A261337 - OEIS

%I #24 Oct 10 2015 14:28:30

%S 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,

%T 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,

%U 5,7,4,6,3,3,3,5,2,7,5,3,3,6,5,8,5,7,4

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

%C 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.

%C Note that a(2n) = a(n). - _Franklin T. Adams-Watters_, Oct 09 2015

%H Anthony Sand, <a href="/A261337/b261337.txt">Table of n, a(n) for n = 0..10000</a>

%e n = 11

%e base = 2

%e 11 mod base = 11 mod 2 = 1

%e int(11/2) = 5

%e base + 1 = 3

%e 5 mod base = 5 mod 3 = 2

%e int(5/3) = 1.

%e base + 2 = 5

%e 1 mod base = 1 mod 5 = 1

%e int(1/5) = 0

%e Therefore incbase(11) = 121 and digsum(11,incbase) = 4.

%e n = 23

%e base = 2

%e 23 mod base = 23 mod 2 = 1

%e int(23/2) = 11

%e base + 1 = 3

%e 11 mod base = 11 mod 3 = 2

%e int(11/3) = 3.

%e base + 2 = 5

%e 3 mod base = 3 mod 5 = 3

%e int(3/5) = 0

%e Therefore incbase(23) = 321 and digsum(23,incbase) = 6.

%o (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,", "); );

%Y Cf. A108731.

%K nonn,base

%O 0,4

%A _Anthony Sand_, Aug 15 2015