A155167 -id:A155167 - cp's OEIS frontend

A191670 Dispersion of A042968 (>1 and congruent to 1 or 2 or 3 mod 4), by antidiagonals.

1, 2, 4, 3, 6, 8, 5, 9, 11, 12, 7, 13, 15, 17, 16, 10, 18, 21, 23, 22, 20, 14, 25, 29, 31, 30, 27, 24, 19, 34, 39, 42, 41, 37, 33, 28, 26, 46, 53, 57, 55, 50, 45, 38, 32, 35, 62, 71, 77, 74, 67, 61, 51, 43, 36, 47, 83, 95, 103, 99, 90, 82, 69, 58, 49, 40, 63

Offset: 1

Clark Kimberling, Jun 11 2011

For a background discussion of dispersions, see A191426.
...
Each of the sequences (4n, n>2), (4n+1, n>0), (3n+2, n>=0), generates a dispersion. Each complement (beginning with its first term >1) also generates a dispersion. The six sequences and dispersions are listed here:
...
A191452=dispersion of A008586 (4k, k>=1)
A191667=dispersion of A016813 (4k+1, k>=1)
A191668=dispersion of A016825 (4k+2, k>=0)
A191669=dispersion of A004767 (4k+3, k>=0)
A191670=dispersion of A042968 (1 or 2 or 3 mod 4 and >=2)
A191671=dispersion of A004772 (0 or 1 or 3 mod 4 and >=2)
A191672=dispersion of A004773 (0 or 1 or 2 mod 4 and >=2)
A191673=dispersion of A004773 (0 or 2 or 3 mod 4 and >=2)
...
EXCEPT for at most 2 initial terms (so that column 1 always starts with 1):
A191452 has 1st col A042968, all else A008486
A191667 has 1st col A004772, all else A016813
A191668 has 1st col A042965, all else A016825
A191669 has 1st col A004773, all else A004767
A191670 has 1st col A008486, all else A042968
A191671 has 1st col A016813, all else A004772
A191672 has 1st col A016825, all else A042965
A191673 has 1st col A004767, all else A004773
...
Regarding the dispersions A191670-A191673, there is a formula for sequences of the type "(a or b or c mod m)", (as in the Mathematica program below):
   If f(n)=(n mod 3), then (a,b,c,a,b,c,a,b,c,...) is given by
   a*f(n+2)+b*f(n+1)+c*f(n), so that "(a or b or c mod m)" is given by
   a*f(n+2)+b*f(n+1)+c*f(n)+m*floor((n-1)/3)), for n>=1.

			Northwest corner:
1....2....3....5....7
4....6....9....13...18
8....11...15...21...29
12...17...23...31...42
16...22...30...41...55

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Row 1: A155167, Row 2: A171861.

Cf. A008486, A042968, A191673, A191452.

(* Program generates the dispersion array T of the increasing sequence f[n] *)
r = 40; r1 = 12; c = 40; c1 = 12;
a = 2; b = 3; c2 = 5; m[n_] := If[Mod[n, 3] == 0, 1, 0];
f[n_] := a*m[n + 2] + b*m[n + 1] + c2*m[n] + 4*Floor[(n - 1)/3]
Table[f[n], {n, 1, 30}] (* A042968 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]] (* A191670 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191670 *)

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

Jon E. Schoenfield, Dec 06 2016

nonn

Inspired by A278586.

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

			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.

A.H.M. Smeets, Constants related to repeated replacement of X with X-ceiling(X/d)

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, ...}?

a:=[0]; aCurr:=0; for n in [1..48] do aCurr:=Floor(aCurr*5/4)+1; a[#a+1]:=aCurr; end for; a;

[n eq 1 select n-1 else Floor(Self(n-1)*5/4)+1: n in [1..70]]; // Vincenzo Librandi, Dec 06 2016

RecurrenceTable[{a[1] == 0, a[n] == Floor[a[n-1] 5/4] + 1}, a, {n, 50}] (* Vincenzo Librandi, Dec 06 2016 *)

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

Jon E. Schoenfield, Dec 06 2016

nonn

Inspired by A278586.

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

			  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.

Robert Israel, Table of n, a(n) for n = 0..10000

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, ...}?

a:=[0]; aCurr:=0; for n in [1..57] do aCurr:=Floor(aCurr*10/9)+1; a[#a+1]:=aCurr; end for; a;

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

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 *)

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

A087192 a(n) = ceiling(a(n-1)*4/3), with a(1) = 1.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 11, 15, 20, 27, 36, 48, 64, 86, 115, 154, 206, 275, 367, 490, 654, 872, 1163, 1551, 2068, 2758, 3678, 4904, 6539, 8719, 11626, 15502, 20670, 27560, 36747, 48996, 65328, 87104, 116139, 154852, 206470, 275294, 367059, 489412, 652550

Offset: 1

Paul D. Hanna, Aug 24 2003

nonn

If you repeatedly base 64 encode a string, starting with a single character, the length of the string at step n is 4*a(n). - Christian Perfect, Jan 06 2016

Robert Israel, Table of n, a(n) for n = 1..7994

Cf. A072493, A087165, A155167.

[n eq 1 select 1 else Ceiling(Self(n-1)*4/3): n in [1..50]]; // Vincenzo Librandi, Aug 17 2017

A[1]:= 1:
for n from 2 to 100 do A[n]:= ceil(4/3*A[n-1]) od:
seq(A[i],i=1..100); # Robert Israel, Aug 17 2017

a[1] = 1; a[n_] := a[n] = Ceiling[4 a[n - 1]/3]; Table[a@ n, {n, 45}] (* Michael De Vlieger, Jan 06 2016 *)

a(n) = if (n==1, 1, ceil(a(n-1)*4/3)) \\ Michel Marcus, Aug 01 2013

from fractions import Fraction
from functools import lru_cache
@lru_cache(maxsize=None)
def A087192(n): return int(Fraction(4*A087192(n-1),3)._ceil_()) if n>1 else 1 # Chai Wah Wu, Sep 07 2023

Partial sums of A072493. Also indices of records in A087165: A087165(a(n))=n.

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

Jon E. Schoenfield, Dec 06 2016

nonn

Inspired by A278586.

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

			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.

A.H.M. Smeets, Constants related to repeated replacement of X with X-ceiling(X/d)

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, ...}?

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

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

Jon E. Schoenfield, Dec 06 2016

nonn

Inspired by A278586.

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

			  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.

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, ...}?

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

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

Jon E. Schoenfield, Dec 06 2016

nonn

Inspired by A278586.

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

			  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.

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, ...}?

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

a(n) = floor(a(n-1)*8/7) + 1.

A279079 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/9) requires n steps to reach 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 19, 22, 25, 29, 33, 38, 43, 49, 56, 64, 73, 83, 94, 106, 120, 136, 154, 174, 196, 221, 249, 281, 317, 357, 402, 453, 510, 574, 646, 727, 818, 921, 1037, 1167, 1313, 1478, 1663, 1871, 2105, 2369, 2666, 3000, 3376, 3799

Offset: 0

Jon E. Schoenfield, Dec 06 2016

nonn

Inspired by A278586.

Lim_{n->inf} a(n)/(9/8)^n = 5.19544896392362185906460915572195169945039729234281...

			  12 -> 12-ceiling(12/9) = 10,
  10 -> 10-ceiling(10/9) = 8,
   8 ->  8-ceiling(8/9)  = 7,
   7 ->  7-ceiling(7/9)  = 6,
...
   1 ->  1-ceiling(1/9)  = 0,
so reaching 0 from 12 requires 10 steps;
  13 -> 13-ceiling(13/9) = 11,
  11 -> 11-ceiling(11/9) = 9,
   9 ->  9-ceiling(9/9)  = 8,
   8 ->  8-ceiling(8/9)  = 7,
   7 ->  7-ceiling(7/9)  = 6,
...
   1 ->  1-ceiling(1/9)  = 0,
so reaching 0 from 13 (or more) requires 11 (or more) steps;
thus, 12 is the largest starting value from which 0 can be reached in 10 steps, so a(10) = 12.

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), (this sequence) (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, ...}?

a:=[0]; aCurr:=0; for n in [1..56] do aCurr:=Floor(aCurr*9/8)+1; a[#a+1]:=aCurr; end for; a;

a(n) = floor(a(n-1)*9/8) + 1.

A357081 Leader at step n of the THROWBACK procedure (see definition in comments).

Original entry on oeis.org

3, 4, 5, 6, 3, 7, 4, 8, 3, 5, 9, 4, 3, 6, 10, 5, 3, 4, 7, 11, 3, 6, 4, 5, 3, 8, 12, 4, 3, 7, 5, 6, 3, 4, 9, 13, 3, 5, 4, 8, 3, 6, 7, 4, 3, 5, 10, 14, 3, 4, 6, 5, 3, 9, 4, 7, 3, 8, 5, 4, 3, 6, 11, 15, 3, 4, 5, 7, 3, 6, 4, 10, 3, 5, 8, 4, 3, 9, 6, 5, 3, 4, 7, 12, 3, 16, 4, 5, 3, 6, 8, 4, 3, 7, 5, 11, 3, 4, 6, 9

Offset: 0

Anthony M. Kozar Jr., Sep 08 2022

The THROWBACK procedure: Start with the infinite sequence of natural numbers beginning with 3, that is 3, 4, 5, 6, 7, 8, ... The first number in the sequence at any step of the procedure is called the "leader". At each step, the leader is moved back in the sequence the number of places equal to its value.
It is conjectured that every number (n >= 3) appears an infinite number of times in this sequence.
The indices of records, ignoring the initial 3, appear to match A155167.
Every fourth term is 3. Values k > 3 occur at nonconstant intervals and the sequence of intervals for each k appears to be cyclic with a period of 3^(k-3). Ignoring the last value, the first 3^(k-3)-1 values of any of these cycles of intervals appear to be a palindrome. E.g., a(n)=5 for n=2,9,15,23,30,37,45,51,58,66,... The intervals between the 5's appear to repeat the pattern 7,6,8,7,7,8,6,7,8 and 7,6,8,7,7,8,6,7 is a palindrome.
Appears to be A087165 with every term incremented by 2. If so, then the recurrence a(n)=3 when n == 0 (mod 4), otherwise a(n) = a(n - ceiling(n/4)) + 1 holds. Also appears to be A087165 with every 1 and every 2 removed. See the comment by Benoit Cloitre on A087165 for another possible way to construct this sequence.
If every 3 is removed then the result appears to be the original sequence with every term incremented by 1.
If the THROWBACK procedure is performed on all natural numbers including 1 and 2, then the sequence of leaders appears to be A001511. Other initial values appear to produce similar patterns to this sequence.

			Before the first step, 3 is the leader, so a(0) = 3. In the first step, 3 is moved back 3 places giving the new sequence 4, 5, 6, 3, 7, 8, ..., so a(1) = 4.

Popular Computing (Calabasas, CA), Coding Fun: Rearranging All The Numbers, Annotated and scanned copy of pages PC55-2, PC55-3, and PC55-1(cover) of Vol. 5 (No. 55, Oct 1977).
Anthony M. Kozar Jr., THROWBACK All of Your Number Sequences

Cf. A087165, A155167.

Cf. A354223, A355080. (Other variants of the THROWBACK procedure).

from collections import deque
from itertools import count, islice
def tgen(): yield from count(3) # generator of sequence to throwback
def agen(): # generator of terms
    g = tgen()
    a = deque([next(g)])
    while True:
        leader = a.popleft()
        yield leader
        while leader > len(a): a.append(next(g))
        a.insert(leader, leader)
print(list(islice(agen(), 100))) # Michael S. Branicky, Sep 11 2022

a(n) = A087165(n+1) + 2 (conjectured).

cp's OEIS Frontend

A191670 Dispersion of A042968 (>1 and congruent to 1 or 2 or 3 mod 4), by antidiagonals.

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

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

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A087192 a(n) = ceiling(a(n-1)*4/3), with a(1) = 1.

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

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Magma

Formula

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

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Author

Keywords

Comments

Examples

Crossrefs

Programs

Magma

Formula

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs