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%I #37 Mar 14 2017 20:47:33
%S 1,1,1,1,2,1,1,3,1,1,1,4,2,2,1,1,5,2,4,1,1,1,6,3,6,3,2,1,1,7,3,9,3,5,
%T 1,1,1,8,4,12,6,8,4,2,1,1,9,4,16,6,14,4,6,1,1,1,10,5,20,10,20,10,10,5,
%U 2,1
%N Triangle related to A152198.
%C Row sums are A027383(n).
%C Diagonal sums are alternately A014739(n) and A001911(n+1).
%C The matrix inverse starts
%C 1;
%C -1,1;
%C 1,-2,1;
%C 1,-1,-1,1;
%C -1,2,0,-2,1;
%C -1,1,2,-2,-1,1;
%C 1,-2,-1,4,-1,-2,1;
%C 1,-1,-3,3,3,-3,-1,1;
%C -1,2,2,-6,0,6,-2,-2,1;
%C -1,1,4,-4,-6,6,4,-4,-1,1;
%C 1,-2,-3,8,2,-12,2,8,-3,-2,1;
%C apparently related to A158854. - _R. J. Mathar_, Apr 08 2013
%C From _Gheorghe Coserea_, Jun 11 2016: (Start)
%C T(n,k) is the number of terms of the sequence A057890 in the interval [2^n,2^(n+1)-1] having binary weight k+1.
%C T(n,k) = A007318(n,k) (mod 2) and the number of odd terms in row n of the triangle is 2^A000120(n).
%C (End)
%H Gheorghe Coserea, <a href="/A207974/b207974.txt">Rows n = 0..200, flattened</a>
%F T(n,k) = T(n-1,k-1) - (-1)^k*T(n-1,k), k>0 ; T(n,0) = 1.
%F T(2n,2k) = T(2n+1,2k) = binomial(n,k) = A007318(n,k).
%F T(2n+1,2k+1) = A110813(n,k).
%F T(2n+2,2k+1) = 2*A135278(n,k).
%F T(n,2k) + T(n,2k+1) = A152201(n,k).
%F T(n,2k) = A152198(n,k).
%F T(n+1,2k+1) = A152201(n,k).
%F T(n,k) = T(n-2,k-2) + T(n-2,k).
%F T(2n,n) = A128014(n+1).
%F T(n,k) = card {p, 2^n <= A057890(p) <= 2^(n+1)-1 and A000120(A057890(p)) = k+1}. - _Gheorghe Coserea_, Jun 09 2016
%F P_n(x) = Sum_{k=0..n} T(n,k)*x^k = ((2+x+(n mod 2)*x^2)*(1+x^2)^(n\2) - 2)/x. - _Gheorghe Coserea_, Mar 14 2017
%e Triangle begins :
%e n\k [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
%e [0] 1;
%e [1] 1, 1;
%e [2] 1, 2, 1;
%e [3] 1, 3, 1, 1;
%e [4] 1, 4, 2, 2, 1;
%e [5] 1, 5, 2, 4, 1, 1;
%e [6] 1, 6, 3, 6, 3, 2, 1;
%e [7] 1, 7, 3, 9, 3, 5, 1, 1;
%e [8] 1, 8, 4, 12, 6, 8, 4, 2, 1;
%e [9] 1, 9, 4, 16, 6, 14, 4, 6, 1, 1;
%e [10] ...
%p A207974 := proc(n,k)
%p if k = 0 then
%p 1;
%p elif k < 0 or k > n then
%p 0 ;
%p else
%p procname(n-1,k-1)-(-1)^k*procname(n-1,k) ;
%p end if;
%p end proc: # _R. J. Mathar_, Apr 08 2013
%o (PARI)
%o seq(N) = {
%o my(t = vector(N+1, n, vector(n, k, k==1 || k == n)));
%o for(n = 2, N+1, for (k = 2, n-1,
%o t[n][k] = t[n-1][k-1] + (-1)^(k%2)*t[n-1][k]));
%o return(t);
%o };
%o concat(seq(10)) \\ _Gheorghe Coserea_, Jun 09 2016
%o (PARI)
%o P(n) = ((2+x+(n%2)*x^2) * (1+x^2)^(n\2) - 2)/x;
%o concat(vector(11, n, Vecrev(P(n-1)))) \\ _Gheorghe Coserea_, Mar 14 2017
%Y Cf. Columns : A000012, A000027, A004526, A002620, A008805, A006918, A058187
%Y Cf. Diagonals : A000012, A000034, A052938, A097362
%Y Cf. A007318, A110813, A135278, A152201
%Y Related to thickness: A000120, A027383, A057890, A274036.
%K easy,nonn,tabl
%O 0,5
%A _Philippe Deléham_, Feb 22 2012
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