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%I #17 Mar 30 2024 07:53:17
%S 1,1,3,1,4,7,1,5,12,15,1,6,17,32,31,1,7,22,49,80,63,1,8,27,66,129,192,
%T 127,1,9,32,83,178,321,448,255,1,10,37,100,227,450,769,1024,511,1,11,
%U 42,117,276,579,1090,1793,2304,1023,1,12,47,134,325,708,1411,2562,4097,5120,2047,1,13,52,151,374,837,1732,3331,5890,9217,11264,4095
%N Array T(n,k) read by antidiagonals: T(1,k) = 2^k-1 and recursively T(n,k) = T(n-1,k) + A000337(k-1), n,k >= 1.
%C Generally, row n of the array is the binomial transform for 0, 1, n, 2n-1, 3n-2, 4n-3, ...
%F T(2,k) = A001787(k), binomial transform of 0, 1, 2, 3, 4, 5, 6, ...
%F T(3,k) = A000337(k), binomial transform of 0, 1, 3, 5, 7, 9, 11, ...
%F T(4,k) = A027992(k-1), binomial transform of 0, 1, 4, 7, 10, 13, 16, 19, 22, 25, ...
%F T(5,k) = binomial transform of 0, 1, 5, 9, 13, 17, 21, 25, 29, ...
%e To the first row, add the terms 0, 1, 5, 17, 49, 129, ... as indicated:
%e 1, 3, 7, 15, 31, 63, ...
%e 0, 1, 5, 17, 49, 129, ... (getting row 2 of the array:
%e 1, 4, 12, 32, 80, 192, ... (= A001787, binomial transform for 1,2,3, ...)
%e Repeat the operation, getting the following array T(n,k):
%e 1, 3, 7, 15, 31, 63, ...
%e 1, 4, 12, 32, 80, 192, ...
%e 1, 5, 17, 49, 129, 321, ...
%e 1, 6, 22, 66, 178, 450, ...
%p A000337 := proc(n)
%p 1+(n-1)*2^n ;
%p end proc:
%p A104746 := proc(n,k)
%p option remember;
%p if n= 1 then
%p 2^k-1 ;
%p else
%p procname(n-1,k)+A000337(k-1) ;
%p end if;
%p end proc:
%p for d from 1 to 12 do
%p for k from 1 to d do
%p n := d-k+1 ;
%p printf("%d,",A104746(n,k)) ;
%p end do:
%p end do; # _R. J. Mathar_, Oct 30 2011
%t A000337[n_] := (n - 1)*2^n + 1;
%t T[1, k_] := 2^k - 1;
%t T[n_, k_] := T[n, k] = T[n - 1, k] + A000337[k - 1];
%t Table[T[n - k + 1, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Mar 30 2024 *)
%Y Cf. A104747 (antidiagonal sums), A001787, A000337, A027992, A059823.
%K nonn,tabl
%O 1,3
%A _Gary W. Adamson_, Mar 23 2005
%E Terms corrected by _R. J. Mathar_, Oct 30 2011