A030179 Quarter-squares squared: A002620^2.

0, 0, 1, 4, 16, 36, 81, 144, 256, 400, 625, 900, 1296, 1764, 2401, 3136, 4096, 5184, 6561, 8100, 10000, 12100, 14641, 17424, 20736, 24336, 28561, 33124, 38416, 44100, 50625, 57600, 65536, 73984, 83521, 93636, 104976, 116964

Offset: 0

Author

N. J. A. Sloane, Jan 10 2002

Keywords

• nonn
• easy

Comments

Conjectured to be crossing number of complete bipartite graph K_{n,n}. Known to be true for n <= 7.

If the Zarankiewicz conjecture is true, then a(n) is also the rectilinear crossing number of K_{n,n}. - Eric W. Weisstein, Apr 24 2017

a(n+1) is the number of 4-tuples (w,x,y,z) with all terms in {0,...,n}, and w,x,y+1,z+1 all even. - Clark Kimberling, May 29 2012

References

• C. Thomassen, Embeddings and minors, pp. 301-349 of R. L. Graham et al., eds., Handbook of Combinatorics, MIT Press.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• G. Xiao, Contfrac.
• Eric Weisstein's World of Mathematics, Complete Bipartite Graph.
• Eric Weisstein's World of Mathematics, Graph Crossing Number.
• Eric Weisstein's World of Mathematics, Rectilinear Crossing Number.
• Eric Weisstein's World of Mathematics, Zarankiewicz's Conjecture.
• Index entries for linear recurrences with constant coefficients, signature (2,2,-6,0,6,-2,-2,1).

Crossrefs

Cf. A000241, A002620, A014540.

Programs

GAP

List([0..40], n-> (2*n^4 -2*n^2 +1 +(-1)^n*(2*n^2 -1))/32); # G. C. Greubel, Dec 28 2019

Magma

[(Floor(n^2/4))^2: n in [0..40]]; // G. C. Greubel, Dec 28 2019

Maple

seq( (2*n^4 -2*n^2 +1 +(-1)^n*(2*n^2 -1))/32, n=0..40); # G. C. Greubel, Dec 28 2019

Mathematica

f[n_]:=Floor[n^2/2]; Table[Nest[f,n,2],{n,5!}]/2 (* Vladimir Joseph Stephan Orlovsky, Mar 10 2010 *)
LinearRecurrence[{2,2,-6,0,6,-2,-2,1}, {0,0,1,4,16,36,81,144}, 40] (* Harvey P. Dale, Apr 26 2011 *)
Floor[Range[0, 30]^2/4]^2 (* Eric W. Weisstein, Apr 24 2017 *)

PARI

a(n) = (n^2\4)^2 \\ Charles R Greathouse IV, Jun 11 2015

Sage

[floor(n^2/4)^2 for n in (0..40)] # G. C. Greubel, Dec 28 2019

Formula

a(n) = floor(n^2/4)^2.

From R. J. Mathar, Jul 08 2010: (Start)

G.f.: x^2*(1+2*x+6*x^2+2*x^3+x^4) / ( (1+x)^3*(1-x)^5 ).

a(n) = 2*a(n-1) +2*a(n-2) -6*a(n-3) +6*a(n-5) -2*a(n-6) -2*a(n-7) +a(n-8). (End)

a(n) = (2*n^4 -2*n^2 +1 +(-1)^n*(2*n^2 -1))/32. - Luce ETIENNE, Aug 11 2014

Sum_{n>=2} 1/a(n) = Pi^4/90 + Pi^2/3 - 3. - Amiram Eldar, Sep 17 2023