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.

A279075 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/5) requires n steps to reach 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 11, 14, 18, 23, 29, 37, 47, 59, 74, 93, 117, 147, 184, 231, 289, 362, 453, 567, 709, 887, 1109, 1387, 1734, 2168, 2711, 3389, 4237, 5297, 6622, 8278, 10348, 12936, 16171, 20214, 25268, 31586, 39483, 49354, 61693, 77117, 96397, 120497
Offset: 0

Views

Author

Jon E. Schoenfield, Dec 06 2016

Keywords

Comments

Inspired by A278586.
Limit_{n->oo} a(n)/(5/4)^n = 2.68723058270145442816383476567331957329199286146873...

Examples

			8 -> 8-ceiling(8/5) = 6,
6 -> 6-ceiling(6/5) = 4,
4 -> 4-ceiling(4/5) = 3,
3 -> 3-ceiling(3/5) = 2,
2 -> 2-ceiling(2/5) = 1,
1 -> 1-ceiling(1/5) = 0,
so reaching 0 from 8 requires 6 steps;
9 -> 9-ceiling(9/5) = 7,
7 -> 7-ceiling(7/5) = 5,
5 -> 5-ceiling(5/5) = 4,
4 -> 4-ceiling(4/5) = 3,
3 -> 3-ceiling(3/5) = 2,
2 -> 2-ceiling(2/5) = 1,
1 -> 1-ceiling(1/5) = 0,
so reaching 0 from 9 (or more) requires 7 (or more) steps;
thus, 8 is the largest starting value from which 0 can be reached in 6 steps, so a(6) = 8.

Links

Crossrefs

Cf. A278586.
See the following sequences for maximum starting value of X such that repeated replacement of X with X-ceiling(X/k) requires n steps to reach 0: A000225 (k=2), A006999 (k=3), A155167 (k=4, apparently; see Formula entry there), (this sequence) (k=5), A279076 (k=6), A279077 (k=7), A279078 (k=8), A279079 (k=9), A279080 (k=10). For each of these values of k, is the sequence the L-sieve transform of {k-1, 2k-1, 3k-1, ...}?

Programs

  • Magma
    a:=[0]; aCurr:=0; for n in [1..48] do aCurr:=Floor(aCurr*5/4)+1; a[#a+1]:=aCurr; end for; a;
  • Magma
    [n eq 1 select n-1 else Floor(Self(n-1)*5/4)+1: n in [1..70]]; // Vincenzo Librandi, Dec 06 2016
  • Mathematica
    RecurrenceTable[{a[1] == 0, a[n] == Floor[a[n-1] 5/4] + 1}, a, {n, 50}] (* Vincenzo Librandi, Dec 06 2016 *)

Formula

a(n) = floor(a(n-1)*5/4) + 1.

A279080 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/10) requires n steps to reach 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 19, 22, 25, 28, 32, 36, 41, 46, 52, 58, 65, 73, 82, 92, 103, 115, 128, 143, 159, 177, 197, 219, 244, 272, 303, 337, 375, 417, 464, 516, 574, 638, 709, 788, 876, 974, 1083, 1204, 1338, 1487, 1653, 1837, 2042, 2269
Offset: 0

Views

Author

Jon E. Schoenfield, Dec 06 2016

Keywords

Comments

Inspired by A278586.
Limit_{n->oo} a(n)/(10/9)^n = 5.60655601136196116133057876294687807265035051745268...

Examples

			  13 -> 13-ceiling(13/10) = 11,
  11 -> 11-ceiling(11/10) = 9,
   9 ->  9-ceiling(9/10)  = 8,
   8 ->  8-ceiling(8/10)  = 7,
...
   1 ->  1-ceiling(1/10)  = 0,
so reaching 0 from 13 requires 11 steps;
  14 -> 14-ceiling(14/10) = 12,
  12 -> 12-ceiling(12/10) = 10,
  10 -> 10-ceiling(10/10) = 9,
   9 ->  9-ceiling(9/10)  = 8,
   8 ->  8-ceiling(8/10)  = 7,
...
   1 ->  1-ceiling(1/10)  = 0,
so reaching 0 from 14 (or more) requires 12 (or more) steps;
thus, 13 is the largest starting value from which 0 can be reached in 11 steps, so a(11) = 13.

Links

Crossrefs

Cf. A278586.
See the following sequences for maximum starting value of X such that repeated replacement of X with X-ceiling(X/k) requires n steps to reach 0: A000225 (k=2), A006999 (k=3), A155167 (k=4, apparently; see Formula entry there), A279075 (k=5), A279076 (k=6), A279077 (k=7), A279078 (k=8), A279079 (k=9), (this sequence) (k=10). For each of these values of k, is the sequence the L-sieve transform of {k-1, 2k-1, 3k-1, ...}?

Programs

  • Magma
    a:=[0]; aCurr:=0; for n in [1..57] do aCurr:=Floor(aCurr*10/9)+1; a[#a+1]:=aCurr; end for; a;
  • Maple
    H:= proc(y) local u,v;
     v:= -y-1 mod 9+1;
     (10*y+v)/9
end proc:
A:= Array(0..100):
A[0]:= 0:
for i from 1 to 100 do A[i]:= H(A[i-1]) od:
convert(A,list); # Robert Israel, Jun 23 2020
  • Mathematica
    With[{s = Array[-1 + Length@ NestWhileList[# - Ceiling[#/10] &, #, # > 0 &] &, 2400, 0]}, Array[-1 + Position[s, #][[-1, 1]] &, Max@ s, 0]] (* Michael De Vlieger, Jun 23 2020 *)

Formula

a(n) = floor(a(n-1)*10/9) + 1.

A279076 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/6) requires n steps to reach 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 9, 11, 14, 17, 21, 26, 32, 39, 47, 57, 69, 83, 100, 121, 146, 176, 212, 255, 307, 369, 443, 532, 639, 767, 921, 1106, 1328, 1594, 1913, 2296, 2756, 3308, 3970, 4765, 5719, 6863, 8236, 9884, 11861, 14234, 17081, 20498, 24598, 29518, 35422
Offset: 0

Views

Author

Jon E. Schoenfield, Dec 06 2016

Keywords

Comments

Inspired by A278586.
Limit_{n->oo} a(n)/(6/5)^n = 3.24387249751177521384734853905517802618171089570674...

Examples

			7 -> 7-ceiling(7/6) = 5,
5 -> 5-ceiling(5/6) = 4,
4 -> 4-ceiling(4/6) = 3,
3 -> 3-ceiling(3/6) = 2,
2 -> 2-ceiling(2/6) = 1,
1 -> 1-ceiling(1/6) = 0,
so reaching 0 from 7 requires 6 steps;
8 -> 8-ceiling(8/6) = 6,
6 -> 6-ceiling(6/6) = 5,
5 -> 5-ceiling(5/6) = 4,
4 -> 4-ceiling(4/6) = 3,
3 -> 3-ceiling(3/6) = 2,
2 -> 2-ceiling(2/6) = 1,
1 -> 1-ceiling(1/6) = 0,
so reaching 0 from 8 (or more) requires 7 (or more) steps;
thus, 7 is the largest starting value from which 0 can be reached in 6 steps, so a(6) = 7.

Links

Crossrefs

Cf. A278586.
See the following sequences for maximum starting value of X such that repeated replacement of X with X-ceiling(X/k) requires n steps to reach 0: A000225 (k=2), A006999 (k=3), A155167 (k=4, apparently; see Formula entry there), A279075 (k=5), (this sequence) (k=6), A279077 (k=7), A279078 (k=8), A279079 (k=9), A279080 (k=10). For each of these values of k, is the sequence the L-sieve transform of {k-1, 2k-1, 3k-1, ...}?

Programs

  • Magma
    a:=[0]; aCurr:=0; for n in [1..51] do aCurr:=Floor(aCurr*6/5)+1; a[#a+1]:=aCurr; end for; a;

Formula

a(n) = floor(a(n-1)*6/5) + 1.

A279077 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/7) requires n steps to reach 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 22, 26, 31, 37, 44, 52, 61, 72, 85, 100, 117, 137, 160, 187, 219, 256, 299, 349, 408, 477, 557, 650, 759, 886, 1034, 1207, 1409, 1644, 1919, 2239, 2613, 3049, 3558, 4152, 4845, 5653, 6596, 7696, 8979, 10476, 12223, 14261
Offset: 0

Views

Author

Jon E. Schoenfield, Dec 06 2016

Keywords

Comments

Inspired by A278586.
Limit_{n->oo} a(n)/(7/6)^n = 4.03710211215303193642791458111196922950551168987041...

Examples

			  10 -> 10-ceiling(10/7) = 8,
   8 ->  8-ceiling(8/7)  = 6,
   6 ->  6-ceiling(6/7)  = 5,
   5 ->  5-ceiling(5/7)  = 4,
   4 ->  4-ceiling(4/7)  = 3,
   3 ->  3-ceiling(3/7)  = 2,
   2 ->  2-ceiling(2/7)  = 1,
   1 ->  1-ceiling(1/7)  = 0,
so reaching 0 from 10 requires 8 steps;
  11 -> 11-ceiling(11/7) = 9,
   9 ->  9-ceiling(9/7)  = 7,
   7 ->  7-ceiling(7/7)  = 6,
   6 ->  6-ceiling(6/7)  = 5,
   5 ->  5-ceiling(5/7)  = 4,
   4 ->  4-ceiling(4/7)  = 3,
   3 ->  3-ceiling(3/7)  = 2,
   2 ->  2-ceiling(2/7)  = 1,
   1 ->  1-ceiling(1/7)  = 0,
so reaching 0 from 11 (or more) requires 9 (or more) steps;
thus, 10 is the largest starting value from which 0 can be reached in 8 steps, so a(8) = 10.

Crossrefs

Cf. A278586.
See the following sequences for maximum starting value of X such that repeated replacement of X with X-ceiling(X/k) requires n steps to reach 0: A000225 (k=2), A006999 (k=3), A155167 (k=4, apparently; see Formula entry there), A279075 (k=5), A279076 (k=6), (this sequence) (k=7), A279078 (k=8), A279079 (k=9), A279080 (k=10). For each of these values of k, is the sequence the L-sieve transform of {k-1, 2k-1, 3k-1, ...}?

Programs

  • Magma
    a:=[0]; aCurr:=0; for n in [1..53] do aCurr:=Floor(aCurr*7/6)+1; a[#a+1]:=aCurr; end for; a;

Formula

a(n) = floor(a(n-1)*7/6) + 1.

A279078 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/8) requires n steps to reach 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 15, 18, 21, 25, 29, 34, 39, 45, 52, 60, 69, 79, 91, 105, 121, 139, 159, 182, 209, 239, 274, 314, 359, 411, 470, 538, 615, 703, 804, 919, 1051, 1202, 1374, 1571, 1796, 2053, 2347, 2683, 3067, 3506, 4007, 4580, 5235, 5983, 6838
Offset: 0

Views

Author

Jon E. Schoenfield, Dec 06 2016

Keywords

Comments

Inspired by A278586.
Limit_{n->oo} a(n)/(8/7)^n = 4.42210347959393228709604412445802201220907917744900...

Examples

			  11 -> 11-ceiling(11/8) = 9,
   9 ->  9-ceiling(9/8)  = 7,
   7 ->  7-ceiling(7/8)  = 6,
   6 ->  6-ceiling(6/8)  = 5,
...
   1 ->  1-ceiling(1/8)  = 0,
so reaching 0 from 11 requires 9 steps;
  12 -> 12-ceiling(12/8) = 10,
  10 -> 10-ceiling(10/8) = 8,
   8 ->  8-ceiling(8/8)  = 7,
   7 ->  7-ceiling(7/8)  = 6,
...
   1 ->  1-ceiling(1/8)  = 0,
so reaching 0 from 12 (or more) requires 10 (or more) steps;
thus, 11 is the largest starting value from which 0 can be reached in 9 steps, so a(9) = 11.

Crossrefs

Cf. A278586.
See the following sequences for maximum starting value of X such that repeated replacement of X with X-ceiling(X/k) requires n steps to reach 0: A000225 (k=2), A006999 (k=3), A155167 (k=4, apparently; see Formula entry there), A279075 (k=5), A279076 (k=6), A279077 (k=7), (this sequence) (k=8), A279079 (k=9), A279080 (k=10). For each of these values of k, is the sequence the L-sieve transform of {k-1, 2k-1, 3k-1, ...}?

Programs

  • Magma
    a:=[0]; aCurr:=0; for n in [1..55] do aCurr:=Floor(aCurr*8/7)+1; a[#a+1]:=aCurr; end for; a;

Formula

a(n) = floor(a(n-1)*8/7) + 1.
Showing 1-5 of 5 results.

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

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