A024634 -id:A024634 - cp's OEIS frontend

A245357 Number of numbers whose base 5/4 expansion (see A024634) has n digits.

5, 5, 5, 5, 5, 10, 10, 15, 15, 20, 25, 30, 40, 50, 60, 75, 95, 120, 150, 185, 235, 290, 365, 455, 570, 710, 890, 1110, 1390, 1735, 2170, 2715, 3390, 4240, 5300, 6625, 8280, 10350, 12940, 16175, 20215, 25270, 31590, 39485, 49355, 61695, 77120, 96400, 120500

Offset: 1

James Van Alstine, Jul 18 2014

			The numbers 10..14 are represented by 430, 431, 432, 433, 434 respectively in base 5/4. These are the only numbers with three digits, and so a(3)=5.

Index entries for sequences related to fractional bases

Cf. A024634, A120160, A081848.

A=[1]
for i in [1..60]:
    A.append(ceil((5-4)/4*sum(A)))
[5*x for x in A]

a(n) = 5*A120160(n).

A120160 a(n) = ceiling(Sum_{i=1..n-1} a(i)/4) for n >= 2 starting with a(1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 5, 6, 8, 10, 12, 15, 19, 24, 30, 37, 47, 58, 73, 91, 114, 142, 178, 222, 278, 347, 434, 543, 678, 848, 1060, 1325, 1656, 2070, 2588, 3235, 4043, 5054, 6318, 7897, 9871, 12339, 15424, 19280, 24100, 30125, 37656, 47070, 58838, 73547

Offset: 1

Graeme McRae, Jun 10 2006

nonn

From Petros Hadjicostas, Jul 21 2020: (Start)
Conjecture 1: a(n) equals the number of multiples of 4 whose representation in base 5/4 (see A024634) has n-1 digits. For example, a(8) = 3 because there are three multiples of 4 with n-1 = 7 digits in their representation in base 5/4: 36 = 4321031, 40 = 4321420, and 44 = 4321424.
Conjecture 2: a(n) equals 1/5 times the number of nonnegative integers with the property that their 5/4-expansion has n digits (assuming that the 5/4-expansion of 0 has 1 digit). For example, a(7)*5 = 10 because the following 10 numbers have 5/4 expansions with n = 7 digits: 35 = 4321030, 36 = 4321031, 37 = 4321032, 38 = 4321033, 39 = 4321034, 40 = 4321420, 41 = 4321421, 42 = 4321422, 43 = 4321423, and 44 = 4321424. (End)
From Petros Hadjicostas, Jul 23 2020: (Start)
Starting at a(11) = 5, we conjecture that this sequence gives all those numbers m for which when we place m persons on a circle, label them 1 through m, start counting at person number 1, and remove every 5th person, the last survivors have numbers in {1, 2, 3, 4}.
When m = 6, 12, 15, 37, 58, 142, 222, 347, ... the last survivor is person 1.
When m = 5, 19, 91, 434, 1656, 2070, 5054, ... the last survivor is person 2.
When m = 8, 10, 24, 30, 73, 114, 178, 278, ... the last survivor is person 3.
When m = 47, 543, 2588, 3235, 6318, 58838, ... the last survivor is person 4. (End)

			From _Petros Hadjicostas_, Jul 23 2020: (Start)
We explain why 6 and 8 belong to the sequence related to the Josephus problem (for the case k = 5) while 7 does not.
When we place m = 6 people on a circle, label them 1 through 6, start counting at person 1, and remove every 5th person, the list of people we remove is 5 -> 4 -> 6 -> 2 -> 3. Thus the last survivor is person 1, so 6 belongs to this sequence.
When we place m = 7 people on a circle, label them 1 through 7, start counting at person 1, and remove every 5th person, the list of people we remove is 5 -> 3 -> 2 -> 4 -> 7 -> 1. Thus the last survivor is person 6 (not in {1, 2, 3, 4}), so 7 does not belong to this sequence.
When we place m = 8 people on a circle, label them 1 through 8, start the counting at person 1, and remove every 5th person, the list of people we remove is 5 -> 2 -> 8 -> 7 -> 1 -> 4 -> 6. Thus the last survivor is 3, so 8 belongs to this sequence. (End)

G. C. Greubel, 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. [The author deals with the representation of n in fractional bases k/(k-1) and its relation to counting-off games (variations of Josephus problem). Here k = 5. See the review in MathSciNet (MR0889384) by R. G. Stoneham.]
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. A011782, A024634, A072493, A073941, A112088, A120170, A120178, A120186, A120194, A120202.

function f(n, a, b)
  t:=0;
    for k in [1..n-1] do
      t+:= a+Ceiling((b+t)/4);
    end for;
  return t;
end function;
g:= func< n, a, b | f(n+1, a, b)-f(n, a, b) >;
A120160:= func< n | n le 4 select 1 else g(n-4, 1, 0) >;
[A120160(n): n in [1..60]]; // G. C. Greubel, Sep 01 2023

f[s_]:= Append[s, Ceiling[Plus @@ s/4]]; Nest[f,1,53] (* Robert G. Wilson v *)
(* Second program *)
f[n_]:= f[n]= 1 + Quotient[Sum[f[k], {k,n-1}], 4];
A120160[n_]:= If[n==1, 1, f[n-1]];
Table[A120160[n], {n, 60}] (* G. C. Greubel, Sep 01 2023 *)

/* PARI program for the general case of Josephus problem. We use the Burde-Thériault algorithm. Here k = 5. 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

@CachedFunction
def A120160(n): return 1 + ceil( sum(A120160(k) for k in range(1,n))//4 )
[A120160(n) for n in range(1,61)] # G. C. Greubel, Sep 01 2023

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

Appended the first missing term. - Tom Edgar, Jul 18 2014

A245335 Sum of digits of n in fractional base 5/4.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 5, 6, 7, 8, 7, 8, 9, 10, 11, 9, 10, 11, 12, 13, 10, 11, 12, 13, 14, 10, 11, 12, 13, 14, 14, 15, 16, 17, 18, 13, 14, 15, 16, 17, 16, 17, 18, 19, 20, 14, 15, 16, 17, 18, 16, 17, 18, 19, 20, 20, 21, 22, 23, 24, 17, 18, 19, 20, 21, 18, 19, 20, 21

Offset: 0

James Van Alstine, Jul 18 2014

The base 5/4 expansion is unique and thus the sum of digits function is well-defined.

			In base 5/4 the number 7 is represented by 42 and so a(7) = 4+2 = 6.

Amiram Eldar, Table of n, a(n) for n = 0..10000
Index entries for sequences related to fractional bases.

Cf. A024634, A007953, A053824.

a[n_] := a[n] = If[n == 0, 0, a[4 * Floor[n/5]] + Mod[n, 5]]; Array[a, 100, 0] (* Amiram Eldar, Jul 31 2025 *)

a(n) = my(ret=0,r); while(n, [n,r]=divrem(n,5); ret+=r; n<<=2); ret; \\ Kevin Ryde, Aug 11 2023

# uses [basepqsum from A245355]
[basepqsum(5,4,y) for y in [0..200]]

a(n) = A007953(A024634(n)). - Kevin Ryde, Aug 11 2023

cp's OEIS Frontend

A245357 Number of numbers whose base 5/4 expansion (see A024634) has n digits.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Sage

Formula

A120160 a(n) = ceiling(Sum_{i=1..n-1} a(i)/4) for n >= 2 starting with a(1) = 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Extensions

A245335 Sum of digits of n in fractional base 5/4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Sage

Formula