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 1-10 of 10 results.

A138662 First term of the first run of at least n consecutive numbers in A137292.

Original entry on oeis.org

1, 3, 3, 469, 1333, 15984, 978131, 978131
Offset: 1

Views

Author

Klaus Brockhaus, Mar 25 2008

Keywords

Examples

			First few terms of A137292 are 1,3,4,5,7,9,10,13,... . The first run of at least two consecutive numbers is 3,4; the first run of at least three consecutive numbers is 3,4,5. Hence a(2) = a(3) = 3.
		

Crossrefs

Cf. A137292.

A138845 First gap of length at least n in A137292, lower end.

Original entry on oeis.org

1, 10, 115, 201, 1015, 1029, 10029, 100021
Offset: 1

Views

Author

Klaus Brockhaus, Mar 31 2008

Keywords

Examples

			First few terms of A137292 are 1,3,4,5,7,8,10,13,14,15,17,... . The first gap of length 1 is between 1 and 3, hence a(1) = 1; the first gap of length 2 is between 10 and 13, hence a(2) = 10.
		

Crossrefs

A136119 Limiting sequence when we start with the positive integers (A000027) and delete in step n >= 1 the term at position n + a(n).

Original entry on oeis.org

1, 3, 4, 5, 7, 8, 10, 11, 13, 14, 15, 17, 18, 20, 21, 22, 24, 25, 27, 28, 29, 31, 32, 34, 35, 37, 38, 39, 41, 42, 44, 45, 46, 48, 49, 51, 52, 54, 55, 56, 58, 59, 61, 62, 63, 65, 66, 68, 69, 71, 72, 73, 75, 76, 78, 79, 80, 82, 83, 85, 86, 87, 89, 90, 92, 93, 95, 96, 97, 99, 100
Offset: 1

Views

Author

Ctibor O. Zizka, Mar 16 2008

Keywords

Comments

Apparently a(n) = A001953(n-1)+1 = floor((n-1/2)*sqrt(2))+1 (confirmed for n < 20000) and a(n+1) - a(n) = A001030(n). From the definitions these conjectures are by no means obvious. Can they be proved? - Klaus Brockhaus, Apr 15 2008 [For an affirmative answer, see the Cloitre link.]
This is the s(n)-Wythoff sequence for s(n)=2n-1; see A184117 for the definition. Complement of A184119. - Clark Kimberling, Jan 09 2011

Examples

			First few steps are:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 1; delete term at position 1+a(1) = 2: 2;
1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 2; delete term at position 2+a(2) = 5: 6;
1,3,4,5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 3; delete term at position 3+a(3) = 7: 9;
1,3,4,5,7,8,10,11,12,13,14,15,16,17,18,19,20,...
n = 4; delete term at position 4+a(4) = 9: 12;
1,3,4,5,7,8,10,11,13,14,15,16,17,18,19,20,...
n = 5; delete term at position 5+a(5) = 12: 16;
1,3,4,5,7,8,10,11,13,14,15,17,18,19,20,...
n = 6; delete term at position 6+a(6) = 14: 19;
1,3,4,5,7,8,10,11,13,14,15,17,18,20,...
		

References

  • B. Cloitre, The golden sieve, preprint 2008

Crossrefs

Cf. A000027, A001953 (floor((n+1/2)*sqrt(2))), A001030 (fixed under 1 -> 21, 2 -> 211), A136110, A137292.
Cf. A242535.
Cf. A000217 (T).

Programs

  • Haskell
    import Data.List (delete)
    a136119 n = a136119_list !! (n-1)
    a136119_list = f [1..] where
       f zs@(y:xs) = y : f (delete (zs !! y) xs)
    -- Reinhard Zumkeller, May 17 2014
    
  • Magma
    [Ceiling((n-1/2)*Sqrt(2)): n in [1..100]]; // Vincenzo Librandi, Jul 01 2019
    
  • Mathematica
    f[0] = Range[100]; f[n_] := f[n] = Module[{pos = n + f[n-1][[n]]}, If[pos > Length[f[n-1]], f[n-1], Delete[f[n-1], pos]]]; f[1]; f[n = 2]; While[f[n] != f[n-1], n++]; f[n] (* Jean-François Alcover, May 08 2019 *)
    T[n_] := n (n + 1)/2; Table[1 + 2 Sqrt[T[n-1]] , {n, 1, 71}] // Floor (* Ralf Steiner, Oct 23 2019 *)
  • PARI
    apply( {A136119(n)=sqrtint(n*(n-1)*2)+1}, [1..99]) \\ M. F. Hasler, Jul 04 2022

Formula

a(n) = ceiling((n-1/2)*sqrt(2)). This can be proved in the same way as the formula given for A099267. There are some generalizations. For instance, it is possible to consider "a(n)+K*n" instead of "a(n)+n" for deleting terms where K=0,1,2,... is fixed. The constant involved in the Beatty sequence for the sequence of deleted terms then depends on K and equals (K + 1 + sqrt((K+1)^2 + 4))/2. K=0 is related to A099267. 1+A001954 is the complement sequence of this sequence A136119. - Benoit Cloitre, Apr 18 2008
a(n) = floor(1 + 2*sqrt(T(n-1))), with triangular numbers T(). - Ralf Steiner, Oct 23 2019
Lim_{n->inf}(a(n)/(n - 1)) = sqrt(2), with {a(n)/(n - 1)} decreasing. - Ralf Steiner, Oct 24 2019

Extensions

Edited and extended by Klaus Brockhaus, Apr 15 2008
An incorrect g.f. removed by Alois P. Heinz, Dec 14 2012

A136259 Stone skipping numbers.

Original entry on oeis.org

1, 3, 4, 5, 9, 13, 18, 19, 31, 32, 33, 38, 39, 55, 56, 57, 58, 59, 94, 95, 96, 97, 103, 104, 156, 157, 239, 244, 245, 249, 253, 254, 255, 256, 257, 258, 275, 276, 277, 419, 420, 609, 610, 787, 788, 789, 790, 791, 792, 1069, 1070, 1664, 1665, 1666, 1667, 1668, 1669, 1670
Offset: 1

Views

Author

Ctibor O. Zizka, Mar 18 2008

Keywords

Comments

The sequence is generated by a sieving method with iterated selection of intervals of the natural numbers as if they were forming a chain of contact points on which a stone could re-bounce once launched at some specific position at the small numbers.
Image a stone with an initial kinetic energy t, which is diminished/dissipated by 1 unit each time it rebounds from the "water surface" of the residual sequence; it rebounds t times and sinks once it has slowed down to t=1. The numbers underneath the arcs of this flight, but not the contact points, are eliminated. We look at the limit of repeatedly skipping stones each time starting at new launching points with larger initial t. In detail:
Start with the set of natural numbers. Let a(0)= t define t. Jump t positions to the right, erase t positions; from the last erased position jump t-1 positions to the right, erase t-1 positions; ...; jump 1 position to the right, erase 1 position. Go to the smallest i>t. Set t=i. Repeat.
Stone skipping sequences are a generalized case of scarce sequences; see A137292.

Examples

			Start with natural numbers
   1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,...
a(0)=1 set t=1 (jump 1 position to the right, erase 1 position) gives
   1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,...
i=3 set t=3 (jump 3 positions to the right, erase 3 positions; from the last erased position jump 2 positions to the right, erase 2 positions; from the last erased position jump 1 position to the right, erase 1 position) gives
   1,3,4,5,9,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,...
i=4 set t=4 (jump 4 positions to the right, erase 4 positions; from the last erased position jump 3 positions to the right, erase 3 positions; from the last erased position jump 2 positions to the right, erase 2 positions;from the last erased position jump 1 position to the right, erase 1 position ) gives
   1,3,4,5,9,13,18,19,23,27,28,...
i=5 set t=5, repeat procedure.
		

Crossrefs

Cf. A137292. Bisections are A238091, A238092.
Cf. A270877.

Programs

  • Maple
    nmax := 3000: a136259 := [seq(i,i=1..nmax)] : s := 1: t := op(s,a136259) : p := 1:
    while op(-1,a136259)>t do p := p+t ; outb := false; while t >= 1 do for eli from 1 to t do if p > nops(a136259) then outb := true; break; fi; a136259 := subsop(p=NULL,a136259) ; od: if outb then break; fi; t := t-1 ; p := p+t-1 ; od: print(a136259) ; s := s+1 ; p := s ; t := op(s,a136259) : od: # R. J. Mathar, Aug 17 2009

Extensions

Edited and corrected by R. J. Mathar, Aug 17 2009

A136272 Waterbird take-off sequence. Complement of A166021.

Original entry on oeis.org

1, 3, 6, 7, 11, 12, 13, 18, 19, 20, 21, 27, 28, 29, 30, 31, 38, 39, 40, 41, 42, 43, 51, 52, 53, 54, 55, 56, 57, 66, 67, 68, 69, 70, 71, 72, 73, 83, 84, 85, 86, 87, 88, 89, 90, 91, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 123, 124, 125, 126, 127, 128, 129
Offset: 1

Views

Author

Ctibor O. Zizka, Mar 19 2008

Keywords

Comments

This kind of sequence can be generalized as follows:
Let F(t), G(t) be arithmetic functions: F(t) the right hand move, G(t) the number of erased positions.
Then starting from the position t=1 do procedure:
JUMP F(t) positions right hand
ERASE G(t) positions
SET t=t+1
repeat procedure from the last erased position.
This sequence has F(t)=t, G(t)=t.
We can use a set of functions F_i(t) and G_i(t) processed in parallel (a flock of birds taking off).

Examples

			1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,...
t=1; from the position 1 go 1 position to the right, erase 1 position:
1..3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,...
t=2; from the last erased position go 2 positions to the right, erase 2 positions:
1..3..,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,...
t=3; from the last erased position go 3 positions to the right, erase 3 positions:
1..3..,6,7....11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,...
t=4; from the last erased position go 4 positions to the right, erase 4 positions:
1..3..,6,7....11,12,13....,18,19,20,21,22,23,24,25,26,27,...
t=5; from the last erased position go 5 positions to the right, erase 5 positions:
1..3..,6,7....11,12,13....,18,19,20,21......27,...
The erased positions form the complement of this sequence: A166021.
		

Crossrefs

Programs

Formula

a(0)=1; let t=1. Start on position t. Jump t positions right hand. Erase t positions. (*P*) Set t=t+1. Start on the last erased position. Jump t positions right hand. Erase t positions. Repeat procedure (*P*).

Extensions

Edited and corrected by Antti Karttunen, Oct 05 2009

A136120 Limiting sequence when we start with the positive integers (A000027) and at step n >= 1 delete the a(n) terms at positions n+a(n) to n-1+2*a(n).

Original entry on oeis.org

1, 3, 4, 5, 9, 10, 15, 16, 22, 23, 24, 25, 26, 36, 37, 48, 49, 50, 51, 52, 53, 69, 70, 87, 88, 89, 90, 91, 92, 93, 116, 117, 141, 142, 167, 168, 194, 195, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 269, 270, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317
Offset: 1

Views

Author

Ctibor O. Zizka, Mar 16 2008

Keywords

Examples

			First few steps are:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,...
n = 1, a(1) = 1; delete terms at positions 2 to 2; this is 2;
1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,...
n = 2,a(2) = 3; delete terms at positions 5 to 7; these are 6,7,8;
1,3,4,5,9,10,11,12,13,14,15,16,17,18,19,20,21,22,...
n = 3, a(3) = 4; delete terms at positions 7 to 10; these are 11,12,13,14;
1,3,4,5,9,10,15,16,17,18,19,20,21,22,23,24,25,26,27,28,...
n = 4, a(4) = 5; delete terms at positions 9 to 13; these are 17,18,19,20,21;
1,3,4,5,9,10,15,16,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36...
n = 5 a(5) = 9; delete terms at positions 14 to 22; these are 27,28,29,30,31,32,33,34,35;
1,3,4,5,9,10,15,16,22,23,24,25,26,36,...
		

Crossrefs

Programs

  • Mathematica
    f[seq_] := Module[{s = seq, n1, n2}, n++; n1 = s[[n]] + n; If[n1 <= len, n2 = Min[n - 1 + 2*s[[n]], len]; len -= n2 - n1 + 1; Drop[s, {n1, n2}], s]]; n = 0; len = 1000; FixedPoint[f, Range[len]] (* Jean-François Alcover, Sep 29 2011 *)

Extensions

Edited and extended by Klaus Brockhaus, Apr 20 2008

A136110 Limiting sequence when we start with the positive integers (A000027) and delete in step n >= 1 the term at position n + tau(a(n)), where tau(k) is the number of divisors of k.

Original entry on oeis.org

1, 3, 4, 6, 7, 9, 12, 13, 15, 17, 18, 22, 23, 24, 28, 29, 30, 32, 33, 36, 37, 38, 43, 44, 47, 49, 51, 52, 55, 56, 58, 59, 62, 65, 66, 68, 70, 72, 73, 74, 78, 79, 80, 84, 85, 86, 88, 90, 92, 94, 96, 97, 98, 104, 105, 106, 108, 109, 111, 116, 118, 119, 121, 122, 126, 129, 130
Offset: 1

Views

Author

Ctibor O. Zizka, Mar 16 2008

Keywords

Examples

			First few steps are:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 1; delete term at position 1+tau(1) = 1+1 =2: 2;
1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 2; delete term at position 2+tau(3) = 1+2 = 3: 5;
1,3,4,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 3; delete term at position 3+tau(4) = 3+3 = 6: 8;
1,3,4,6,7,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 4; delete term at position 4+tau(6) = 4+4 = 8: 11;
1,3,4,6,7,9,10,12,13,14,15,16,17,18,19,20,...
n = 5; delete term at position 5+tau(7) = 5+2 = 7: 10;
1,3,4,6,7,9,12,13,14,15,16,17,18,19,20,...
n = 6; delete term at position 6+tau(9) = 6+3 = 9: 14;
1,3,4,6,7,9,12,13,15,16,17,18,19,20,...
		

Crossrefs

Cf. A000005 (number of divisors), A000027, A137292, A138899, A138900.

Extensions

Edited and extended by Klaus Brockhaus, Apr 03 2008
Moved references to the Links section R. J. Mathar, Oct 23 2009

A137832 Limiting sequence when we start with the positive integers (A000027) and delete in step n >= 1 the last digit in the term at position n + a(n).

Original entry on oeis.org

1, 3, 4, 5, 7, 8, 10, 11, 1, 1, 1, 16, 1, 1, 1, 21, 2, 23, 2, 25, 2, 27, 2, 29, 3, 31, 3, 3, 34, 3, 3, 37, 3, 3, 40, 4, 43, 44, 45, 47, 48, 49, 5, 51, 52, 53, 5, 5, 56, 57, 5, 5, 6, 61, 6, 6, 6, 65, 6, 67, 69, 7, 7, 72, 7, 7, 75, 76, 7, 7, 79, 8, 82, 83, 84, 8
Offset: 1

Views

Author

Ctibor O. Zizka, Apr 29 2008

Keywords

Examples

			First few steps are:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 1; delete the last digit in the term at position 1+a(1) = 2: 2;
1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 2; delete the last digit in the term at position 2+a(2) = 5: 6;
1,3,4,5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 3; delete the last digit in the term at position 3+a(3) = 7: 9;
1,3,4,5,7,8,10,11,12,13,14,15,16,17,18,19,20,...
n = 4; delete the last digit in the term at position 4+a(4) = 9: 2;
1,3,4,5,7,8,10,11,1,13,14,15,16,17,18,19,20,...
n = 5; delete the last digit in the term at position 5+a(5) = 12: 5;
1,3,4,5,7,8,10,11,1,13,14,1,16,17,18,19,20,...
n = 6; delete the last digit in the term at position 6+a(6) = 14: 8;
1,3,4,5,7,8,10,11,1,13,14,1,16,17,1,19,20,...
		

Crossrefs

A137838 Limiting sequence when we start with the positive integers (A000027) and delete in step n >= 1 the first digit in the term at position n + a(n).

Original entry on oeis.org

1, 3, 4, 5, 7, 8, 10, 11, 2, 13, 4, 5, 16, 7, 8, 19, 21, 2, 23, 4, 5, 6, 7, 8, 29, 0
Offset: 1

Views

Author

Ctibor O. Zizka, Apr 29 2008

Keywords

Examples

			First few steps are:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 1; delete the first digit in the term at position 1+a(1) = 2: 2;
1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 2; delete the first digit in the term at position 2+a(2) = 5: 6;
1,3,4,5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
n = 3; delete the first digit in the term at position 3+a(3) = 7: 9;
1,3,4,5,7,8,10,11,12,13,14,15,16,17,18,19,20,...
n = 4; delete the first digit in the term at position 4+a(4) = 9: 1;
1,3,4,5,7,8,10,11,2,13,14,15,16,17,18,19,20,...
n = 5; delete the first digit in the term at position 5+a(5) = 12: 1;
1,3,4,5,7,8,10,11,2,13,14,5,16,17,18,19,20,...
n = 6; delete the first digit in the term at position 6+a(6) = 14: 1;
1,3,4,5,7,8,10,11,2,13,14,5,16,7,18,19,20,...
		

Crossrefs

A138095 Limiting sequence when we start with positive integers (A000027) and at step n >= 1 add: to the term at position n + 1*a(n) the value 1, to the term at position n + 2*a(n) the value 1, ..., to the term at position n + n*a(n) the value 1.

Original entry on oeis.org

1, 3, 3, 4, 6, 7, 7, 10, 10, 10, 12, 14, 14, 15, 15, 17, 18, 19, 20, 23, 22, 22, 25, 24, 25, 27, 29, 30, 32, 32, 31, 32, 34, 35, 39, 36, 38, 39, 41, 42, 43, 43, 45, 46, 46, 46, 48, 52, 51, 53, 51, 52, 55, 55, 56, 59, 57, 60, 63, 62, 62, 64, 63, 65, 66, 68, 69, 70, 72, 71
Offset: 1

Views

Author

Ctibor O. Zizka, May 03 2008

Keywords

Examples

			        n | 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
  --------+-----------------------------------------------------------
  Step  1 |    3
  Step  2 |             6        9
  Step  3 |                7       10       13
  Step  4 |                     10          14          17          21
  Step  5 |                              12                18
  Step  6 |                                    14                   22
  Step  7 |                                       15
  Step  8 |                                                   19
  Step  9 |                                                      20
  Step 10 |                                                         23
  --------+-----------------------------------------------------------
     a(n) | 1  3  3  4  6  7  7 10 10 10 12 14 14 15 15 17 18 19 20 23
		

Crossrefs

Programs

  • PARI
    lista(nn) = my(va = [1..nn]); for (n=1, nn, for (k=1, n, my(j = n+k*va[n]); if (j <= #va, va[j]++); )); va; \\ Michel Marcus, Aug 09 2022

Extensions

Corrected and extended by Michel Marcus, Aug 09 2022
Showing 1-10 of 10 results.