A126268 -id:A126268 - cp's OEIS frontend

A126277 Triangle generated from Eulerian numbers.

1, 1, 2, 1, 3, 3, 1, 4, 7, 4, 1, 5, 11, 15, 5, 1, 6, 15, 26, 31, 6, 1, 7, 19, 37, 57, 63, 7, 1, 8, 23, 48, 83, 120, 127, 8, 1, 9, 27, 59, 109, 177, 247, 255, 9, 1, 10, 31, 70, 135, 234, 367, 502, 511, 10

Offset: 1

Gary w. Adamson, Dec 23 2006

N-th diagonal starting from the right = binomial transform of [1, N, q, q, q, ...) where q = 2*N - 2. Given the infinite set of triangles "T" composed of partial column sums of the polygonal numbers, the N-th diagonal starting from the right = row sums of triangle "T": (T=3 = A104712; T=4 = A125165; T=5 = A125232; T=6 = A125233; T=7 = A125234, T=8 = A125235; and so on). For example, 3rd diagonal from the right = the offset Eulerian numbers, (1, 4, 11, 26, 57, 120, ...) = row sums of Triangle A104712 having partial column sums of the triangular numbers: 1; 3, 1; 6, 4, 1; 10, 10, 5, 1; 15, 20, 15, 6, 1; ... Row sums = A124671: (1, 3, 7, 16, 37, 85, 191, ...).

			First few rows of the triangle:
  1;
  1,  2;
  1,  3,  3;
  1,  4,  7,  4;
  1,  5, 11, 15,   5;
  1,  6, 15, 26,  31,   6;
  1,  7, 19, 37,  57,  63,   7;
  1,  8, 23, 48,  83, 120, 127,   8;
  1,  9, 27, 59, 109, 177, 247, 255,   9;
  1, 10, 31, 70, 135, 234, 367, 502, 511, 10;
  ...
T(7,4) = 37 = A000295(4) + T(6,4) = 11 + 26.

G. C. Greubel, Rows n=1..100 of triangle, flattened

Cf. A126268, A000295, A104712, A125165, A104712, A125232 - A125235.

T[n_,1]:=1; T[n_,n_]:=n; T[n_,k_]:= T[n-1,k] + 2^k - k - 1; Table[T[n,k], {n,1,15}, {k,1,n}]//Flatten (* G. C. Greubel, Oct 23 2018 *)

{T(n,k) = if(k==1, 1, if(k==n, n, 2^k - k - 1 + T(n-1,k)))};
for(n=1,10, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 23 2018

Given right border = (1,2,3,...), T(n,k) = A000295(k) + T(n-1,k); where A000295 = the Eulerian numbers starting (0, 1, 4, 11, 26, 57, ...).

A126269 Numbers n such that hcl(n,n) < hcl(n,n-1) where hcl(n,i) is the Huffman code length; see comments.

Original entry on oeis.org

3, 4, 9, 10, 21, 22, 45, 46, 93, 94, 189, 190, 381, 382, 765, 766, 1533, 1534, 3069, 3070, 6141, 6142, 12285, 12286, 24573, 24574, 49149, 49150

Offset: 3

Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Dec 22 2006

Consider a string which consists of n distinct symbols such that symbol(i) has frequency i (i=1,2,...,n). Then hcl(n,i) is the Huffman code length of symbol(i).

			Possible Huffman codes for n = 3,4,5 are:
1 : 00
2 : 01
3 : 1
--------
1 : 100
2 : 101
3 : 11
4 : 0
--------
1 : 000
2 : 001
3 : 01
4 : 10
5 : 11
hcl(3,3)=1 < 2=hcl(3,2) and hcl(4,4)=1 < 2=hcl(4,3); so 3,4 are in the sequence.
hcl(5,5)=2=hcl(5,4) so 5 is not in the sequence.

Sean A. Irvine, Java program (github)

Cf. A126268 A126014 A068156 A033484.

Conjecture: a(2k) = A033484(k-1) and a(2k-1) = A068156(k-1), k >= 2.
Conjectures from Colin Barker, Aug 06 2019: (Start)
G.f.: x^3*(3 + 4*x - 2*x^3) / ((1 - x)*(1 + x)*(1 - 2*x^2)).
a(n) = 3*a(n-2) - 2*a(n-4) for n>6.
a(n) = -5/2 + (-1)^n/2 + 3*2^((1/2)*(n-5))*(2-2*(-1)^n + sqrt(2) + (-1)^n*sqrt(2)) for n>2.
(End)

More terms from Sean A. Irvine, Aug 05 2019

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A126277 Triangle generated from Eulerian numbers.

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