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%I #14 Mar 24 2017 00:47:50
%S 0,0,0,0,1,0,0,1,1,0,0,2,2,2,0,0,1,2,2,1,0,0,3,3,4,3,3,0,0,1,3,3,3,3,
%T 1,0,0,3,3,5,4,5,3,3,0,0,2,4,4,5,5,4,4,2,0,0,3,4,6,5,7,5,6,4,3,0,0,1,
%U 3,4,5,5,5,5,4,3,1,0,0,5,5,7,7,9,7,9,7,7,5,5,0,0,1,5,5,6,7,7,7,7,6,5,5,1,0
%N Number of irreducible factors of Gauss polynomials.
%C T(n,k) is the number of irreducible factors of the (separable) polynomial [n]!/([k]![n-k]!). Here [n]! denotes the product of the first n quantum integers, the n-th quantum integer being defined as (1-q^n)/(1-q).
%C T(n,k) gives the number of positive integers m <= n such that (n mod m) < (k mod m). - _Tom Edgar_, Aug 21 2014
%H Stan Wagon and Herbert S. Wilf, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v1i1r3">When are subset sums equidistributed modulo m?</a>, The Electronic Journal of Combinatorics, Vol. 1, 1994 (#R3).
%F T(n, k) = T(n-1, k-1) + d(n) - d(k), where d(n) is the number of divisors of n.
%e The triangle T(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ...
%e 0: 0
%e 1: 0 0
%e 2: 0 1 0
%e 3: 0 1 1 0
%e 4: 0 2 2 2 0
%e 5: 0 1 2 2 1 0
%e 6: 0 3 3 4 3 3 0
%e 7: 0 1 3 3 3 3 1 0
%e 8: 0 3 3 5 4 5 3 3 0
%e 9: 0 2 4 4 5 5 4 4 2 0
%e 10: 0 3 4 6 5 7 5 6 4 3 0
%e 11: 0 1 3 4 5 5 5 5 4 3 1 0
%e 12: 0 5 5 7 7 9 7 9 7 7 5 5 0
%e 13: 0 1 5 5 6 7 7 7 7 6 5 5 1 0
%e ... Formatted by _Wolfdieter Lang_, Dec 07 2012
%e T(8,3) equals the number of irreducible factors of (1-q^8)(1-q^7)(1-q^6)/((1-q^3)(1-q^2)(1-q)), which is a product of 5 cyclotomic polynomials in q, namely the 2nd, 4th, 6th, 7th and 8th. Thus T(8,3)=5.
%Y Cf. A008967, A047971, A219237, A000005.
%K easy,nonn,tabl
%O 0,12
%A _Paul Boddington_, Jan 09 2004
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