A134346 - OEIS

%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