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%I #33 Mar 09 2018 21:23:15
%S 1,2,5,9,15,21,29,38,48,60,72,86,101,117,135,153,173,194,216,240,264,
%T 290,317,345,375,405,437,470,504,540,576,614,653,693,735,777,821,866,
%U 912,960,1008,1058,1109,1161,1215,1269,1325,1382,1440,1500,1560,1622,1685,1749,1815,1881,1949,2018,2088,2160,2232,2306,2381
%N a(n) = floor(3/5 * n^2), with a(1)=1.
%C Apart from the initial term this is the elliptic troublemaker sequence R_n(2,5) in the notation of Stange (see Table 1, p.16). For other elliptic troublemaker sequences see the cross references below. - _Peter Bala_, Aug 08 2013
%H Ray Chandler, <a href="/A184535/b184535.txt">Table of n, a(n) for n = 1..10000</a>
%H K. E. Stange, <a href="http://arxiv.org/abs/1108.3051">Integral points on elliptic curves and explicit valuations of division polynomials</a> arXiv:1108.3051v3 [math.NT]
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/File:Deciamond.png">Maximum number of deciamonds in a hexagon.</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1, 0, 0, 1, -2, 1).
%F a(n) = floor(1/{(5+n^3)^(1/3)}), where {}=fractional part.
%F a(n)= +2*a(n-1) -a(n-2) +a(n-5) -2*a(n-6) +a(n-7), for n>8, with g.f. 1-x^2*(1+x)*(2*x^2-x+2)/ ((x^4+x^3+x^2+x+1) *(x-1)^3), so a(n) is (3n^2-2)/5 plus a fifth of A164116 for n>1. [_Bruno Berselli_, Jan 30 2011. See the following Bala's comment.]
%F From _Peter Bala_, Aug 08 2013: (Start)
%F a(n) = floor(3/5*n^2) for n >= 2.
%F The sequence b(n) := floor(3/5*n^2) - 3/5*n^2, n >= 1, is periodic with period [-3/5, -2/5, -2/5, -3/5, 0] of length 5. The generating function and recurrence equation given above easily follow from these observations.
%F The sequence c(n) := 5/2*( (2*n/5 - floor(2*n/5))^2 - (2*n/5 - floor(2*n/5)) ) is also periodic with period 5, and calculation shows it has the same period as the sequence b(n). Thus b(n) = c(n), yielding the alternative formula a(n) = 3/5*n^2 + 5/2*( (2*n/5 - floor(2*n/5))^2 - (2*n/5 - floor(2*n/5)) ), which is one of the formulas for the elliptic troublemaker sequence R_n(2,5) given in Stange (see Section 7, equation (21)). (End)
%p 1,seq(floor(3/5*n^2), n=2..10^3); # _Muniru A Asiru_, Feb 04 2018
%t p[n_] := FractionalPart[(n^3 + 5)^(1/3)]; q[n_] := Floor[1/p[n]]; Table[q[n], {n, 1, 120}]
%t Join[{1},LinearRecurrence[{2, -1, 0, 0, 1, -2, 1},{2, 5, 9, 15, 21, 29, 38},62]] (* _Ray Chandler_, Aug 31 2015 *)
%o (GAP) Concatenation([1], List([2..10^3], n->Int(3/5 * n^2))); # _Muniru A Asiru_, Feb 04 2018
%o (PARI) a(n) = if(n==1, 1, 3*n^2\5); \\ _Altug Alkan_, Mar 03 2018
%Y Cf. A183532, A279169.
%Y Elliptic troublemaker sequences: A000212 (= R_n(1,3) = R_n(2,3)), A002620 (= R_n(1,2)), A007590 (= R_n(2,4)), A030511 (= R_n(2,6) = R_n(4,6))), A033436 (= R_n(1,4) = R_n(3,4)), A033437 (= R_n(1,5) = R_n(4,5)), A033438 (= R_n(1,6) = R_n(5,6)), A184535 (= R_n(2,5) = R_n(3,5)).
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jan 16 2011
%E Better name from _Peter Bala_, Aug 08 2013
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