A054556 a(n) = 4*n^2 - 9*n + 6.
1, 4, 15, 34, 61, 96, 139, 190, 249, 316, 391, 474, 565, 664, 771, 886, 1009, 1140, 1279, 1426, 1581, 1744, 1915, 2094, 2281, 2476, 2679, 2890, 3109, 3336, 3571, 3814, 4065, 4324, 4591, 4866, 5149, 5440, 5739, 6046, 6361, 6684, 7015, 7354, 7701, 8056, 8419, 8790
Offset: 1
Links
- Ivan Panchenko, Table of n, a(n) for n = 1..1000
- Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
- Robert G. Wilson v, Cover of the March 1964 issue of Scientific American
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Cf. A266883: m*(4*m+1)+1 for m = 0,-1,1,-2,2,-3,3,...
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Programs
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Magma
[4*n^2-9*n+6 : n in [1..50]]; // Vincenzo Librandi, Mar 10 2018
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Maple
a:=n->4*n^2-9*n+6: seq(a(n),n=1..50); # Muniru A Asiru, Mar 09 2018
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Mathematica
a[n_] := 4*n^2 - 9*n + 6; Array[a, 40] (* Vladimir Joseph Stephan Orlovsky, Sep 01 2008 *) LinearRecurrence[{3,-3,1},{1,4,15},50] (* Harvey P. Dale, Sep 06 2015 *) CoefficientList[Series[-(6x^2 + x + 1)/(x - 1)^3, {x, 0, 49}], x] (* Robert G. Wilson v, Mar 12 2018 *)
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PARI
a(n)=4*n^2-9*n+6 \\ Charles R Greathouse IV, Sep 24 2015
Formula
a(n)^2 = Sum_{i = 0..2*(4*n-5)} (4*n^2-13*n+9+i)^2*(-1)^i = ((n-1)*(4*n-5)+1)^2. - Bruno Berselli, Apr 29 2010
From Harvey P. Dale, Aug 21 2011: (Start)
a(0)=1, a(1)=4, a(2)=15; for n > 2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -x*(6*x^2+x+1)/(x-1)^3. (End)
From Franck Maminirina Ramaharo, Mar 09 2018: (Start)
a(n) = binomial(2*n - 2, 2) + 2*(n - 1)^2 + 1.
E.g.f.: exp(x)*(6 - 5*x + 4*x^2) - 6. - Stefano Spezia, Apr 24 2024
Extensions
Edited by Frank Ellermann, Feb 24 2002
Incorrect formula deleted by N. J. A. Sloane, Aug 02 2009
Comments