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%I #42 Apr 21 2017 04:01:22
%S 1,1,2,1,3,3,1,7,5,4,1,18,13,7,5,1,47,38,19,9,6,1,123,117,58,25,11,7,
%T 1,322,370,187,78,31,13,8,1,843,1186,622,257,98,37,15,9,1,2207,3827,
%U 2110,874,327,118,43,17,10,1,5778,12389,7252,3034,1126,397,138,49,19,11
%N Sequence read from antidiagonals of rectangular array with entry in row n and column q given by T(n,q) = 2^(2*n)*(Sum_{j=1..n+1} (cos(j*Pi/(2*q+1)))^(2*n)), n >= 0, q >= 1.
%C Row indices n begin with 0, column indices q begin with 1.
%H Andrew Howroyd, <a href="/A186740/b186740.txt">Table of n, a(n) for n = 0..1275</a>
%H S. Barbero, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Barbero/barbero15.html">Dickson Polynomials, Chebyshev Polynomials, and Some Conjectures of Jeffery</a>, Journal of Integer Sequences, 17 (2014), #14.3.8.
%F Conjecture: G.f. for column q is F_q(x) = (Sum_{r=0..q-1} ((q-r)*(-1)^r*binomial(2*q-r,r)*x^r)) / (Sum_{s=0..q} ((-1)^s*binomial(2*q-s,s)*x^s)), q >= 1.
%F Conjecture: G.f. for n-th row is of the form G_n(x) = H_n(x)/(1-x)^2, where H_n(x) is a polynomial in x.
%e Array begins:
%e 1 2 3 4 5 6 7 8 9 ...
%e 1 3 5 7 9 11 13 15 17 ...
%e 1 7 13 19 25 31 37 43 49 ...
%e 1 18 38 58 78 98 118 138 158 ...
%e 1 47 117 187 257 327 397 467 537 ...
%e 1 123 370 622 874 1126 1378 1630 1882 ...
%e 1 322 1186 2110 3034 3958 4882 5806 6730 ...
%e 1 843 3827 7252 10684 14116 17548 20980 24412 ...
%e 1 2207 12389 25147 38017 50887 63757 76627 89497 ...
%e ...
%e As a triangle:
%e 1,
%e 1, 2,
%e 1, 3, 3,
%e 1, 7, 5, 4,
%e 1, 18, 13, 7, 5,
%e 1, 47, 38, 19, 9, 6,
%e ...
%Y Conjecture: Transpose of array A185095.
%Y Conjecture: Columns 0,1,2 (up to an offset) are A000012, A005248, A198636 (proved, see the Barbero, et al., reference there).
%Y Conjecture: Rows 0,1,2,3,4 (up to an offset) are A000027, A005408, A016921, A114698, A114646.
%Y Cf. A209235.
%K nonn,tabl
%O 0,3
%A _L. Edson Jeffery_, Jan 21 2012
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