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%I #53 Jun 04 2021 23:18:23
%S 1,1,0,1,1,0,1,0,0,0,1,1,1,0,0,1,0,0,0,0,0,1,1,0,1,0,0,0,1,0,1,0,0,0,
%T 0,0,1,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,
%U 0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,1,0,1,0,0,0,0,0,0,0,0,0
%N Left summatory matrix, T, by antidiagonals upwards.
%C If S = (s(1), s(2), ...) is a sequence written as a column vector, then T*S is the summatory sequence of S; i.e., its n-th term is Sum_{k|n} s(k). T is the inverse of the left Moebius transformation matrix, A077050. Except for the first term in some cases, column 1 of T^(-2) is A007427, column 1 of T^(-1) is A008683, Column c of T^2 is A000005, column 1 of T^3 is A007425.
%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 right summatory matrix, and A077051 the left one. - _Franklin T. Adams-Watters_, Apr 08 2009
%C From _Gary W. Adamson_, Apr 28 2010: (Start)
%C As defined with antidiagonals of the array = the triangle shown in the example section. Row sums of this triangle = A032741 (with a different offset): 1, 1, 2, 1, 3, 1, 3, ...
%C Let the triangle = M. Then lim_{n->inf} M^n = A002033, the left-shifted vector considered as a sequence: (1, 1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 8, ...). (End)
%H Michael De Vlieger, <a href="/A077049/b077049.txt">Table of n, a(n) for n = 1..11325</a> (Rows 1 <= n <= 150).
%H Clark Kimberling, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Kimberling/kimberling24.html">Matrix Transformations of Integer Sequences</a>, J. Integer Seqs., Vol. 6, 2003.
%F T(n,k)=1 if k|n, otherwise T(n,k)=0, k >= 1, n >= 1.
%F From _Boris Putievskiy_, May 08 2013: (Start)
%F As table T(n,k) = floor(k/n) - floor((k-1)/n).
%F As linear sequence a(n) = floor(A004736(n)/A002260(n)) - floor((A004736(n)-1)/A002260(n)); a(n) = floor(j/i)-floor((j-1)/i), where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2). (End)
%e T(4,2) = 1 since 2 divides 4. Northwest corner:
%e 1 0 0 0 0 0
%e 1 1 0 0 0 0
%e 1 0 1 0 0 0
%e 1 1 0 1 0 0
%e 1 0 0 0 1 0
%e 1 1 1 0 0 1
%e From _Gary W. Adamson_, Apr 28 2010: (Start)
%e First few rows of the triangle (when T is read by antidiagonals upwards):
%e 1;
%e 1, 0;
%e 1, 1, 0;
%e 1, 0, 0, 0;
%e 1, 1, 1, 0, 0;
%e 1, 0, 0, 0, 0, 0;
%e 1, 1, 0, 1, 0, 0, 0;
%e 1, 0, 1, 0, 0, 0, 0, 0;
%e 1, 1, 0, 0, 1, 0, 0, 0, 0;
%e ... (End)
%p A077049 := proc(n,k)
%p if modp(n,k) = 0 then
%p 1;
%p else
%p 0 ;
%p end if;
%p end proc:
%p for d from 2 to 10 do
%p for k from 1 to d-1 do
%p n := d-k ;
%p printf("%d,",A077049(n,k)) ;
%p end do:
%p end do: # _R. J. Mathar_, Jul 22 2017
%t With[{nn = 14}, DeleteCases[#, -1] & /@ Transpose@ Table[Take[#, nn] &@ Flatten@ Join[ConstantArray[-1, k - 1], ConstantArray[Reverse@ IntegerDigits[2^(k - 1), 2], Ceiling[(nn - k + 1)/k]]], {k, nn}]] // Flatten (* _Michael De Vlieger_, Jul 22 2017 *)
%o (PARI) nn=10; matrix(nn, nn, n, k, if (n % k, 0, 1)) \\ _Michel Marcus_, May 21 2015
%o (Python)
%o def T(n, k):
%o return 1 if n%k==0 else 0
%o for n in range(1, 11): print([T(n - k + 1, k) for k in range(1, n + 1)]) # _Indranil Ghosh_, Jul 22 2017
%Y Cf. A051731, A077050, A077051, A077052, A000005 (row sums).
%Y Cf. A032741, A002033. - _Gary W. Adamson_, Apr 28 2010
%K nonn,tabl
%O 1,1
%A _Clark Kimberling_, Oct 22 2002
%E Name edited by _Petros Hadjicostas_, Jul 27 2019
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