login
Coordination sequence for 4-dimensional I-centered cubic orthogonal lattice.
1

%I #27 Sep 08 2022 08:44:36

%S 1,10,44,126,280,530,900,1414,2096,2970,4060,5390,6984,8866,11060,

%T 13590,16480,19754,23436,27550,32120,37170,42724,48806,55440,62650,

%U 70460,78894,87976,97730,108180,119350,131264,143946,157420,171710,186840,202834,219716,237510

%N Coordination sequence for 4-dimensional I-centered cubic orthogonal lattice.

%C Let f(x) = x^2 + x + 1 then sequence gives f(f(n+1)) - f(f(n)), n >= 0.

%H Colin Barker, <a href="/A008532/b008532.txt">Table of n, a(n) for n = 0..1000</a>

%H M. O'Keeffe, <a href="http://dx.doi.org/10.1524/zkri.1995.210.12.905">Coordination sequences for lattices</a>, Zeit. f. Krist., 210 (1995), 905-908.

%H M. O'Keeffe, <a href="/A008527/a008527.pdf">Coordination sequences for lattices</a>, Zeit. f. Krist., 210 (1995), 905-908. [Annotated scanned copy]

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = 4*n^3 + 6*n, n >= 1.

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>4. - _Colin Barker_, Mar 03 2015

%F G.f.: (1+x)^2*(1+4*x+x^2)/(1-x)^4. - _Colin Barker_, Mar 03 2015

%F a(0) = 1; for n > 0, a(n) = A005898(n-1) + A005898(n) = (n-1)^3 + 2n^3 + (n+1)^3. - _Doug Bell_, Aug 18 2015

%F E.g.f.: 1 + 2*x*(5 + 6*x + 2*x^2)*exp(x). - _G. C. Greubel_, Aug 21 2015

%p 1, seq( 4*k^3+6*k, k=1..40);

%t Table[If[n==0,1,2*n*(3+2*n^2)], {n,0,40}] (* _G. C. Greubel_, Nov 10 2019 *)

%o (PARI) Vec((x+1)^2*(x^2+4*x+1)/(x-1)^4 + O(x^40)) \\ _Colin Barker_, Mar 03 2015

%o (PARI) vector(46, n, if(n==1,1, 2*(n-1)*(3 +2*(n-1)^2) ) ) \\ _G. C. Greubel_, Nov 10 2019

%o (Magma) [1] cat [2*n*(3+2*n^2): n in [1..45]]; // _G. C. Greubel_, Nov 10 2019

%o (Sage) [1]+[2*n*(3+2*n^2) for n in (1..45)]; # _G. C. Greubel_, Nov 10 2019

%o (GAP) Concatenation([1], List([1..45], n-> 2*n*(3+2*n^2) )); # _G. C. Greubel_, Nov 10 2019

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_