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.

Showing 1-8 of 8 results.

A050926 Binary representation of A007908(n).

Original entry on oeis.org

1, 1100, 1111011, 10011010010, 11000000111001, 11110001001000000, 100101101011010000111, 101111000110000101001110, 111010110111100110100010101, 1011011111110111000001110000111110
Offset: 1

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 30 1999

Keywords

Crossrefs

Formula

a(n) = A007088(A007908(n)). - Seiichi Manyama, Apr 23 2022

A097580 Base 3 representation of the concatenation of the first n numbers with the most significant digits first.

Original entry on oeis.org

1, 110, 11120, 1200201, 121221020, 20021100110, 2022201111201, 212020020002100, 22121022020212200, 1011212101120110200001, 11101000122011021220211010, 121012010100112220022220220120
Offset: 1

Views

Author

Cino Hilliard, Aug 29 2004

Keywords

Comments

Consider numbers of the form 1, 12, 123, 1234, ..., N. Find the highest power of 3^p such that 3^p <= N. Then p = [log(N)/log(3)] and for 0 <= qi <= 2 [N/3^p] = q1 + r1 [r1/3^(p-1)] = q2 + r2 ........................ rp/3^1 = qp + rp+1 rp+1/3^0 = qp+1 0 For N = 1234, p = [log(1234)/log(3)] = 6 division quot rem 1234/3^6 = 1 505 505/3^5 = 2 19 19/3^4 = 0 19 19/3^3 = 0 19 19/3^2 = 2 1 1/3^1 = 0 1 1/3^0 = 1 0 The sequence of quotients, top down, form the entry in the table for 1234. Obviously this algorithm works for any N.

Examples

			The 4th concatenation of the integers > 0 is 1234. base(10,3,1234) = 1200201 the 4th entry in the table.
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[IntegerDigits[FromDigits[Flatten[Table[ IntegerDigits[n],{n,i}]]],3]],{i,12}] (* Harvey P. Dale, May 23 2011 *)

Formula

a(n) = A007089(A007908(n)). - Seiichi Manyama, Apr 23 2022

A097582 Base 7 representation of the concatenation of the first n numbers with the most significant digits first.

Original entry on oeis.org

1, 15, 234, 3412, 50664, 1022634, 13331215, 206636142, 3026236221, 614636352655, 155123512633260, 35001313215554565, 10403265603212022112, 2132066345452131466644, 434014101450663623262042
Offset: 1

Views

Author

Cino Hilliard, Aug 29 2004

Keywords

Comments

Consider numbers of the form 1, 12, 123, 1234, ..., N. Find the highest power of 7^p such that 7^p < N. Then p = [log(N)/log(7)] and for 0 <= qi <= 6 [N/7^p] = q1 + r1 [r1/7^(p-1)] = q2 + r2 ........................ rp/7^1 = qp + rp+1 rp+1/7^0 = qp+1 0 For N = 1234, p = [log(1234)/log(7)] = 3 division quot rem 1234/7^3 = 3 205 205/7^2 = 4 9 9/7^1 = 1 2 2/7^0 = 2 0 The sequence of quotients, top down, forms the entry in the table for 1234. Obviously this algorithm works for any N.

Crossrefs

Formula

a(n) = A007093(A007908(n)). - Seiichi Manyama, Apr 23 2022

A353116 Base-8 representation of A000422(n).

Original entry on oeis.org

1, 25, 501, 10341, 152061, 2375761, 35145661, 516277661, 7267464261, 121672446261, 20125401442261, 3342313131342261, 564731226736442261, 116343242472501442261, 20325073566675147442261, 3327107422474572347442261
Offset: 1

Views

Author

Seiichi Manyama, Apr 24 2022

Keywords

Crossrefs

Programs

  • Ruby
    def A(k, n)
      (1..n).map{|i| (1..i).to_a.reverse.join.to_i.to_s(k).to_i}
    end
    p A(8, 20)

Formula

a(n) = A007094(A000422(n)).

A353104 Base-4 representation of A007908(n).

Original entry on oeis.org

1, 30, 1323, 103102, 3000321, 132021000, 10231122013, 233012011032, 13112330310111, 23133313001300332, 101331301332300201003, 130010202012003303220320, 223313011101131330223313231, 1010202010033102302310203002202
Offset: 1

Views

Author

Seiichi Manyama, Apr 23 2022

Keywords

Crossrefs

Programs

  • Ruby
    def A(k, n)
      (1..n).map{|i| (1..i).to_a.join.to_i.to_s(k).to_i}
    end
    p A(4, 20)

Formula

a(n) = A007090(A007908(n)).

A353105 Base-5 representation of A007908(n).

Original entry on oeis.org

1, 22, 443, 14414, 343340, 12422311, 304001232, 11130030203, 223101104124, 200240443211120, 130211343340003021, 112140204001002213422, 100421133100401442024323, 40324014240311242321340224, 31241112311230113034201201130
Offset: 1

Views

Author

Seiichi Manyama, Apr 23 2022

Keywords

Crossrefs

Programs

  • Ruby
    def A(k, n)
      (1..n).map{|i| (1..i).to_a.join.to_i.to_s(k).to_i}
    end
    p A(5, 20)

Formula

a(n) = A007091(A007908(n)).

A353106 Base-6 representation of A007908(n).

Original entry on oeis.org

1, 20, 323, 5414, 133053, 2351320, 42243331, 1120335530, 20130035113, 5401014424514, 2343052550252003, 1114323133240053240, 321321022332303252301, 132140014431255340214310, 42015444551405453142112503
Offset: 1

Views

Author

Seiichi Manyama, Apr 23 2022

Keywords

Crossrefs

Programs

  • Ruby
    def A(k, n)
      (1..n).map{|i| (1..i).to_a.join.to_i.to_s(k).to_i}
    end
    p A(6, 20)

Formula

a(n) = A007092(A007908(n)).

A353107 Base-9 representation of A007908(n).

Original entry on oeis.org

1, 13, 146, 1621, 17836, 207313, 2281451, 25206070, 277266780, 34771513601, 4330564256733, 535110486286816, 65858371163036861, 8202185121837583406, 1113465620754570813253, 134741562223525280514741, 16425841240157671153405780
Offset: 1

Views

Author

Seiichi Manyama, Apr 23 2022

Keywords

Crossrefs

Programs

  • Ruby
    def A(k, n)
      (1..n).map{|i| (1..i).to_a.join.to_i.to_s(k).to_i}
    end
    p A(9, 20)

Formula

a(n) = A007095(A007908(n)).
Showing 1-8 of 8 results.