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Right summatory matrix, T, by antidiagonals.
5

%I #16 Jun 24 2017 00:55:39

%S 1,0,1,0,1,1,0,0,0,1,0,0,1,1,1,0,0,0,0,0,1,0,0,0,1,0,1,1,0,0,0,0,0,1,

%T 0,1,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,1,1,1,1,0,0,

%U 0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,0,1,0,1

%N Right summatory matrix, T, by antidiagonals.

%C If S=(s(1),s(2),...) is a sequence written as a row vector, then S*T is the summatory sequence of S; i.e. its n-th term is Sum{s(k): k|n}. T is the transpose of the left summatory matrix, A077049; T is the inverse of the right Moebius transformation matrix. See A077049 for further properties.

%C This is essentially the same as A051731, which includes only the triangle. Note that the standard in the OEIS is left to right antidiagonals, which would make this the left summatory matrix, and A077049 the right one. - _Franklin T. Adams-Watters_, Apr 08 2009

%C Sum of column k is A000005. - _Geoffrey Critzer_, Mar 29 2015

%F T(n, k)=1 if n|k else T(n, k)=0.

%F From _Boris Putievskiy_, May 08 2013: (Start)

%F As table T(n,k)= floor(n/k)-floor((n-1)/k).

%F As linear sequence a(n) = floor(A002260(n)/A004736(n)) - floor((A002260(n)-1)/A004736(n));

%F a(n) = floor(i/j) - floor((i-1)/j), where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2). (End)

%F G.f. for row n: x^n/(1-x^n). - _Geoffrey Critzer_, Mar 29 2015

%e Northwest corner:

%e 1 1 1 1 1 1

%e 0 1 0 1 0 1

%e 0 0 1 0 0 1

%e 0 0 0 1 0 0

%e 0 0 0 0 1 0

%e 0 0 0 0 0 1

%t (* returns the northwest corner *) nn = 20; Table[PadRight[Drop[CoefficientList[Series[x^n/(1 - x^n), {x, 0, nn}], x],1], nn], {n, 1, nn}] // Grid (* _Geoffrey Critzer_, Mar 29 2015 *)

%Y Cf. A077049, A077050, A077052.

%K nonn,tabl

%O 1,1

%A _Clark Kimberling_, Oct 22 2002