A141539 Square array A(n,k) of numbers of length n binary words with at least k "0" between any two "1" digits (n,k >= 0), read by antidiagonals.

1, 1, 2, 1, 2, 4, 1, 2, 3, 8, 1, 2, 3, 5, 16, 1, 2, 3, 4, 8, 32, 1, 2, 3, 4, 6, 13, 64, 1, 2, 3, 4, 5, 9, 21, 128, 1, 2, 3, 4, 5, 7, 13, 34, 256, 1, 2, 3, 4, 5, 6, 10, 19, 55, 512, 1, 2, 3, 4, 5, 6, 8, 14, 28, 89, 1024, 1, 2, 3, 4, 5, 6, 7, 11, 19, 41, 144, 2048, 1, 2, 3, 4, 5, 6, 7, 9, 15, 26, 60, 233, 4096

Offset

0,3

Comments

A(n,k+1) = A(n,k) - A143291(n,k).

From Gary W. Adamson, Dec 19 2009: (Start)

Alternative method generated from variants of an infinite lower triangle T(n) = A000012 = (1; 1,1; 1,1,1; ...) such that T(n) has the leftmost column shifted up n times. Then take lim_{k->infinity} T(n)^k, obtaining a left-shifted vector considered as rows of an array (deleting the first 1) as follows:

1, 2, 4, 8, 16, 32, 64, 128, 256, ... = powers of 2

1, 1, 2, 3, 5, 8, 13, 21, 34, ... = Fibonacci numbers

1, 1, 1, 2, 3, 4, 6, 9, 13, ... = A000930

1, 1, 1, 1, 2, 3, 4, 5, 7, ... = A003269

... with the next rows A003520, A005708, A005709, ... such that beginning with the Fibonacci row, the succession of rows are recursive sequences generated from a(n) = a(n-1) + a(n-2); a(n) = a(n-1) + a(n-3), ... a(n) = a(n-1) + a(n-k); k = 2,3,4,... Last, columns going up from the topmost 1 become rows of triangle A141539. (End)

Links

Alois P. Heinz, Table of n, a(n) for n = 0..140, flattened

Formula

G.f. of column k: x^(-k)/(1-x-x^(k+1)).

A(n,k) = 2^n if k=0, otherwise A(n,k) = n+1 if n<=k, otherwise A(n,k) = A(n-1,k) + A(n-k-1,k).

Example

A(4,2) = 6, because 6 binary words of length 4 have at least 2 "0" between any two "1" digits: 0000, 0001, 0010, 0100, 1000, 1001.
Square array A(n,k) begins:
    1,  1,  1,  1,  1,  1,  1,  1, ...
    2,  2,  2,  2,  2,  2,  2,  2, ...
    4,  3,  3,  3,  3,  3,  3,  3, ...
    8,  5,  4,  4,  4,  4,  4,  4, ...
   16,  8,  6,  5,  5,  5,  5,  5, ...
   32, 13,  9,  7,  6,  6,  6,  6, ...
   64, 21, 13, 10,  8,  7,  7,  7, ...
  128, 34, 19, 14, 11,  9,  8,  8, ...

Maple

A:= proc(n, k) option remember;
      if k=0 then 2^n
    elif n<=k and n>=0 then n+1
    elif n>0 then A(n-1, k) +A(n-k-1, k)
    else          A(n+1+k, k) -A(n+k, k)
      fi
    end:
seq(seq(A(n, d-n), n=0..d), d=0..15);

Mathematica

a[n_, k_] := a[n, k] = Which[k == 0, 2^n, n <= k && n >= 0, n+1, n > 0, a[n-1, k] + a[n-k-1, k], True, a[n+1+k, k] - a[n+k, k]]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 15}] // Flatten (* Jean-François Alcover, Dec 17 2013, translated from Maple *)

Crossrefs

Cf. column k=0: A000079, k=1: A000045(n+2), k=2: A000930(n+2), A068921, A078012(n+5), k=3: A003269(n+4), A017898(n+7), k=4: A003520(n+4), A017899(n+9), k=5: A005708(n+5), A017900(n+11), k=6: A005709(n+6), A017901(n+13), k=7: A005710(n+7), A017902(n+15), k=8: A005711(n+7), A017903(n+17), k=9: A017904(n+19), k=10: A017905(n+21), k=11: A017906(n+23), k=12: A017907(n+25), k=13: A017908(n+27), k=14: A017909(n+29).

Main diagonal gives A000027(n+1).

A(2n,n) gives A000217(n+1)

A(3n,n) gives A008778.

A(3n,2n) gives A034856(n+1).

A(2n,3n) gives A005408.

A(2^n-1,n) gives A376697.

See also A143291.

Sequence in context: A081532 A174843 A253572 * A340547 A376033 A327844

Adjacent sequences: A141536 A141537 A141538 * A141540 A141541 A141542

Keyword

nonn,tabl

Author

Alois P. Heinz, Aug 15 2008

Status

approved