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a(n) is the number of regions into which three-dimensional Euclidean space is divided by n-1 planes parallel to each face of a regular tetrahedron with edge length n, dividing the edges into unit segments.
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%I #56 Dec 02 2023 03:35:52

%S 1,15,65,174,365,661,1085,1660,2409,3355,4521,5930,7605,9569,11845,

%T 14456,17425,20775,24529,28710,33341,38445,44045,50164,56825,64051,

%U 71865,80290,89349,99065,109461,120560,132385,144959,158305,172446,187405,203205,219869,237420,255881,275275,295625,316954

%N a(n) is the number of regions into which three-dimensional Euclidean space is divided by n-1 planes parallel to each face of a regular tetrahedron with edge length n, dividing the edges into unit segments.

%C This design neither includes planes passing through the vertex of the tetrahedron parallel to the opposite face, nor planes that extend the faces.

%H Nicolay Avilov, <a href="https://www.diofant.ru/problem/3435/">Problem 1680. Tetrahedron and planes</a> (in Russian).

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

%F a(n) = (23*n^3 - 30*n^2 + 13*n)/6 [from Anatoly Kazmerchuk].

%F G.f.: x*(1 + 11*x + 11*x^2)/(1 - x)^4. - _Stefano Spezia_, Jul 22 2023

%e a(1) = 1, there are no planes and all space is one part;

%e a(2) = 1 + 4 + 4 + 6 = 15 because in this case there are four planes defining a tetrahedron. These four planes divide the space into 15 parts, namely:

%e 1 part - the inside of the tetrahedron;

%e 4 parts are adjacent to the faces of the tetrahedron;

%e 4 parts are adjacent to the vertices of the tetrahedron;

%e 6 parts are adjacent to the edges of the tetrahedron;

%e a(3) = (23*27 - 30*9 + 13*3)/6 = 65.

%t LinearRecurrence[{4, -6, 4, -1}, {1, 15, 65, 174}, 42] (* _Robert P. P. McKone_, Aug 25 2023 *)

%Y Cf. A000125, A153445, A248598.

%K nonn,easy

%O 1,2

%A _Nicolay Avilov_, Jul 22 2023

%E a(1) = 1 inserted by _Nicolay Avilov_, Oct 19 2023

%E Name corrected by _Talmon Silver_, Oct 29 2023

%E a(44) = 316954 added by _Talmon Silver_, Dec 01 2023