A261337 Digit-sums in an incremental base that adjusts itself as the digits of n are generated from right to left.

0, 1, 1, 2, 1, 3, 2, 2, 1, 3, 3, 4, 2, 3, 2, 4, 1, 5, 3, 2, 3, 5, 4, 6, 2, 3, 3, 3, 2, 7, 4, 4, 1, 4, 5, 4, 3, 3, 2, 5, 3, 5, 5, 4, 4, 6, 6, 6, 2, 5, 3, 4, 3, 7, 3, 2, 2, 5, 7, 8, 4, 5, 4, 6, 1, 5, 4, 6, 5, 7, 4, 6, 3, 3, 3, 5, 2, 7, 5, 3, 3, 6, 5, 8, 5, 7, 4

Offset

0,4

Comments

In a standard base, the digits are generated from right to left by finding (n modulo base) and dividing by the base, until n = 0. In this incremental base, the base is first set equal to 2, then increases according to the digits generated by (n modulo base). For example, 5 = 21 in this base because 5 mod 2 = 1, int(5/2) = 2, 2 mod (2+1 = 3) = 2 and int(2/3) = 0. When n is a power of 2, the base remains 2 throughout, because all digits generated from right to left are 0 until the final digit.

Note that a(2n) = a(n). - Franklin T. Adams-Watters, Oct 09 2015

Links

Anthony Sand, Table of n, a(n) for n = 0..10000

Example

n = 11
base = 2
11 mod base = 11 mod 2 = 1
int(11/2) = 5
base + 1 = 3
5 mod base = 5 mod 3 = 2
int(5/3) = 1.
base + 2 = 5
1 mod base = 1 mod 5 = 1
int(1/5) = 0
Therefore incbase(11) = 121 and digsum(11,incbase) = 4.
n = 23
base = 2
23 mod base = 23 mod 2 = 1
int(23/2) = 11
base + 1 = 3
11 mod base = 11 mod 3 = 2
int(11/3) = 3.
base + 2 = 5
3 mod base = 3 mod 5 = 3
int(3/5) = 0
Therefore incbase(23) = 321 and digsum(23,incbase) = 6.

Prog

(PARI) n=0; nmx=1000; d=vector(20); bs=vector(20); while(n < nmx, n++; b=2; nn=n; di=0; while(nn>0, di++; d[di] = nn % b; nn \= b; b += d[di]; ); s = sum(i=1, di, d[i]); print1(s, ", "); );

Crossrefs

Cf. A108731.

Sequence in context: A157810 A072339 A342507 * A374998 A375000 A337195

Adjacent sequences: A261334 A261335 A261336 * A261338 A261339 A261340

Keyword

nonn,base

Author

Anthony Sand, Aug 15 2015

Status

approved