cp's OEIS Frontend

A077051 Right summatory matrix, T, by antidiagonals.

%I A077051 #16 Jun 24 2017 00:55:39
%S A077051 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 A077051 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 A077051 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 A077051 Right summatory matrix, T, by antidiagonals.
%C A077051 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 A077051 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 A077051 Sum of column k is A000005. - _Geoffrey Critzer_, Mar 29 2015
%F A077051 T(n, k)=1 if n|k else T(n, k)=0.
%F A077051 From _Boris Putievskiy_, May 08 2013: (Start)
%F A077051 As table T(n,k)= floor(n/k)-floor((n-1)/k).
%F A077051 As linear sequence a(n) = floor(A002260(n)/A004736(n)) - floor((A002260(n)-1)/A004736(n));
%F A077051 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 A077051 G.f. for row n: x^n/(1-x^n). - _Geoffrey Critzer_, Mar 29 2015
%e A077051 Northwest corner:
%e A077051 1 1 1 1 1 1
%e A077051 0 1 0 1 0 1
%e A077051 0 0 1 0 0 1
%e A077051 0 0 0 1 0 0
%e A077051 0 0 0 0 1 0
%e A077051 0 0 0 0 0 1
%t A077051 (* 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 A077051 Cf. A077049, A077050, A077052.
%K A077051 nonn,tabl
%O A077051 1,1
%A A077051 _Clark Kimberling_, Oct 22 2002