A245321 Sum of digits of n written in fractional base 6/5.

0, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10, 9, 10, 11, 12, 13, 14, 12, 13, 14, 15, 16, 17, 14, 15, 16, 17, 18, 19, 15, 16, 17, 18, 19, 20, 15, 16, 17, 18, 19, 20, 20, 21, 22, 23, 24, 25, 19, 20, 21, 22, 23, 24, 23, 24, 25, 26, 27, 28, 21, 22, 23, 24, 25, 26, 24, 25

Offset: 0

Author

Tom Edgar, Jul 18 2014

Keywords

• nonn
• base
• easy

Comments

The base 6/5 expansion is unique and thus the sum of digits function is well-defined.

Examples

			In base 6/5 the number 15 is represented by 543 and so a(15) = 5 + 4 + 3 = 12.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 0..10000 (terms 0..1000 from G. C. Greubel)
• Index entries for sequences related to fractional bases.

Crossrefs

Cf. A000120, A007953, A024638, A053827, A244040.

Programs

Maple

a:= proc(n) `if`(n<1, 0, irem(n, 6, 'q')+a(5*q)) end:
seq(a(n), n=0..100);  # Alois P. Heinz, Aug 19 2019

Mathematica

a[n_]:= a[n] = If[n==0, 0, a[5*Floor[n/6]] + Mod[n,6]]; Table[a[n], {n, 0, 70}] (* G. C. Greubel, Aug 19 2019 *)

PARI

a(n) = if(n == 0, 0, a(n\6 * 5) + n % 6); \\ Amiram Eldar, Jul 31 2025

Sage

def basepqsum(p,q,n):
    L=[n]
    i=1
    while L[i-1]>=p:
        x=L[i-1]
        L[i-1]=x.mod(p)
        L.append(q*(x//p))
        i+=1
    return sum(L)
[basepqsum(6,5,i) for i in [0..70]]

Formula

a(n) = A007953(A024638(n)).