A184535 a(n) = floor(3/5 * n^2), with a(1)=1.

1, 2, 5, 9, 15, 21, 29, 38, 48, 60, 72, 86, 101, 117, 135, 153, 173, 194, 216, 240, 264, 290, 317, 345, 375, 405, 437, 470, 504, 540, 576, 614, 653, 693, 735, 777, 821, 866, 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

Offset: 1

Author

Clark Kimberling, Jan 16 2011

Keywords

• nonn
• easy

Comments

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

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000
• K. E. Stange, Integral points on elliptic curves and explicit valuations of division polynomials arXiv:1108.3051v3 [math.NT]
• Wikipedia, Maximum number of deciamonds in a hexagon.
• Index entries for linear recurrences with constant coefficients, signature (2, -1, 0, 0, 1, -2, 1).

Crossrefs

Cf. A183532, A279169.

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)).

Programs

GAP

Concatenation([1], List([2..10^3], n->Int(3/5 * n^2))); # Muniru A Asiru, Feb 04 2018

Maple

1,seq(floor(3/5*n^2), n=2..10^3); # Muniru A Asiru, Feb 04 2018

Mathematica

p[n_] := FractionalPart[(n^3 + 5)^(1/3)]; q[n_] := Floor[1/p[n]]; Table[q[n], {n, 1, 120}]
Join[{1},LinearRecurrence[{2, -1, 0, 0, 1, -2, 1},{2, 5, 9, 15, 21, 29, 38},62]] (* Ray Chandler, Aug 31 2015 *)

PARI

a(n) = if(n==1, 1, 3*n^2\5); \\ Altug Alkan, Mar 03 2018

Python

def A184535(n): return 3*n**2//5 if n>1 else 1 # Chai Wah Wu, Aug 04 2025

Formula

a(n) = floor(1/{(5+n^3)^(1/3)}), where {}=fractional part.

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.]

From Peter Bala, Aug 08 2013: (Start)

a(n) = floor(3/5*n^2) for n >= 2.

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.

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)

Extensions

Better name from Peter Bala, Aug 08 2013