%I #60 Dec 20 2023 08:02:17
%S 1,3,3,7,14,7,15,45,45,15,31,124,186,124,31,63,315,630,630,315,63,127,
%T 762,1905,2540,1905,762,127,255,1785,5355,8925,8925,5355,1785,255,511,
%U 4088,14308,28616,35770,28616,14308,4088,511
%N Triangle read by rows: T(n,k) = (2^(n+1)-1)*binomial(n,k).
%C Inverse binomial transform: A134347.
%C From _Wolfdieter Lang_, Jul 27 2022: (Start)
%C Also the triangle t with offset 1 and elements t(n, m) = T(n-1, m-1) read by rows, giving in row n >= 1 the sums of the entries of A356028 of like m.
%C Also triangle t with offset 1 read by rows, giving in row n >= 1 the sum of the numbers from 1, 2, ..., 2^n - 1 with binary weight m, for m = 1, 2, ..., n. [Observation by _Kevin Ryde_.] (End)
%C T(n,k) is the sum of the entries in the (k+2)-th column of the Christmas tree pattern (A367562) of order n+1. - _Paolo Xausa_, Dec 20 2023
%H Paolo Xausa, <a href="/A134346/b134346.txt">Table of n, a(n) for n = 0..11475</a> (rows 0..150 of the triangle, flattened).
%F T(n, m) = A000225(n+1)*A007318(n, m).
%F From _Wolfdieter Lang_, Aug 21 2022: (Start)
%F T(n, k) = 0 for n < k, T(n, 0) = 2^(n+1) - 1, and
%F T(n, k) = T(n-1, k) + T(n-1, k-1) + binomial(n, k)*2^n, or
%F T(n, k) = 2*(T(n-1, k) + T(n-1, k-1)) + binomial(n-1, k-1).
%F (Proof for T(n-1, m-1) = t(n, m), offset 1, by separating in the list of the binary code of the numbers 1, 2, ..., 2^n-1 of length n and weight m the sublists with first entry 1 and 0. The total number of elements of the list for n and m is binomial(n, m).) (End)
%F T(n, k) = [x^k] ((1/2 - x)^(k - n - 1) - (1 - x)^(k - n - 1)). - _Peter Luschny_, Aug 22 2022
%e First few rows of the triangle:
%e n\k 0 1 2 3 4 5 6 7 8 9 ...
%e 0: 1
%e 1: 3 3
%e 2: 7 14 7
%e 3: 15 45 45 15
%e 4: 31 124 186 124 31
%e 5: 63 315 630 630 315 63
%e 6: 127 762 1905 2540 1905 762 127
%e 7: 255 1785 5355 8925 8925 5355 1785 255
%e 8: 511 4088 14308 28616 35770 28616 14308 4088 511
%e 9: 1023 9207 36828 85932 128898 128898 85932 36828 9207 1023
%e ... reformatted by _Wolfdieter Lang_, Aug 21 2022
%e ----------------------------------------------------------------------------------
%e T(3, 1) = 12 + 10 + 9 + 6 + 5 + 3 = 45. (From A356028 row n = 4, m = 2.)
%e Recurrences: T(4, 1) = 45 + 15 + 4*16 = 2*(45 + 15) +4 = 124. - _Wolfdieter Lang_, Jul 27 2022
%p A134346 := proc(n,k)
%p (2^(n+1)-1)*binomial(n,k) ;
%p end proc:
%p seq(seq( A134346(n,k),k=0..n),n=0..10) ; # _R. J. Mathar_, Aug 15 2022
%p ser := series((1/2 - x)^(k - n - 1) - (1 - x)^(k - n - 1), x, 10):
%p seq(seq(coeff(ser, x, k), k = 0..n), n = 0..9); # _Peter Luschny_, Aug 22 2022
%t A134346[n_,k_]:=(2^(n+1)-1)Binomial[n,k];
%t Table[A134346[n,k],{n,0,10},{k,0,n}] (* _Paolo Xausa_, Dec 20 2023 *)
%o (PARI) T(n,k) = my(b=binomial(n,k)); b<<(n+1) - b; \\ _Kevin Ryde_, Aug 15 2022
%Y Cf. A000225, A006516(n+1) (row sums), A124929, A134347, A356028, A356117.
%Y Cf. A367508, A367562.
%K nonn,tabl,easy
%O 0,2
%A _Gary W. Adamson_, Oct 21 2007
%E Name simplified by _R. J. Mathar_, Aug 15 2022