cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A153281 Triangle read by rows, A130321 * A127647. Also, number of subsets of [n+2] with consecutive integers that start at k.

Original entry on oeis.org

1, 2, 1, 4, 2, 2, 8, 4, 4, 3, 16, 8, 8, 6, 5, 32, 16, 16, 12, 10, 8, 64, 32, 32, 24, 20, 16, 13, 128, 64, 64, 48, 40, 32, 26, 21, 256, 128, 128, 96, 80, 64, 52, 42, 34, 512, 256, 256, 192, 160, 128, 104, 84, 68, 55
Offset: 0

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Author

Gary W. Adamson, Dec 23 2008

Keywords

Comments

Row sums = A008466(k-2): (1, 3, 8, 19, 43, 94, ...).
T(n,k) is the number of subsets of {1,...,n+2} that contain consecutive integers and that have k as the first integer in the first consecutive string. (See the example below.) Hence rows sums of T(n,k) give the number of subsets of {1,...,n+2} that contain consecutive integers. Also, T(n,k) = F(k)*2^(n+1-k), where F(k) is the k-th Fibonacci number, since there are F(k) subsets of {1,...,k-2} that contain no consecutive integers and there are 2^(n+1-k) subsets of {k+2,...,n+2}. [Dennis P. Walsh, Dec 21 2011]

Examples

			First few rows of the triangle:
    1;
    2,   1;
    4,   2,   2;
    8,   4,   4,   3;
   16,   8,   8,   6,   5;
   32,  16,  16,  12,  10,   8;
   64,  32,  32,  24,  20,  16,  13;
  128,  64,  64,  48,  40,  32,  26,  21;
  256, 128, 128,  96,  80,  64,  52,  42,  34;
  512, 256, 256, 192, 160, 128, 104,  84,  68,  55;
  ...
Row 4 = (16, 8, 8, 6, 5) = termwise products of (16, 8, 4, 2, 1) and (1, 1, 2, 3, 5).
For n=5 and k=3, T(5,3)=16 since there are 16 subsets of {1,2,3,4,5,6,7} containing consecutive integers with 3 as the first integer in the first consecutive string, namely,
  {1,3,4}, {1,3,4,5}, {1,3,4,6}, {1,3,4,7}, {1,3,4,5,6}, {1,3,4,5,7}, {1,3,4,6,7}, {1,3,4,5,6,7}, {3,4}, {3,4,5}, {3,4,6}, {3,4,7}, {3,4,5,6}, {3,4,5,7}, {3,4,6,7}, and {3,4,5,6,7}. [_Dennis P. Walsh_, Dec 21 2011]
		

Crossrefs

Programs

  • Maple
    with(combinat, fibonacci):
    seq(seq(2^(n+1-k)*fibonacci(k),k=1..(n+1)),n=0..10);
  • Mathematica
    Table[2^(n+1-k) Fibonacci[k],{n,0,10},{k,n+1}]//Flatten (* Harvey P. Dale, Apr 26 2020 *)

Formula

Triangle read by rows, A130321 * A127647. A130321 = an infinite lower triangular matrix with powers of 2: (A000079) in every column: (1, 2, 4, 8, ...).
A127647 = an infinite lower triangular matrix with the Fibonacci numbers, A000045 as the main diagonal and the rest zeros.
T(n,k)=2^(n+1-k)*F(k) where F(k) is the k-th Fibonacci number. [Dennis Walsh, Dec 21 2011]