

A174391


Triangle, read by row, of constant term of X^k modulo nth cyclotomic polynomial.


0



1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1
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OFFSET

1,1


COMMENTS

a(n,k) (for n>=1 and 0<=k<n) is the constant term in the polynomial remainder of X^k modulo the nth cyclotomic polynomial; this sequence reads the triangle a(n,k) by rows (a(1,0)=1, a(2,0)=1, a(2,1)=1, a(3,0)=1, a(3,1)=0, a(3,2)=1, etc.). For each n, the finite sequence a(n,k), seen as a cyclic sequence of length n, has the property that the sum of every d'th term, for d dividing n, is zero, starting anywhere.


LINKS

Table of n, a(n) for n=1..98.


EXAMPLE

For n=6, the 6th cyclotomic polynomial is Phi_6(X) = X^2X+1, and the remainders of 1, X, X^2,..., X^5 mod Phi_6 are 1, X, X1, 1, X, X+1, so a(6,k)=1,0,1,1,0,1 for k from 0 to 5. (This gives the terms 16 to 22 of this sequence.)


PROG

(Sage) R.<x> = QQ['x']; [[((x^k)%(R.cyclotomic_polynomial(n))).subs(0) for k in range(n)] for n in range(1, 31)]


CROSSREFS

Sequence in context: A140865 A114000 A131218 * A343910 A329682 A113998
Adjacent sequences: A174388 A174389 A174390 * A174392 A174393 A174394


KEYWORD

sign,tabl


AUTHOR

David A. Madore, Mar 18 2010


STATUS

approved



