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A008763
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Expansion of g.f.: x^4/((1-x)*(1-x^2)^2*(1-x^3)).
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17
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0, 0, 0, 0, 1, 1, 3, 4, 7, 9, 14, 17, 24, 29, 38, 45, 57, 66, 81, 93, 111, 126, 148, 166, 192, 214, 244, 270, 305, 335, 375, 410, 455, 495, 546, 591, 648, 699, 762, 819, 889, 952, 1029, 1099, 1183, 1260, 1352, 1436, 1536, 1628, 1736, 1836, 1953, 2061, 2187, 2304, 2439
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OFFSET
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0,7
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COMMENTS
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Number of 2 X 2 square partitions of n.
1/((1-x^2)*(1-x^4)^2*(1-x^6)) is the Molien series for 4-dimensional representation of a certain group of order 192 [Nebe, Rains, Sloane, Chap. 7].
Number of ways of writing n as n = p+q+r+s so that p >= q, p >= r, q >= s, r >= s with p, q, r, s >= 1. That is, we can partition n as
pq
rs
with p >= q, p >= r, q >= s, r >= s.
The coefficient of s(2n) in s(n,n) * s(n,n) * s(n,n) * s(n,n) is a(n+4), where s(n) is the Schur function corresponding to the trivial representation, s(n,n) is a Schur function corresponding to the two row partition and * represents the inner or Kronecker product of symmetric functions. - Mike Zabrocki, Dec 22 2005
Let F() be the Fibonacci sequence A000045. Let f([x, y, z, w]) = F(x) * F(y) * F(z) * F(w). Let N([x, y, z, w]) = x^2 + y^2 + z^2 + w^2. Let Q(k) = set of all ordered quadruples of integers [x, y, z, w] such that 1 <= x <= y <= z <= w and N([x, y, z, w]) = k. Let P(n) = set of all unordered triples {q1, q2, q3} of elements of some Q(k) such that max(w1, w2, w3) = n and f(q1) + f(q2) = f(q3). Then a(n-1) is the number of elements of P(n). - Michael Somos, Jan 21 2015
Number of partitions of 2n+2 into 4 parts with alternating parity from smallest to largest (or vice versa). - Wesley Ivan Hurt, Jan 19 2021
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REFERENCES
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G. E. Andrews, MacMahon's Partition Analysis II: Fundamental Theorems, Annals Combinatorics, 4 (2000), 327-338.
G. E. Andrews, P. Paule and A. Riese, MacMahon's Partition Analysis VIII: Plane partition diamonds, Advances Applied Math., 27 (2001), 231-242 (Cor. 2.1, n=1).
S. P. Humphries, Braid groups, infinite Lie algebras of Cartan type and rings of invariants, Topology and its Applications, 95 (3) (1999) pp. 173-205.
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LINKS
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FORMULA
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Let f4(n) = number of partitions n = p+q+r+s into exactly 4 parts, with p >= q >= r >= s >= 1 (see A026810, A001400) and let g4(n) be the number with q > r (so that g4(n) = f4(n-2)). Then a(n) = f4(n) + g4(n).
a(n) = (1/144)*( 2*n^3 + 9*n*((-1)^n - 1) - 16*((n is 2 mod 3) - (n is 1 mod 3)) ).
a(n) = (1/72)*(n+3)*(n+2)*(n+1)-(1/12)*(n+2)*(n+1)+(5/144)*(n+1)+(1/16)*(n+1)*(-1)^n+(1/16)*(-1)^(n+1)+(7/144)+(2*sqrt(3)/27)*sin(2*Pi*n/3). - Richard Choulet, Nov 27 2008
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - 2*a(n-4) - a(n-5) + 2*a(n-6) + a(n-7) - a(n-8), n>7. - Harvey P. Dale, Mar 04 2012
a(n) = floor((9*(n+1)*(-1)^n + 2*n^3 - 9*n + 65)/144). - Tani Akinari, Nov 06 2012
Euler transform of length 3 sequence [1, 2, 1]. - Michael Somos, Jun 26 2017
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EXAMPLE
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a(7) = 4:
41 32 31 22
11 11 21 21
G.f. = x^4 + x^5 + 3*x^6 + 4*x^7 + 7*x^8 + 9*x^9 + 14*x^10 + 17*x^11 + ...
a(5-1) = 1 because P(5) has only one triple {[1,1,1,5], [2,2,2,4], [1,3,3,3]} of elements from Q(28) where f([1,1,1,5]) = 5, f([2,2,2,4]) = 3, f(1,3,3,3]) = 8, and 5 + 3 = 8. - Michael Somos, Jan 21 2015
a(6-1) = 1 because P(6) has only one triple {[1,1,2,6], [2,2,3,5], [1,3,4,4]} of elements from Q(42) where f([1,1,2,6]) = 8, f([2,2,3,5]) = 10, f([1,3,4,4]) = 18 and 8 + 10 = 18. - Michael Somos, Jan 21 2015
a(7-1) = 3 because P(7) has three triples. The triple {[1,1,1,7], [2,4,4,4], [3,3,3,5]} from Q(52) where f([1,1,1,7]) = 13, f([2,4,4,4]) = 27, f([3,3,3,5]) = 40 and 13 + 27 = 40. The triple {[1,2,2,7], [2,3,3,6], [1,4,4,5]} from Q(58) where f([1,2,2,7]) = 13, f([2,3,3,6]) = 32, f([1,4,4,5]) = 45 and 13 + 32 = 45. The triple {[1,1,3,7], [2,2,4,6], [1,3,5,5]} from Q(60) where f([1,1,3,7]) = 26, f([2,2,4,6]) = 24, f([1,3,5,5]) = 50 and 26 + 24 = 50. - Michael Somos, Jan 21 2015
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MAPLE
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a:= n-> (Matrix(8, (i, j)-> if (i=j-1) then 1 elif j=1 then [1, 2, -1, -2, -1, 2, 1, -1][i] else 0 fi)^n)[1, 5]: seq(a(n), n=0..60); # Alois P. Heinz, Jul 31 2008
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MATHEMATICA
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CoefficientList[Series[x^4/((1-x)*(1-x^2)^2*(1-x^3)), {x, 0, 60}], x] (* Jean-François Alcover, Mar 30 2011 *)
LinearRecurrence[{1, 2, -1, -2, -1, 2, 1, -1}, {0, 0, 0, 0, 1, 1, 3, 4}, 60] (* Harvey P. Dale, Mar 04 2012 *)
a[ n_]:= Quotient[9(n+1)(-1)^n +2n^3 -9n +65, 144]; (* Michael Somos, Jan 21 2015 *)
a[ n_]:= Sign[n] SeriesCoefficient[ x^4/((1-x)(1-x^2)^2(1-x^3)), {x, 0, Abs@n}]; (* Michael Somos, Jan 21 2015 *)
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PROG
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(Magma) K:=Rationals(); M:=MatrixAlgebra(K, 4); q1:=DiagonalMatrix(M, [1, -1, 1, -1]); p1:=DiagonalMatrix(M, [1, 1, -1, -1]); q2:=DiagonalMatrix(M, [1, 1, 1, -1]); h:=M![1, 1, 1, 1, 1, 1, -1, -1, 1, -1, 1, -1, 1, -1, -1, 1]/2; H:=MatrixGroup<4, K|q1, q2, h, p1>; MolienSeries(H);
(Magma) R<x>:=PowerSeriesRing(Integers(), 60); [0, 0, 0, 0] cat Coefficients(R!( x^4/((1-x)*(1-x^2)^2*(1-x^3)) )); // G. C. Greubel, Sep 10 2019
(PARI) {a(n) = (9*(n+1)*(-1)^n + 2*n^3 - 9*n + 65) \ 144}; /* Michael Somos, Jan 21 2015 */
(PARI) a(n)=([0, 1, 0, 0, 0, 0, 0, 0; 0, 0, 1, 0, 0, 0, 0, 0; 0, 0, 0, 1, 0, 0, 0, 0; 0, 0, 0, 0, 1, 0, 0, 0; 0, 0, 0, 0, 0, 1, 0, 0; 0, 0, 0, 0, 0, 0, 1, 0; 0, 0, 0, 0, 0, 0, 0, 1; -1, 1, 2, -1, -2, -1, 2, 1]^n*[0; 0; 0; 0; 1; 1; 3; 4])[1, 1] \\ Charles R Greathouse IV, Feb 06 2017
(Sage)
def AA008763_list(prec):
P.<x> = PowerSeriesRing(ZZ, prec)
return P(x^4/((1-x)*(1-x^2)^2*(1-x^3))).list()
(GAP) a:=[0, 0, 0, 0, 1, 1, 3, 4];; for n in [9..60] do a[n]:=a[n-1]+2*a[n-2]-a[n-3]-2*a[n-4]-a[n-5]+2*a[n-6]+a[n-7]-a[n-8]; od; a; # G. C. Greubel, Sep 10 2019
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CROSSREFS
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See A266769 for a version without the four leading zeros.
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KEYWORD
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nonn,nice,easy
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AUTHOR
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EXTENSIONS
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Entry revised Dec 25 2003
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STATUS
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approved
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