A024631 n written in fractional base 4/3.

0, 1, 2, 3, 30, 31, 32, 33, 320, 321, 322, 323, 3210, 3211, 3212, 3213, 32100, 32101, 32102, 32103, 32130, 32131, 32132, 32133, 321020, 321021, 321022, 321023, 321310, 321311, 321312, 321313, 3210200, 3210201, 3210202, 3210203, 3210230, 3210231

Offset

0,3

Links

G. C. Greubel, Table of n, a(n) for n = 0..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 = 4. See the review in MathSciNet (MR0889384) by R. G. Stoneham.]

Index entries for sequences related to the Josephus Problem

Formula

To represent a number in base b, if a digit is greater than or equal to b, subtract b and carry 1. In fractional base a/b, subtract a and carry b.

Maple

a:= proc(n) `if`(n<1, 0, irem(n, 4, 'q')+a(3*q)*10) end:
seq(a(n), n=0..45);  # Alois P. Heinz, Aug 20 2019

Mathematica

p:= 4; q:= 3; a[n_]:= a[n]= If[n==0, 0, 10*a[q*Floor[n/p]] + Mod[n, p]]; Table[a[n], {n, 0, 40}] (* G. C. Greubel, Aug 20 2019 *)

Prog

(PARI) a(n) = my(p=4, q=3); if(n==0, 0, 10*a(q*(n\p)) + (n%p));
vector(40, n, n--; a(n)) \\ G. C. Greubel, Aug 20 2019
(Sage)
def basepqExpansion(p, q, n):
    L, i = [n], 1
    while L[i-1] >= p:
        x=L[i-1]
        L[i-1]=x.mod(p)
        L.append(q*(x//p))
        i+=1
    return Integer(''.join(str(x) for x in reversed(L)))
[basepqExpansion(4, 3, n) for n in [0..40]] # G. C. Greubel, Aug 20 2019

Crossrefs

Cf. A244041 (sum of digits).

Cf. A024629 (base 3/2).

Sequence in context: A273467 A278861 A303158 * A032814 A233587 A354860

Adjacent sequences: A024628 A024629 A024630 * A024632 A024633 A024634

Keyword

nonn,base,easy

Author

David W. Wilson

Status

approved