A184005 a(n) = n - 1 + ceiling(3*n^2/4); complement of A184004.
1, 4, 9, 15, 23, 32, 43, 55, 69, 84, 101, 119, 139, 160, 183, 207, 233, 260, 289, 319, 351, 384, 419, 455, 493, 532, 573, 615, 659, 704, 751, 799, 849, 900, 953, 1007, 1063, 1120, 1179, 1239, 1301, 1364, 1429, 1495, 1563, 1632, 1703, 1775, 1849, 1924, 2001, 2079, 2159, 2240, 2323, 2407, 2493, 2580, 2669, 2759
Offset: 1
Links
- G. C. Greubel, Table of n, a(n) for n = 1..5000
- Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Crossrefs
Cf. A184004.
Programs
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Magma
[(6*n^2 + 8*n - (-1)^n - 7)/8: n in [1..80]]; // Vincenzo Librandi, Feb 09 2011
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Mathematica
a=4/3; b=0; Table[n+Floor[(a*n+b)^(1/2)],{n,80}] Table[n-1+Ceiling[(n*n-b)/a],{n,60}] Table[n - 1 + Ceiling[3 n^2/4], {n, 60}] (* or *) CoefficientList[ Series[x (1 + 2 x + x^2 - x^3)/((1 + x) (1 - x)^3), {x, 0, 60}], x] (* or *) Table[Round[(6 n^2 + 8 n - 7)/8], {n, 60}] (* Michael De Vlieger, Mar 23 2016 *) LinearRecurrence[{2,0,-2,1},{1,4,9,15},60] (* Harvey P. Dale, Sep 16 2016 *) -
PARI
my(x='x+O('x^200)); Vec(x*(1+2*x+x^2-x^3)/((1+x)*(1-x)^3)) \\ Altug Alkan, Mar 23 2016 -
Python
def A184005(n): return n+((m:=3*n**2)>>2)-(not m&3) # Chai Wah Wu, Oct 01 2024
Formula
a(n) = 2*a(n-1) - 2*a(n-3) + 1*a(n-4).
From Bruno Berselli, Jan 25 2011: (Start)
G.f.: x*(1 + 2*x + x^2 - x^3)/((1 + x)*(1 - x)^3).
a(n) = (6*n^2 + 8*n - (-1)^n - 7)/8. (End)
a(n) = round((6*n^2 + 8*n - 7)/8). - Bruno Berselli, Jan 25 2011
From Paul Curtz, Feb 09 2011: (Start)
a(n) - a(n-1) = A007494(n).
a(n) - a(n-2) = 3*n - 1 = A016789(n-1).
a(n) - a(n-4) = 6*n - 8 = A016957(n-2).
a(n) - a(n-8) = 12*n - 40 = A017617(n-4).
a(n) - a(n-16) = 24*n - 176 = 8*A016789(n-8).
a(n) - a(n-32) = 48*n - 736 = 16*A016789(n-16). (End)
a(n) = n^2 - floor((n-2)^2/4). - Bruno Berselli, Jan 17 2017
E.g.f.: (1/8)*(8 + (6*x^2 + 14*x -7)*exp(x) - exp(-x)). - G. C. Greubel, Jul 22 2017