt_, Aug 27 2014
%t PadRight[{},110,{0,1,1,1,0,2}] (* _Harvey P. Dale_, Jan 28 2013 *)
%t LinearRecurrence[{-1, 0, 1, 1},{0, 1, 1, 1},105] (* _Ray Chandler_, Aug 26 2015 *)
%o (Magma) &cat[[0, 1, 1, 1, 0, 2]^^20]; // _Wesley Ivan Hurt_, Jun 19 2016
%Y Cf. A178142. - _Vladimir Shevelev_, May 21 2010
%Y Cf. A115357.
%Y Cf. A001221, A059841, A065331, A079978, A089128, A169611.
%Y Number of distinct prime factors <= p: this sequence (p=3), A178146 (p=5), A210679 (p=7).
%K nonn,easy,less
%O 1,6
%A _Juri-Stepan Gerasimov_, Dec 04 2009, Dec 07 2009
%E Edited by _Charles R Greathouse IV_, Mar 23 2010
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