A073941 -id:A073941 - cp's OEIS frontend

A120170 a(n) = ceiling( Sum_{i=1..n-1} a(i)/5 ), a(1)=1.

1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 6, 7, 8, 10, 12, 14, 17, 21, 25, 30, 36, 43, 52, 62, 74, 89, 107, 128, 154, 185, 222, 266, 319, 383, 460, 552, 662, 795, 954, 1144, 1373, 1648, 1977, 2373, 2847, 3417, 4100, 4920, 5904, 7085, 8502, 10202, 12243, 14691, 17630

Offset: 1

Graeme McRae, Jun 10 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A011782, A073941, A072493, A112088, A120160, A120178, A120186, A120194, A120202.

function f(n, a, b)
  t:=0;
    for k in [1..n-1] do
      t+:= a+Ceiling ((b+t)/5);
    end for;
  return t;
end function;
g:= func< n, a, b | f(n+1, a, b)-f(n, a, b) >;
A120170:= func< n | n eq 1 select 1 else g(n-1, 1, -4) >;
[A120170(n): n in [1..60]]; // G. C. Greubel, Dec 25 2023

f[s_] := Append[s, Ceiling[Plus @@ s/5]]; Nest[f, {1}, 57] (* Robert G. Wilson v, Jul 07 2006 *)

@CachedFunction
def a(n):
    if (n==1): return 1
    else: return ceil(sum(a(k)/5 for k in (1..n-1)))
[a(n) for n in (1..60)] # G. C. Greubel, Aug 19 2019

Edited and extended by Robert G. Wilson v, Jul 07 2006

A120186 a(n) = ceiling( Sum_{i=1..n-1} a(i)/7 ), a(1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 23, 27, 30, 35, 40, 45, 52, 59, 68, 77, 88, 101, 115, 132, 151, 172, 197, 225, 257, 294, 336, 384, 439, 501, 573, 655, 748, 855, 977, 1117, 1277, 1459, 1667, 1906, 2178, 2489, 2845

Offset: 1

Graeme McRae, Jun 10 2006

nonn

Cf. A112088, A072493, A011782, A073941, A072493, A120160, A120170, A120178, A120186, A120194, A120202.

f[s_] := Append[s, Ceiling[Plus @@ s/7]]; Nest[f, {1}, 64] (* Robert G. Wilson v, Jul 07 2006 *)

Edited and extended by Robert G. Wilson v, Jul 07 2006

A120194 a(n) = ceiling( Sum_{i=1..n-1} a(i)/8 ), a(1)=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 18, 20, 22, 25, 28, 32, 36, 40, 45, 51, 57, 64, 72, 81, 91, 103, 116, 130, 146, 165, 185, 208, 234, 264, 297, 334, 376, 423, 475, 535, 602, 677, 762, 857, 964, 1084, 1220, 1372, 1544

Offset: 1

Graeme McRae, Jun 10 2006

nonn

Cf. A112088, A011782, A073941, A072493, A120160, A120170, A120178, A120186, A120194, A120202.

f[s_] := Append[s, Ceiling[Plus @@ s/8]]; Nest[f, {1}, 67] (* Robert G. Wilson v *)

Edited and extended by Robert G. Wilson v, Jul 07 2006

A120202 a(n) = ceiling( Sum_{i=1..n-1} a(i)/9), a(1)=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 25, 28, 31, 34, 38, 42, 47, 52, 58, 64, 71, 79, 88, 98, 109, 121, 134, 149, 166, 184, 205, 227, 253, 281, 312, 347, 385, 428, 476, 528, 587, 652, 725, 805, 895

Offset: 1

Graeme McRae, Jun 10 2006

nonn

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A112088, A011782, A073941, A072493, A120160, A120170, A120178, A120186, A120194.

N:= 100: # to get a(1) to a(N)
S:= 0: A[1]:= 1:
for n from 2 to N do
  S:= S + A[n-1];
  A[n]:= ceil(S/9);
od:
seq(A[n],n=1..N); # Robert Israel, Jul 14 2014

a[s_] := Append[s, Ceiling[Plus @@ s/9]]; Nest[a, {1}, 70] (* Robert G. Wilson v, Jul 07 2006 *)

Edited and extended by Robert G. Wilson v, Jul 07 2006

Typo in name corrected by Tom Edgar, Jul 14 2014

A120178 a(n)=ceiling( sum_{i=1..n-1} a(i)/6), a(1)=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 20, 23, 27, 32, 37, 43, 50, 59, 69, 80, 93, 109, 127, 148, 173, 202, 235, 275, 320, 374, 436, 509, 594, 693, 808, 943, 1100, 1283, 1497, 1747, 2038, 2377, 2774, 3236, 3775, 4404, 5138, 5995, 6994

Offset: 1

Graeme McRae, Jun 10 2006

nonn

Cf. A112088, A072493, A011782, A073941, A072493, A120160, A120170, A120186, A120194, A120202.

f[s_] := Append[s, Ceiling[Plus @@ s/5]]; Nest[f, {1}, 61] (* Robert G. Wilson v *)

Edited and extended by Robert G. Wilson v, Jul 07 2006

A303500 The smallest positive even integer that can be written with n digits in base 3/2.

Original entry on oeis.org

2, 21, 210, 2101, 21011, 210110, 2101100, 21011000, 210110001, 2101100011, 21011000110, 210110001101, 2101100011010, 21011000110100, 210110001101001, 2101100011010011, 21011000110100110, 210110001101001101

Offset: 0

Tanya Khovanova and PRIMES STEP Senior group, May 09 2018

a(n) is a prefix of a(n+1).

The smallest, not necessarily even, integer in base 3/2 with n digits is a(n-1) with 0 added at the end.

			The number 5 in base 3/2 is 22, and the number 6 is 210. Therefore, 210 is the smallest even integer with 3 digits in base 3/2.

B. Chen, R. Chen, J. Guo, S. Lee et al., On Base 3/2 and its Sequences, arXiv:1808.04304 [math.NT], 2018.

See A024629 for the base-3/2 expansion of n.

Cf. also A304024, A304025, A070885, A304272, A081848, A246435, A005428, A073941.

roll32 := proc(L)
    local piv,L1 ;
    piv := 1;
    L1 := subsop(piv=op(piv,L)+1,L) ;
    while op(piv,L1) >= 3 do
        L1 := [seq(0,i=1..piv), op(piv+1,L1)+1, seq(op(i,L1),i=piv+2..nops(L1))] ;
        piv := piv+1 ;
    end do:
    L1 ;
end proc:
from32 := proc(L)
    add( op(i,L)*(3/2)^(i-1),i=1..nops(L)) ;
end proc:
A303500 := proc(n)
    local dgs ;
    dgs := [seq(0,i=1..n-1),1] ;
    while not type(from32(dgs),'even') do
        dgs := roll32(dgs) ;
    end do:
    dgs := ListTools[Reverse](dgs) ;
    digcatL(%) ;
end proc: # R. J. Mathar, Jun 25 2018

a(n) = A024629(A305498(n)). - R. J. Mathar, Jun 25 2018

A304024 a(n) is the largest integer with n digits in base 3/2.

Original entry on oeis.org

2, 22, 212, 2122, 21222, 212212, 2122112, 21221112, 212211122, 2122111222, 21221112212, 212211122122, 2122111221212, 21221112212112, 212211122121122, 2122111221211222, 21221112212112212, 212211122121122122

Offset: 0

Tanya Khovanova and PRIMES STEP Senior group, May 04 2018

Every number starts and ends with 2 and contains only twos and ones.
Removing the last digit produces sequence A304272 of the largest even integers in base 3/2.
The value of this sequence in base 10 is A304025.
When adding 1 to the value of this sequence we get A070885.
The largest integer with a given number of digits in base 3/2 can be produced directly from the smallest number, sequence A304023, by replacing 21 at the beginning and 0 at the end with 2, and by shifting the rest up by 1, see sequence A304023.

			The number 5 in base 3/2 is 22, and the number 6 is 210. Therefore, 22 is the largest two-digit integer.

Cf. A005428, A024629, A070885, A073941, A081848, A246435, A303500, A304023, A304025, A304272.

first(n) = {my(res=vector(n), c = 2); res[1]=2; for(i=2, n, res[i] = 10 * res[i-1] + 2; if(c % 2 == 1, res[i] -= 10); c = 3 * c / 2 + if(c%2==0, 2, 1/2)); res} \\ David A. Corneth, May 11 2018

a(1) = 2, for n > 1, a(n) = 10 * a(n - 1) + 2 if A304025(n - 1) is even. Otherwise, a(n) = 10 * a(n - 1) - 8. - David A. Corneth, May 11 2018

A304025 a(n) is the largest integer that can be written with n digits in base 3/2.

Original entry on oeis.org

2, 5, 8, 14, 23, 35, 53, 80, 122, 185, 278, 419, 629, 944, 1418, 2129, 3194, 4793, 7190, 10787, 16181, 24272, 36410, 54617, 81926, 122891, 184337, 276506, 414761, 622142, 933215, 1399823, 2099735, 3149603, 4724405, 7086608, 10629914

Offset: 1

Tanya Khovanova and PRIMES STEP Senior group, May 04 2018

A070885 is the smallest integer that can be written with n digits in base 3/2.

This sequence represented in base 3/2 is A304024.

			The number 5 in base 3/2 is 22, and the number 6 is 210. Therefore, 5 is the largest integer needing two digits in base 3/2.

B. Chen, R. Chen, J. Guo, S. Lee et al., On Base 3/2 and its Sequences, arXiv:1808.04304 [math.NT], 2018.

Cf. A005428, A024629, A070885, A073941, A081848, A246435, A303500, A304024, A304272.

first(n) = {my(res = vector(n)); res[1] = 2; for(i = 2, n, res[i] = 3 * res[i-1] / 2 + if(res[i-1] % 2==0, 2, 1/2));res} \\ David A. Corneth, May 11 2018

a(n) = A070885(n+1) - 1.

A304272 The largest even integer that can be written with n digits in base 3/2.

Original entry on oeis.org

2, 21, 212, 2122, 21221, 212211, 2122111, 21221112, 212211122, 2122111221, 21221112212, 212211122121, 2122111221211, 21221112212112, 212211122121122, 2122111221211221, 21221112212112212, 212211122121122121, 2122111221211221212, 21221112212112212121, 212211122121122121211, 2122111221211221212112

Offset: 1

Tanya Khovanova and PRIMES STEP Senior group, May 09 2018

a(n) is a prefix of a(n+1).

The largest, not necessarily even, integer in base 3/2 with n digits is a(n-1) with 2 added at the end.

			The number 4 in base 3/2 is 21, and number 6 is 210. Therefore, 21 is the largest even integer with 2 digits in base 3/2.

B. Chen, R. Chen, J. Guo, S. Lee et al., On Base 3/2 and its Sequences, arXiv:1808.04304 [math.NT], 2018.

Cf. A304024, A304025, A070885, A303500, A024629, A081848, A246435, A005428, A073941.

Table[StringTake["212211122121122121211221211212112", n], {n, 32}]

A005427 Josephus problem: numbers m such that, when m people are arranged on a circle and numbered 1 through m, the final survivor when we remove every 4th person is one of the first three people.

Original entry on oeis.org

5, 7, 9, 12, 16, 22, 29, 39, 52, 69, 92, 123, 164, 218, 291, 388, 517, 690, 920, 1226, 1635, 2180, 2907, 3876, 5168, 6890, 9187, 12249, 16332, 21776, 29035, 38713, 51618, 68824, 91765, 122353, 163138, 217517, 290023, 386697, 515596, 687461, 916615, 1222153, 1629538, 2172717, 2896956, 3862608, 5150144, 6866859, 9155812, 12207749, 16276999, 21702665, 28936887, 38582516, 51443354

Offset: 1

N. J. A. Sloane, Simon Plouffe

nonn

Is this the same as A072493 with its first 8 terms removed? See also the similar conjecture concerning A005428 and A073941.
From Petros Hadjicostas, Jul 20 2020: (Start)
We describe the counting-off game of Burde (1987) using language from Schuh (1968). Suppose m people are labeled with the numbers 1 through m (say clockwise). (Burde uses the numbers 0 through m-1 probably because he relates this problem to the representation of m in the fractional base k/(k-1) = 4/3. He actually modifies the (4/3)-representation of m to include negative coefficients. See the coefficients f(n;k) below.)
Suppose we start the counting at the person labeled 1, and we remove every 4th person. This sequence gives those numbers m for which the last survivor is one of the first three people.
When m = 5, 9, 12, 16, 218, 517, ... the last survivor is the first person.
When m = 7, 29, 69, 92, 291, 388, ... the last survivor is the second person.
When m = 22, 39, 52, 123, 164, 690, ... the last survivor is the third person.
If we know m = a(n) and the number, say i(n), of the last survivor (when there are a(n) people on the circle), we may find a(n+1) and the number i(n+1) of the new last survivor (when there are a(n+1) people on the circle) in the following way:
(a) If 0 = a(n) mod 3, then a(n+1) = (4/3)*a(n), and i(n+1) = i(n).
(b) If 1 = a(n) mod 3 and i(n) = 1, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 3.
(c) If 1 = a(n) mod 3 and i(n) = 2, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 1.
(d) If 1 = a(n) mod 3 and i(n) = 3, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 2.
(e) If 2 = a(n) mod 3 and i(n) = 1, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 2.
(f) If 2 = a(n) mod 3 and i(n) = 2, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 3.
(g) If 2 = a(n) mod 3 and i(n) = 3, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 1. (End)
From Petros Hadjicostas, Jul 22 2020: (Start)
In general, for k >= 2, it seems that when m people are placed on a circle, labeled 1 through m, and every k-th person is removed (starting the counting at person 1), we may determine those m for which the last survivor is in {1, 2, ..., k-1} in the following way.
Define the sequence (T(n;k): n >= 1) by T(n;k) = ceiling(Sum_{s=1..n-1} T(s;k)/(k-1)) for n >= 2 starting with T(1; k) = 1. Then the list of those m's for which the last survivor is in {1, 2, ..., k-1} consists of all the numbers T(n;k) >= k (thus, we exclude the cases m = 1, ..., k-1 that may be repeated more than once in the sequence (T(n;k): n >= 1)).
I do not have a general proof of this conjecture though I strongly believe that Schuh's (1968) way of solving the case k = 3 (see pp. 373-375 and 377-379, where he provides two methods of solution) may provide clues for proving the conjecture.
We have T(n; k=2) = A011782(n+1), T(n; k=3) = A073941(n), T(n; k=4) = A072493(n), T(n; k=5) = A120160(n), T(n; k=6) = A120170(n), T(n; k=7) = A120178(n), T(n; k=8) = A120186(n), T(n; k=9) = A120194(n), and T(n; k=10) = A120202(n).
We also have T(n+1;k) = floor((k/(k-1))*T(n;k)) or ceiling((k/(k-1)*T(n;k)).
To identify the last survivor that results when we place T(n; k) people on the circle (with T(n;k) >= k) in the above Josephus problem, we use a modification of Burde's algorithm due to Thériault (2000).
We use the following recursions but we start at T(k;k) (rather than at the smallest n for which T(n;k) >= k). Define the sequence (S(n;k): n >= 1) by S(n;k) = T(n+k-1; k) for n >= 1. (It is easy to prove that S(1;k) = T(k;k) = 1.)
Define also the sequences (j(n;k): n >= 1) and (f(n;k): n >= 1) by j(1;k) = 1, f(1;k) = 0, f(n+1;k) = ((j(n;k) - S(n;k) - 1) mod (k-1)) + 1 - j(n;k) and j(n+1;k) = j(n;k) + f(n+1;k) for n >= 2.
Then for all n s.t. S(n;k) >= k, j(n;k) is the number of the last survivor of the Josephus problem where every k-th person is removed (provided we start the counting at number 1). It will always be the case that j(n;k) is in {1,2,...,k-1}.
We actually have S(n+1; k) = (k*S(n;k) + f(n+1;k))/(k-1) for n >= 1.
Notice that the Burde-Thériault algorithm is a generalization of Schuh's method. (End)

			From _Petros Hadjicostas_, Jul 22 2020: (Start)
We explain why 5 and 7 are in the sequence but 6 is not.
If we put m = 5 people on the circle, label them 1 through 5, start the counting at person 1, and remove every 4th person, then the list of people who are removed is 4 -> 3 -> 5 -> 2. Thus, the last survivor is 1, so m = 5 is included in this sequence.
If we put m = 6 people on a circle, label them 1 through 6, start the counting at person 1, and remove every 4th person, then the list of people who are removed is 4 -> 2 -> 1 -> 3 -> 6. Thus, the last survivor is 5 (not 1, 2, or 3), so m = 6 is not included in this sequence.
If we put m = 7 people on a circle, label them 1 through 7, start the counting at person 1, and remove every 4th person, then the list of people who are removed is 4 -> 1 -> 6 -> 5 -> 7 -> 3. Thus, the last survivor is 2, so m = 7 is included in this sequence.
Strictly speaking, m = 2 and m = 3 should have been included as well (since clearly the last survivor would be 1 or 2 or 3). In addition, m = 4 should have been included as well because the list of people removed is 4 -> 1 -> 3. The case of number 1 does create a problem since there is no survivor. Note that the numbers 1, 2, 3, 4 are all included in A072493. (End)

Fred Schuh, The Master Book of Mathematical Recreations, Dover, New York, 1968. [This book is cited in Burde (1987). Table 18, p. 374, is related to a very similar sequence (A073941). Thus, definitely, the counting-off games described in the book are related to a similar counting-off game in Burde (1987).]
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
K. Burde, Das Problem der Abzählreime und Zahlentwicklungen mit gebrochenen Basen [The problem of counting rhymes and number expansions with fractional bases], J. Number Theory 26(2) (1987), 192-209.
Nicolas Thériault, Generalizations of the Josephus problem, Utilitas Mathematica, 58 (2000), 161-173 (MR1801309).
Nicolas Thériault, Generalizations of the Josephus problem, Utilitas Mathematica, 58 (2000), 161-173 (MR1801309).
Index entries for sequences related to the Josephus Problem

Cf. A005428, A072493.

Similar sequences: A011782 (k = 2), A073941 (k = 3), A072493 (k = 4), A120160 (k = 5), A120170 (k = 6), A120178 (k = 7), A120186 (k = 8), A120194 (k = 9), A120202 (k = 10).

f[s_] := Append[s, Ceiling[5 + Plus@@(s/3)]]; Nest[f, {5}, 100] (* Vladimir Joseph Stephan Orlovsky, Jan 08 2011 *)

/* Gives an n X 2 matrix w s.t. w[,1] are the terms of this sequence and w[,2] are the corresponding numbers of the last survivors (1, 2 or 3). */
lista(nn) = {my(w = matrix(nn,2)); w[1,1] = 5; w[1,2] = 1; for(n=1, nn-1,
if(0 == w[n,1] % 3, w[n+1,1] = w[n,1]*4/3; w[n+1,2] = w[n,2]);
if(1 == w[n,1] % 3 && w[n,2] == 1, w[n+1,1] = ceil(w[n,1]*4/3);  w[n+1,2] = w[n,2] + 2);
if(1 == w[n,1] % 3 && w[n,2] == 2, w[n+1,1] = floor(w[n,1]*4/3); w[n+1,2] = w[n,2] - 1);
if(1 == w[n,1] % 3 && w[n,2] == 3, w[n+1,1] = floor(w[n,1]*4/3); w[n+1,2] = w[n,2] - 1);
if(2 == w[n,1] % 3 && w[n,2] == 1, w[n+1,1] = ceil(w[n,1]*4/3);  w[n+1,2] = w[n,2] + 1);
if(2 == w[n,1] % 3 && w[n,2] == 2, w[n+1,1] = ceil(w[n,1]*4/3);  w[n+1,2] = w[n,2] + 1);
if(2 == w[n,1] % 3 && w[n,2] == 3, w[n+1,1] = floor(w[n,1]*4/3); w[n+1,2] = w[n,2] - 2);
); Vec(w[,1]);} \\ Petros Hadjicostas, Jul 21 2020

/* Second PARI program for the general case of Josephus problem. We use the Burde-Thériault algorithm, not the formula T(n;k) = ceiling(Sum_{s=1..n-1} T(s;k)/(k-1)). We start with T(k;k) = 1 (and omit all previous 1's). Burde starts with the smallest T(n;k) >= k whose corresponding last survivor is 1. This, however, can be very large. To get the corresponding last survivors, modify the program to get the vector j. */
lista(nn,k) = {my(j=vector(nn)); my(f=vector(nn)); my(N=vector(nn));
j[1]=1; f[1]=0; N[1] = 1;
for(n=1, nn-1, f[n+1] = ((j[n]-N[n]-1) % (k-1)) + 1 - j[n];
j[n+1] = j[n] + f[n+1]; N[n+1] = (k*N[n] + f[n+1])/(k-1););
for(n=1, nn, if(N[n] > k-1, print1(N[n],",")));} \\ Petros Hadjicostas, Jul 23 2020

a(n) = 5 + ceiling(Sum_{k=1..n-1} a(k)/3). - Petros Hadjicostas, Jul 21 2020

More terms (from the Burde paper, p. 208) from R. J. Mathar, Sep 26 2006

Name edited by Petros Hadjicostas, Jul 20 2020

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A120170 a(n) = ceiling( Sum_{i=1..n-1} a(i)/5 ), a(1)=1.

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A120186 a(n) = ceiling( Sum_{i=1..n-1} a(i)/7 ), a(1) = 1.

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A120194 a(n) = ceiling( Sum_{i=1..n-1} a(i)/8 ), a(1)=1.

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A120202 a(n) = ceiling( Sum_{i=1..n-1} a(i)/9), a(1)=1.

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A120178 a(n)=ceiling( sum_{i=1..n-1} a(i)/6), a(1)=1.

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A303500 The smallest positive even integer that can be written with n digits in base 3/2.

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A304024 a(n) is the largest integer with n digits in base 3/2.

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A304025 a(n) is the largest integer that can be written with n digits in base 3/2.

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A304272 The largest even integer that can be written with n digits in base 3/2.

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