A275799 -id:A275799 - cp's OEIS frontend

A054772 Triangle T(n,k) of n X n binary matrices with k=0..n^2 ones, up to rotational symmetry.

1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 10, 22, 34, 34, 22, 10, 3, 1, 1, 4, 32, 140, 464, 1092, 2016, 2860, 3238, 2860, 2016, 1092, 464, 140, 32, 4, 1, 1, 7, 78, 578, 3182, 13302, 44330, 120230, 270525, 510875, 817388, 1114548, 1300316, 1300316, 1114548, 817388

Offset: 0

Vladeta Jovovic, May 18 2000

Row sums give A047937.
From Wolfdieter Lang, Oct 01 2016: (Start)
The formula is obtained from Pólya's counting theorem. See, e.g., the Harary-Palmer reference.
The cycle index for a square grid of n X n  squares G(n), n >= 1, under the cyclic group C_4 is
  (s[1]^(n^2)+s[2]^(n^2/2)+2*s[4]^(n^2/4))/4 if n is even,
  s[1]*(s[1]^(n^2-1) + s[2]^((n^2-1)/2) + 2*s[4]^((n^2-1)/4))/4 if n is odd. (Numerate the squares from 1 .. n^2 and compute for the C_4 rotations the cycle structure of the permutation from the symmetric group S(n^2)).
The figure counting series is c(x) = 1+x for coloring, say black and white (in the matrix case binary entries).
Therefore the counting series is C(n,x) = G(n) with substitution s[2^j] = c(x^(2*j)) = 1 + x^(2^j) for j=0,1,2. Row n gives the coefficients of C(n,x) in rising (or falling) order. (End)
A pedantic note: One should not use 0,1 matrices for this T(n,k) model because 1 (also |) is not C_4 invariant. Square grids with coloring of the squares, say black and white, or central entries o and + are better suited. - Wolfdieter Lang, Oct 02 2016

			[1],[1,1],[1,1,2,1,1],[1,3,10,22,34,34,22,10,3,1],...;
There are 10 inequivalent 3 X 3 binary matrices with 2 ones, up to rotational symmetry:
[0 0 0] [0 0 0] [0 0 0] [0 0 0] [0 0 0]
[0 0 0] [0 0 0] [0 0 0] [0 0 1] [0 0 1]
[0 1 1] [1 0 1] [1 1 0] [0 1 0] [1 0 0]
-------
[0 0 0] [0 0 0] [0 0 0] [0 0 0] [0 0 1]
[0 1 0] [0 1 0] [1 0 0] [1 0 1] [0 0 0]
[0 0 1] [0 1 0] [0 0 1] [0 0 0] [1 0 0].
- reformatted. _Wolfdieter Lang_, Oct 01 2016
See a remark above: use o for 0 and + for 1.
n=3: Cycle index G(3) = s[1]*(s[1]^8 + s[2]^4 + 2*s[4]^2)/4. C(3,x) = (1+x)*((1+x)^8 + (1+x^2)^4 + 2*(1+x^4)^2)/4 = 1 + 3*x + 10*x^2 + 22*x^3 + 34*x^4 + 34*x^5 + 22*x^6 + 10*x^7 + 3*x^8 + x^9. - _Wolfdieter Lang_, Oct 01 2016

F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 42, (2.4.6).

Cf. A054252, columns k=0..4: A000012, A004652, A212714, A011863, A275799.

See the comment above: T(n,k) = [x^k]C(n,x), with the counting series C(n,x) obtained from the cycle index for the n X n grid under C_4 rotations G(n;s[1],s[2],s[4]) with s[2^j] = 1 + x^(2^j) for j=0,1,2. - Wolfdieter Lang, Oct 01 2016

A277226 Number of inequivalent (modulo C_4 rotations) square n X n grids with squares coming in two colors and four squares have one of the colors.

Original entry on oeis.org

1, 34, 464, 3182, 14769, 53044, 158976, 416140, 980625, 2124310, 4295376, 8199674, 14907809, 25992232, 43700224, 71167704, 112680801, 173990730, 262690000, 388656070, 564571601, 806527964, 1134722304, 1574255332, 2156041329, 2917838014, 3905408976, 5173826770, 6788930625

Offset: 2

Wolfdieter Lang, Oct 06 2016

See the k=4 column of table A054772(n, k), with more explanations there.

Colin Barker, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (6,-12,2,27,-36,0,36,-27,-2,12,-6,1).

Cf. A054772, A000012 (k=0), A004652 (k=1), A212714 (k=2), A275799 (k=3).

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x^2*(1+28*x+272*x^2+804*x^3+1150*x^4+804*x^5 +272*x^6+28*x^7+x^8)/((1-x)^9*(1+x)^3))); // G. C. Greubel, Oct 22 2018

CoefficientList[Series[x^2*(1+28*x+272*x^2+804*x^3+1150*x^4+804*x^5 +272*x^6+28*x^7+x^8)/((1-x)^9*(1+x)^3), {x, 0, 50}], x] (* G. C. Greubel, Oct 22 2018 *)

Vec(x^2*(1+28*x+272*x^2+804*x^3+1150*x^4+804*x^5+272*x^6+28*x^7 +x^8)/((1-x)^9*(1+x)^3) + O(x^40)) \\ Colin Barker, Oct 16 2016

a(n) = A054772(n, 4) = A054772(n, n^2-4), n >= 2.
From Colin Barker, Oct 09 2016: (Start)
G.f.: x^2*(1+28*x+272*x^2+804*x^3+1150*x^4+804*x^5+272*x^6+28*x^7+x^8) / ((1-x)^9*(1+x)^3).
a(n) = (n^8-6*n^6+14*n^4)/96 for n even.
a(n) = (n^8-6*n^6+14*n^4-6*n^2-3)/96 for n odd. (End)
From Stefan Hollos, Oct 16 2016: (Start)
a(n) = (C(n^2,4) + C(n^2/2,2) + n^2/2)/4 for n even,
a(n) = (C(n^2,4) + C((n^2-1)/2,2) + (n^2-1)/2)/4 for n odd. (End)

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A054772 Triangle T(n,k) of n X n binary matrices with k=0..n^2 ones, up to rotational symmetry.

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A277226 Number of inequivalent (modulo C_4 rotations) square n X n grids with squares coming in two colors and four squares have one of the colors.

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