A135738 Least positive integer with even digit sum in bases 2..n.

3, 6, 10, 10, 54, 54, 54, 54, 130, 130, 130, 130, 390, 390, 2000, 2000, 3238, 3238, 4080, 4080, 7326, 7326, 16584, 16584, 17310, 17310, 17310, 17310, 17310, 17310, 17310, 17310, 231000, 231000, 231000, 231000, 466352, 466352, 466352, 466352, 3020830

Offset

2,1

Comments

The sequence is obviously increasing. It seems that a(2n+1) = a(2n) for n > 1. Is there a simple proof? Is there a simple way to construct a(n)? Notice the pattern in base N, e.g., 130 = 10000010_2 = 11211_3 = 2002_4 = 1010_5 = 334_6 = 244_7 = 202_8 = 154_9 = 109_11 = {10}{10}_12 = {10}0_13.

Links

Table of n, a(n) for n=2..42.

"Davar55" on mersenneforum.org, Puzzles / "Sum of digits".

Example

a(2)=3 since 1=1_2, 2=10_2, so 3=11_2 is the number > 0 with even digit sum (1+1) in base 2.
a(3)=6 since 4=100_2, 5=12_3, so 6=20_3=110_2 is the least N > 0 with even digit sum in base 2 and in base 3.
a(4)=a(5)=10=1010_2=101_3=22_4=20_5 is the least N > 0 having even digit sum in bases 2 through 4 and has so also in base 5.

Prog

(PARI) digitsum(n, b=10, s)={n=[n]; while(n=divrem(n[1], b), s+=n[2]); s}
A135738(Bmax, n=1)={until(!n++, for(b=2, Bmax, digitsum(n, b)%2&next(2)); return(n))} /* n-th element of the sequence */
t=1; for(b=2, 100, print(b, ":", t=A135738(b, t))) /* display the list */

Crossrefs

Cf. A000120, A053735, A053737, A053824, A053827-A053836, A007953.

Sequence in context: A130483 A115012 A200723 * A307574 A169911 A073851

Adjacent sequences: A135735 A135736 A135737 * A135739 A135740 A135741

Keyword

nonn,base

Author

M. F. Hasler, Dec 06 2007

Extensions

Corrected example a(3)=5 to a(3)=6 David Yablon (davar55(AT)yahoo.com), Mar 19 2010

Status

approved