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 1-5 of 5 results.

A105035 Positions of record values in A104234.

Original entry on oeis.org

0, 1, 5, 2037
Offset: 1

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Author

N. J. A. Sloane, Apr 04 2005

Keywords

Comments

Is this 2^g(n-1) - g(n-1), where g = A034797?
Yes, this formula can be proved by induction on n. - Max Alekseyev, Mar 16 2023
a(5) = 2^2059 - 2059 with 620 decimal digits is too large to be included here.

Links

  • David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].

Formula

a(n) = 2^A034797(n-1) - A034797(n-1). - Max Alekseyev, Mar 16 2023

Extensions

Offset corrected by Max Alekseyev, Mar 17 2023

A102370 "Sloping binary numbers": write numbers in binary under each other (right-justified), read diagonals in upward direction, convert to decimal.

Original entry on oeis.org

0, 3, 6, 5, 4, 15, 10, 9, 8, 11, 14, 13, 28, 23, 18, 17, 16, 19, 22, 21, 20, 31, 26, 25, 24, 27, 30, 61, 44, 39, 34, 33, 32, 35, 38, 37, 36, 47, 42, 41, 40, 43, 46, 45, 60, 55, 50, 49, 48, 51, 54, 53, 52, 63, 58, 57, 56, 59, 126, 93, 76, 71, 66, 65, 64, 67, 70, 69
Offset: 0

Views

Author

Philippe Deléham, Feb 13 2005

Keywords

Comments

All terms are distinct, but certain terms (see A102371) are missing. But see A103122.
Trajectory of 1 is 1, 3, 5, 15, 17, 19, 21, 31, 33, ..., see A103192.

Examples

			........0
........1
.......10
.......11
......100
......101
......110
......111
.....1000
.........
The upward-sloping diagonals are:
0
11
110
101
100
1111
1010
.......
giving 0, 3, 6, 5, 4, 15, 10, ...
The sequence has a natural decomposition into blocks (see the paper): 0; 3; 6, 5, 4; 15, 10, 9, 8, 11, 14, 13; 28, 23, 18, 17, 16, 19, 22, 21, 20, 31, 26, 25, 24, 27, 30; 61, ...
Reading the array of binary numbers along diagonals with slope 1 gives this sequence, slope 2 gives A105085, slope 0 gives A001477 and slope -1 gives A105033.

Links

Crossrefs

Related sequences (1): A103542 (binary version), A102371 (complement), A103185, A103528, A103529, A103530, A103318, A034797, A103543, A103581, A103582, A103583.
Related sequences (2): A103584, A103585, A103586, A103587, A103127, A103192 (trajectory of 1), A103122, A103588, A103589, A103202 (sorted), A103205 (base 10 version).
Related sequences (3): A103747 (trajectory of 2), A103621, A103745, A103615, A103842, A103863, A104234, A104235, A103813, A105023, A105024, A105025, A105026, A105027, A105028.
Related sequences (4): A105029, A105030, A105031, A105032, A105033, A105034, A105035, A105108.
Related sequences (5): A105229, A105271, A104378, A104401, A104403, A104489, A104490, A104853, A104893, A104894, A105085.

Programs

  • Haskell
    a102370 n = a102370_list !! n
a102370_list = 0 : map (a105027 . toInteger) a062289_list
-- Reinhard Zumkeller, Jul 21 2012
  • Maple
    A102370:=proc(n) local t1,l; t1:=n; for l from 1 to n do if n+l mod 2^l = 0 then t1:=t1+2^l; fi; od: t1; end;
  • Mathematica
    f[n_] := Block[{k = 1, s = 0, l = Max[2, Floor[Log[2, n + 1] + 2]]}, While[k < l, If[ Mod[n + k, 2^k] == 0, s = s + 2^k]; k++ ]; s]; Table[ f[n] + n, {n, 0, 71}] (* Robert G. Wilson v, Mar 21 2005 *)
  • PARI
    A102370(n)=n-1+sum(k=0,ceil(log(n+1)/log(2)),if((n+k)%2^k,0,2^k)) \\ Benoit Cloitre, Mar 20 2005
  • PARI
    {a(n) = if( n<1, 0, sum( k=0, length( binary( n)), bitand( n + k, 2^k)))} /* Michael Somos, Mar 26 2012 */
  • Python
    def a(n): return 0 if n<1 else sum([(n + k)&(2**k) for k in range(len(bin(n)[2:]) + 1)]) # Indranil Ghosh, May 03 2017

Formula

a(n) = n + Sum_{ k >= 1 such that n + k == 0 mod 2^k } 2^k. (Cf. A103185.) In particular, a(n) >= n. - N. J. A. Sloane, Mar 18 2005
a(n) = A105027(A062289(n)) for n > 0. - Reinhard Zumkeller, Jul 21 2012

Extensions

More terms from Benoit Cloitre, Mar 20 2005

A103318 Number of solutions i in range [0,n-1] to i == 0 mod 2^(n-i).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2, 3, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2, 3, 3, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3
Offset: 1

Views

Author

N. J. A. Sloane, Mar 21 2005

Keywords

Comments

i=0 is always a solution.
a(n) is the number of 1's in (A103745(n) written in base 2). - Philippe Deléham, Apr 02 2005

Examples

			For n = 11 solutions are i = 0, 8 and 10. Four solutions occur for the first time at n = 2059: they are i = 0, 2048, 2056, 2058. Five solutions occur for the first time at n = 2^2059 + 2059 (see A034797).

Links

  • David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].

Crossrefs

For records see A034797. Cf. A103745.

Programs

  • Maple
    f:= proc (n) local t1, l; t1 := 0; for l to n do if `mod`(n-l,2^l) = 0 then t1 := t1+1 end if end do; t1 end proc;
  • Mathematica
    f[n_] := Block[{c = 1, k = Max[1, n - Floor[ Log[2, n] + 2]]}, While[k < n, If[ Mod[k, 2^(n - k)] == 0, c++ ]; k++ ]; c]; Table[ f[n], {n, 105}] (* Robert G. Wilson v, Mar 21 2005 *)

Formula

a(n) = A104234(2^n - n). - Philippe Deléham, Apr 21 2005

A105031 Binary equivalents of A103185.

Original entry on oeis.org

0, 1, 10, 1, 0, 101, 10, 1, 0, 1, 10, 1, 1000, 101, 10, 1, 0, 1, 10, 1, 0, 101, 10, 1, 0, 1, 10, 10001, 1000, 101, 10, 1, 0, 1, 10, 1, 0, 101, 10, 1, 0, 1, 10, 1, 1000, 101, 10, 1, 0, 1, 10, 1, 0, 101, 10, 1, 0, 1, 100010, 10001, 1000, 101, 10, 1, 0, 1, 10, 1, 0, 101, 10, 1, 0, 1, 10
Offset: 0

Views

Author

N. J. A. Sloane, David Applegate, Benoit Cloitre and Philippe Deléham, Apr 03 2005

Keywords

Links

Crossrefs

Cf. A102370, A104234, A105032.

Extensions

More terms from Benoit Cloitre, Apr 04 2005

A289281 Square array whose rows m >= 2 hold the limit under iterations of the morphism { x -> (x, ..., x+k-1) if k|x ; x -> x+1 otherwise }, starting with (0); read by falling antidiagonals.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 2, 2, 1, 0, 3, 2, 2, 1, 0, 2, 3, 3, 2, 1, 0, 3, 3, 2, 3, 2, 1, 0, 4, 3, 3, 4, 3, 2, 1, 0, 2, 4, 4, 2, 4, 3, 2, 1, 0, 3, 5, 3, 3, 5, 4, 3, 2, 1, 0, 4, 3, 4, 4, 2, 5, 4, 3, 2, 1, 0, 4, 4, 4, 5, 3, 6, 5, 4, 3, 2, 1, 0, 5, 5, 5, 3, 4, 2, 6, 5, 4, 3, 2, 1, 0, 2, 3, 6, 4, 5, 3, 7, 6, 5, 4, 3, 2, 1, 0, 3, 4, 7, 5, 6, 4, 2, 7, 6, 5, 4, 3, 2, 1, 0, 4, 5, 4, 5, 3, 5, 3, 8, 7, 6, 5, 4
Offset: 2

Views

Author

M. F. Hasler, Jul 01 2017

Keywords

Comments

The generalization of A104234 (row 2) and A288577 (row 3) to arbitrary m.

Examples

			The array starts (first row: m=2)
  [ 0 1 2 2 3 2 3 4 2 3  4  4  5  2  3  4  4  5  4  5  6  2  3  4  4 ...]
  [ 0 1 2 2 3 3 3 4 5 3  4  5  3  4  5  5  6  3  4  5  5  6  3  4  5 ...]
  [ 0 1 2 3 2 3 4 3 4 4  5  6  7  4  4  5  6  7  4  5  6  7  6  7  8 ...]
  [ 0 1 2 3 4 2 3 4 5 3  4  5  5  6  7  8  9  4  5  5  6  7  8  9  5 ...]
  [ 0 1 2 3 4 5 2 3 4 5  6  3  4  5  6  6  7  8  9 10 11  4  5  6  6 ...]
  [ 0 1 2 3 4 5 6 2 3 4  5  6  7  3  4  5  6  7  7  8  9 10 11 12 13 ...]
  [ 0 1 2 3 4 5 6 7 2 3  4  5  6  7  8  3  4  5  6  7  8  8  9 10 11 ...]
  [ 0 1 2 3 4 5 6 7 8 2  3  4  5  6  7  8  9  3  4  5  6  7  8  9  9 ...]
  [ 0 1 2 3 4 5 6 7 8 9  2  3  4  5  6  7  8  9 10  3  4  5  6  7  8 ...]
  [ 0 1 2 3 4 5 6 7 8 9 10  2  3  4  5  6  7  8  9 10 11  3  4  5  6 ...]
  [ 0 1 2 3 4 5 6 7 8 9 10 11  2  3  4  5  6  7  8  9 10 11 12  3  4 ...]
  [ 0 1 2 3 4 5 6 7 8 9 10 11 12  2  3  4  5  6  7  8  9 10 11 12 13 ...]
  ...
It is easy to prove that row m starts with (0, ..., m-1; 2, ..., m; 3, ..., m; m, ..., 2m-1; ...).

Links

Crossrefs

Cf. A104234 (row 2), A288577 (row 3).

Programs

  • PARI
    A289281_row(n=30,k=2,a=[0])={while(#a
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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