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%I #22 Jul 31 2018 16:00:28
%S 1,1,1,2,2,0,2,4,2,0,3,6,4,2,1,3,9,10,6,3,1,4,12,16,16,12,4,0,4,16,28,
%T 32,28,16,4,0,5,20,40,60,66,44,16,4,1,5,25,60,100,126,110,60,20,5,1,6,
%U 30,80,160,236,236,160,80,30,6,0,6,36,110,240,396,472,396,240,110,36,6,0
%N Triangle T(m,n) = number of subsets of {1,...,m} with n elements having an odd sum, 1 <= n <= m.
%C One could extend the triangle to include values for m=0 and/or n=0, but these correspond to empty sets and would always be 0. The first odd value for odd m and 1<n<m is T(13,5) = 651.
%H Alois P. Heinz, <a href="/A159916/b159916.txt">Rows n = 1..141, flattened</a>
%H Johann Cigler, <a href="https://arxiv.org/abs/1711.03340">Some remarks on Rogers-Szegö polynomials and Losanitsch's triangle</a>, arXiv:1711.03340 [math.CO], 2017.
%H Johann Cigler, <a href="https://homepage.univie.ac.at/johann.cigler/preprints/losanitsch3.pdf">Some Pascal-like triangles</a>, 2018.
%H Project Euler, <a href="http://projecteuler.net/index.php?section=problems&id=242">Problem 242: Odd Triplets</a>, April 25, 2009.
%F T(m,m) = A133872(m-1), T(m,1) = A004526(m+1).
%F T(n,k) = A007318(n,k) - A282011(n,k). - _Alois P. Heinz_, Feb 06 2017
%e The triangle starts:
%e (m=1) 1,
%e (m=2) 1,1,
%e (m=3) 2,2,0,
%e (m=4) 2,4,2,0,
%e (m=5) 3,6,4,2,1,
%e ...
%e T(5,3)=4, since the set {1,2,3,4,5} has four 3-element subsets having an odd sum of elements, namely {1,2,4}, {1,3,5}, {2,3,4} and {2,4,5}.
%p b:= proc(n, s) option remember; expand(
%p `if`(n=0, s, b(n-1, s)+x*b(n-1, irem(s+n, 2))))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, 0)):
%p seq(T(n), n=1..15); # _Alois P. Heinz_, Feb 04 2017
%t b[n_, s_] := b[n, s] = Expand[If[n==0, s, b[n-1, s] + x*b[n-1, Mod[s+n, 2]] ]];
%t T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[n, 0]];
%t Table[T[n], {n, 1, 15}] // Flatten (* _Jean-François Alcover_, Nov 17 2017, after _Alois P. Heinz_ *)
%o (PARI) T(n,k)=sum( i=2^k-1,2^n-2^(n-k), norml2(binary(i))==k & sum(j=0,n\2, bittest(i,2*j))%2 )
%Y Cf. A004526, A007318, A133872, A282011.
%Y T(2n,n) gives A110145.
%K nonn,tabl
%O 1,4
%A _M. F. Hasler_, Apr 30 2009
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