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A290540
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Determinant of circulant matrix of order 10 with entries in the first row that are (-1)^(j-1)*Sum_{k>=0} (-1)^k*binomial(n, 10*k+j-1), for j=1..10.
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1
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1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -2276485387658524, -523547340003805770400, -39617190432735671861429500, -2896792542975174202888623380000, -95819032881785191861991031568287500, -1018409199709889673458815786392849200000
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OFFSET
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0,11
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COMMENTS
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a(n) = 0 for n == 9 (mod 10).
A generalization. For an even N >= 2, consider the determinant of circulant matrix of order N with entries in the first row (-1)^(j-1)K_j(n), j=1..N, where K_j(n) = Sum_{k>=0} (-1)^k*binomial(n, N*k+j-1). Then it is 0 for n == N-1 (mod N). This statement follows from an easily proved identity K_j(N*t + N - 1) = (-1)^t*K_(N - j + 1)(N*t + N - 1) and a known calculation formula for the determinant of circulant matrix [Wikipedia]. Besides, it is 0 for n=1..N-2. We also conjecture that every such sequence contains infinitely many blocks of N-1 negative and N-1 positive terms separated by 0's.
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LINKS
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MAPLE
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f:= n -> LinearAlgebra:-Determinant(Matrix(10, 10, shape=
Circulant[seq((-1)^j*add((-1)^k*binomial(n, 10*k+j),
k=0..(n-j)/10), j=0..9)])):
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MATHEMATICA
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ro[n_] := Table[(-1)^(j-1) Sum[(-1)^k Binomial[n, 10k+j-1], {k, 0, n/10}], {j, 1, 10}];
M[n_] := Table[RotateRight[ro[n], m], {m, 0, 9}];
a[n_] := Det[M[n]];
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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