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 1-9 of 9 results.

A050928 Sum of digits of A050926(n).

Original entry on oeis.org

1, 2, 6, 5, 6, 6, 11, 12, 16, 21, 21, 19, 34, 23, 41, 35, 43, 52, 49, 55, 58, 49, 61, 69, 67, 81, 79, 77, 86, 74, 85, 91, 92, 117, 102, 99, 105, 110, 101, 114, 127, 128, 119, 125, 126, 153, 138, 160, 153, 157
Offset: 1

Views

Author

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

Keywords

Crossrefs

Cf. A050926.

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

A097583 Octal representation of the concatenation of the first n decimal numbers with the most significant digits first.

Original entry on oeis.org

1, 14, 173, 2322, 30071, 361100, 4553207, 57060516, 726746425, 133767016076, 21756176604103, 3404420603635070, 536705213574536755, 104420417226264430242, 15305164771273206577527
Offset: 1

Views

Author

Cino Hilliard, Aug 29 2004

Keywords

Examples

			1234 decimal is 2322 octal the 4th entry in the table.
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[IntegerDigits[FromDigits[Flatten[IntegerDigits/@ Range[ n]]],8]],{n,20}] (* Harvey P. Dale, Aug 11 2021 *)

Formula

a(n) = A007094(A007908(n)). - Seiichi Manyama, Apr 23 2022

A353110 Binary representation of A000422(n).

Original entry on oeis.org

1, 10101, 101000001, 1000011100001, 1101010000110001, 10011111101111110001, 11101001100101110110001, 101001110010111111110110001, 111010110111100110100010110001, 1010001110111010100100110010110001
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(2, 10)

Formula

a(n) = A007088(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 1-9 of 9 results.