A376697 -id:A376697 - cp's OEIS frontend

A376033 Number A(n,k) of binary words of length n avoiding distance (i+1) between "1" digits if the i-th bit is set in the binary representation of k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

1, 1, 2, 1, 2, 4, 1, 2, 3, 8, 1, 2, 4, 5, 16, 1, 2, 3, 6, 8, 32, 1, 2, 4, 4, 9, 13, 64, 1, 2, 3, 8, 6, 15, 21, 128, 1, 2, 4, 5, 12, 9, 25, 34, 256, 1, 2, 3, 6, 7, 18, 13, 40, 55, 512, 1, 2, 4, 4, 8, 11, 27, 19, 64, 89, 1024, 1, 2, 3, 8, 5, 11, 16, 45, 28, 104, 144, 2048

Offset: 0

Alois P. Heinz, Sep 09 2024

Also the number of subsets of [n] avoiding distance (i+1) between elements if the i-th bit is set in the binary representation of k. A(6,3) = 13: {}, {1}, {2}, {3}, {4}, {5}, {6}, {1,4}, {1,5}, {1,6}, {2,5}, {2,6}, {3,6}.
Each column sequence satisfies a linear recurrence with constant coefficients.
The sequence of row n is periodic with period A011782(n) = ceiling(2^(n-1)).

			A(6,6) = 17: 000000, 000001, 000010, 000011, 000100, 000110, 001000, 001100, 010000, 010001, 011000, 100000, 100001, 100010, 100011, 110000, 110001 because 6 = 110_2 and no two "1" digits have distance 2 or 3.
A(6,7) = 10: 000000, 000001, 000010, 000100, 001000, 010000, 010001, 100000, 100001, 100010.
A(7,7) = 14: 0000000, 0000001, 0000010, 0000100, 0001000, 0010000, 0010001, 0100000, 0100001, 0100010, 1000000, 1000001, 1000010, 1000100.
Square array A(n,k) begins:
     1,  1,   1,  1,   1,  1,  1,  1,   1,  1, ...
     2,  2,   2,  2,   2,  2,  2,  2,   2,  2, ...
     4,  3,   4,  3,   4,  3,  4,  3,   4,  3, ...
     8,  5,   6,  4,   8,  5,  6,  4,   8,  5, ...
    16,  8,   9,  6,  12,  7,  8,  5,  16,  8, ...
    32, 13,  15,  9,  18, 11, 11,  7,  24, 11, ...
    64, 21,  25, 13,  27, 16, 17, 10,  36, 17, ...
   128, 34,  40, 19,  45, 25, 27, 14,  54, 25, ...
   256, 55,  64, 28,  75, 37, 41, 19,  81, 37, ...
   512, 89, 104, 41, 125, 57, 60, 26, 135, 57, ...

Alois P. Heinz, Antidiagonals n = 0..200, flattened

Columns k=0-20 give: A000079, A000045(n+2), A006498(n+2), A000930(n+2), A006500, A130137, A079972(n+3), A003269(n+4), A031923(n+1), A263710(n+1), A224809(n+4), A317669(n+4), A351873, A351874, A121832(n+4), A003520(n+4), A208742, A374737, A375977, A375980, A375978.
Rows n=0-2 give: A000012, A007395(k+1), A010702(k+1).
Main diagonal gives A376091.
A(n,2^k-1) gives A141539.
A(2^n-1,2^n-1) gives A376697.
A(n,2^k) gives A209435.
Cf. A011782, A062383, A098011, A376921.

h:= proc(n) option remember; `if`(n=0, 1, 2^(1+ilog2(n))) end:
b:= proc(n, k, t) option remember; `if`(n=0, 1, add(`if`(j=1 and
      Bits[And](t, k)>0, 0, b(n-1, k, irem(2*t+j, h(k)))), j=0..1))
    end:
A:= (n, k)-> b(n, k, 0):
seq(seq(A(n, d-n), n=0..d), d=0..12);

step(v,b)={vector(#v, i, my(j=(i-1)>>1); if(bittest(i-1,0), if(bitand(b,j)==0, v[1+j], 0), v[1+j] + v[1+#v/2+j]));}
col(n,k)={my(v=vector(2^(1+logint(k,2))), r=vector(1+n)); v[1]=r[1]=1; for(i=1, n, v=step(v,k); r[1+i]=vecsum(v)); r}
A(n,k)=if(k==0, 2^n, col(n,k)[n+1]) \\ Andrew Howroyd, Oct 03 2024

A(n,k) = A(n,k+ceiling(2^(n-1))).
A(n,ceiling(2^(n-1))-1) = n+1.
A(n,ceiling(2^(n-2))) = ceiling(3*2^(n-2)) = A098011(n+2).

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.

Original entry on oeis.org

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

Alois P. Heinz, Aug 15 2008

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)

			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, ...

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

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.

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);

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 *)

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).

A376091 Number of binary words of length n avoiding distance (i+1) between "1" digits if the i-th bit is set in the binary representation of n.

Original entry on oeis.org

1, 2, 4, 4, 12, 11, 17, 14, 81, 57, 81, 61, 260, 126, 236, 106, 5000, 1623, 2653, 1181, 6848, 4751, 2838, 1286, 42024, 7526, 14272, 6416, 55012, 10422, 21992, 3970, 12595401, 1148865, 2411809, 268605, 2146689, 656872, 1018489, 186997, 25401600, 5147033, 1567504

Offset: 0

Alois P. Heinz, Sep 09 2024

Also the number of subsets of [n] avoiding distance (i+1) between elements if the i-th bit is set in the binary representation of n. a(6) = 17: {}, {1}, {2}, {3}, {4}, {5}, {6}, {1,2}, {1,5}, {1,6}, {2,3}, {2,6}, {3,4}, {4,5}, {5,6}, {1,2,6}, {1,5,6}.

			a(6) = 17: 000000, 000001, 000010, 000011, 000100, 000110, 001000, 001100, 010000, 010001, 011000, 100000, 100001, 100010, 100011, 110000, 110001 because 6 = 110_2 and no two "1" digits have distance 2 or 3.

Alois P. Heinz, Table of n, a(n) for n = 0..1023

Main diagonal of A376033.

Cf. A062383, A376697.

h:= proc(n) option remember; `if`(n=0, 1, 2^(1+ilog2(n))) end:
b:= proc(n, k, t) option remember; `if`(n=0, 1, add(`if`(j=1 and
      Bits[And](t, k)>0, 0, b(n-1, k, irem(2*t+j, h(k)))), j=0..1))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..50);

a(2^n-1) = A376697(n).

cp's OEIS Frontend

A376033 Number A(n,k) of binary words of length n avoiding distance (i+1) between "1" digits if the i-th bit is set in the binary representation of k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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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.

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A376091 Number of binary words of length n avoiding distance (i+1) between "1" digits if the i-th bit is set in the binary representation of n.

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