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.

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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.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Sep 09 2024

Keywords

Comments

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)).

Examples

			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, ...

Links

Crossrefs

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.

Programs

  • Maple
    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);
  • PARI
    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

Formula

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).

A209434 Table T(n,m), read by antidiagonals, is the number of subsets of {1,...,n} which do not contain two elements whose difference is m+1.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 5, 4, 2, 1, 8, 6, 4, 2, 1, 13, 9, 8, 4, 2, 1, 21, 15, 12, 8, 4, 2, 1, 34, 25, 18, 16, 8, 4, 2, 1, 55, 40, 27, 24, 16, 8, 4, 2, 1, 89, 64, 45, 36, 32, 16, 8, 4, 2, 1, 144, 104, 75, 54, 48, 32, 16, 8, 4, 2, 1, 233, 169, 125, 81, 72, 64, 32
Offset: 0

Views

Author

David Nacin, Mar 09 2012

Keywords

Comments

1st column is the Fibonacci sequence.

Examples

			Table begins:
1,   1,   1,   1,   1,   1,   1,   1,   1,   1,    1,    ...
2,   2,   2,   2,   2,   2,   2,   2,   2,   2,    2,    ...
3,   4,   4,   4,   4,   4,   4,   4,   4,   4,    4,    ...
5,   6,   8,   8,   8,   8,   8,   8,   8,   8,    8,    ...
8,   9,   12,  16,  16,  16,  16,  16,  16,  16,   16,   ...
13,  15,  18,  24,  32,  32,  32,  32,  32,  32,   32,   ...
21,  25,  27,  36,  48,  64,  64,  64,  64,  64,   64,   ...
34,  40,  45,  54,  72,  96,  128, 128, 128, 128,  128,  ...
55,  64,  75,  81,  108, 144, 192, 256, 256, 256,  256,  ...
89,  104, 125, 135, 162, 216, 288, 384, 512, 512,  512,  ...
144, 169, 200, 225, 243, 324, 432, 576, 768, 1024, 1024, ...
............................................................

References

  • M. El-Mikkawy, T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1.

Links

Crossrefs

Cf. A209435, A209436, A209437. Columns: A006498, A006500, A031923, A208742, A208743, A009641

Programs

  • Mathematica
    a[n_, m_] := Product[Fibonacci[Floor[(n + i)/(m + 1) + 2]], {i, 0, m}]; Flatten[Table[a[j - i, i], {j, 0, 30}, {i, 0, j}]]

Formula

T(n,m) = Product_{i=0 to m} F(floor[(n + i)/(m + 1) + 2]) where F(n) is the n-th Fibonacci number.

A208743 Number of subsets of the set {1,2,...,n} which do not contain two elements whose difference is 6.

Original entry on oeis.org

2, 4, 8, 16, 32, 64, 96, 144, 216, 324, 486, 729, 1215, 2025, 3375, 5625, 9375, 15625, 25000, 40000, 64000, 102400, 163840, 262144, 425984, 692224, 1124864, 1827904, 2970344, 4826809, 7797153, 12595401, 20346417, 32867289, 53093313, 85766121, 138859434
Offset: 1

Views

Author

David Nacin, Mar 01 2012

Keywords

Examples

			If n=7 then we must count all subsets not containing both 1 and 7.  There are 2^5 subsets containing 1 and 7, giving us 2^7 - 2^5 = 48.  Thus a(7) = 96.

References

  • M. El-Mikkawy, T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1 k=6.

Links

Crossrefs

Cf. A006498, A006500, A208741, A208742.

Programs

  • Mathematica
    Table[Fibonacci[Floor[n/6] + 3]^Mod[n, 6] * Fibonacci[Floor[n/6] + 2]^(6 - Mod[n, 6]), {n, 1, 80}]
LinearRecurrence[{1, 1, 0, 0, 0, -5, 5, 5, 0, 0, 0, 15, -15, -15, 0, 0, 0, 15, -15, -15, 0, 0, 0, -5, 5, 5, 0, 0, 0, -1, 1, 1}, {2, 4, 8, 16, 32, 64, 96, 144, 216, 324, 486, 729, 1215, 2025, 3375, 5625, 9375, 15625, 25000, 40000, 64000, 102400, 163840, 262144, 425984, 692224, 1124864, 1827904, 2970344, 4826809, 7797153, 12595401}, 80]
  • PARI
    Vec(x*(2 + 2*x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 10*x^6 - 6*x^7 - 14*x^8 - 16*x^9 - 14*x^10 - x^11 - 30*x^12 - 29*x^13 - 15*x^14 - 15*x^15 - 15*x^16 - 20*x^17 - 30*x^18 - 10*x^19 + 5*x^20 + 5*x^21 + 5*x^22 + 4*x^23 + 10*x^24 + 6*x^25 + x^26 + x^27 + x^28 + x^29 + 2*x^30 + x^31) / ((1 + x^2)*(1 - x - x^2)*(1 - x^2 + x^4)*(1 + x^3 - x^6)*(1 - x^3 - x^6)*(1 + 7*x^6 + x^12)) + O(x^30)) \\ Colin Barker, Feb 23 2017

Formula

a(n) = F(floor(n/6) + 3)^(n mod 6)*F(floor(n/6) + 2)^(6 - (n mod 6)) where F(n) is the n-th Fibonacci number.
a(n) = a(n-1) + a(n-2) - 5*a(n-6) + 5*a(n-7) + 5*a(n-8) + 15*a(n-12) - 15*a(n-13) - 15*a(n-14) + 15*a(n-18) - 15*a(n-19) - 15*a(n-20) - 5*a(n-24) + 5*a(n-25) + 5*a(n-26) - a(n-30) + a(n-31) + a(n-32).
G.f.: x*(2 + 2*x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 10*x^6 - 6*x^7 - 14*x^8 - 16*x^9 - 14*x^10 - x^11 - 30*x^12 - 29*x^13 - 15*x^14 - 15*x^15 - 15*x^16 - 20*x^17 - 30*x^18 - 10*x^19 + 5*x^20 + 5*x^21 + 5*x^22 + 4*x^23 + 10*x^24 + 6*x^25 + x^26 + x^27 + x^28 + x^29 + 2*x^30 + x^31) / ((1 + x^2)*(1 - x - x^2)*(1 - x^2 + x^4)*(1 + x^3 - x^6)*(1 - x^3 - x^6)*(1 + 7*x^6 + x^12)). - Colin Barker, Feb 23 2017
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