A358655 a(n) is the number of distinct scalar products which can be formed by pairs of signed permutations (V, W) of [n].
1, 2, 7, 24, 61, 111, 183, 281, 409, 571, 771, 1013, 1301, 1639, 2031, 2481, 2993, 3571, 4219, 4941, 5741, 6623, 7591, 8649, 9801, 11051, 12403, 13861, 15429, 17111, 18911, 20833, 22881, 25059, 27371, 29821, 32413, 35151, 38039, 41081, 44281, 47643
Offset: 0
Examples
a(2) = 7 Columns in the table below: A: Result of the scalar product. B: Count of combinations for this result. C: An example. A B C 5 1 [ 2, 1]*[ 2, 1] 4 1 [ 1, 2]*[ 2, 1] 3 1 [ 2, -1]*[ 2, 1] 0 2 [ 1, -2]*[ 2, 1] -3 1 [-2, 1]*[ 2, 1] -4 1 [-1, -2]*[ 2, 1] -5 1 [-2, -1]*[ 2, 1] We have 7 rows. The sum over B is A000165(2). For a(2) all vectors C are part of the vertices of an octagon.
Links
- Paolo Xausa, Table of n, a(n) for n = 0..10000
- Thomas Scheuerle, The permutohedron for a(3) a truncated cuboctahedron.
- Thomas Scheuerle, The truncated cuboctahedron of a(3) with 24 planes intersecting in at least one vertex.
- Thomas Scheuerle, The octagon of a(2) with 7 planes intersecting in at least one vertex.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Programs
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Mathematica
LinearRecurrence[{4, -6, 4, -1}, {1, 2, 7, 24, 61, 111, 183, 281}, 50] (* Paolo Xausa, Oct 02 2024 *)
Formula
a(n) = (2*n^3 + 3*n^2 + n + 3)/3 = A188475(n), for n > 3 (because valid if A000165(n)/2 > A188475(n)).
From Stefano Spezia, Nov 28 2022: (Start)
G.f.: (1 - 2*x + 5*x^2 + 4*x^3 - 15*x^5 + 16*x^6 - 5*x^7)/(1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 7. (End)
Comments