[seqfan] Re: a(n), a(n+1), [a(n)+a(n+1)] share no digits

franktaw at netscape.net franktaw at netscape.net
Fri Sep 4 05:31:05 CEST 2009


I've been looking at this in other bases.  In bases 7 and 8, the 
sequence is infinite.  If my program is correct, the base 7 sequence 
starts:

1,2,3,4,5,6,8,16,24,30,40,60,114,171,212,285,403,800,
1200,1486,2000,2804,5602,8403,10404,14005,19611,
39216,58824,72830,98040

which in base 7 is:

1,2,3,4,5,6,
11,22,33,42,55,
114,222,333,422,555,
1114,2222,3333,4222,5555,
11114,22222,33333,42222,55555,
111114,222222,333333,422222,555555

Base 8 starts:

1,2,3,4,5,6,7,9,18,27,34,45,50,59,73,146,206,260,341,388,
471,1609,2340,2706,3127,6235,21845,24836,30167,
102985,149796,173202,200119,399067,1398101,
1589508,1930711,6591049,9586980,11084946,12807607,
25540315,89478485,101728516,123565527,

which in base 8 is:

1,2,3,4,5,6,7,
11,22,33,42,55,62,73,
111,222,316,404,525,604,727,
3111,4444,5222,6067,
14133,52525,60404,72727,
311111,444444,522222,606667,
1413333,5252525,6040404,7272727,
31111111,44444444,52222222,60666667,
141333333,525252525,604040404,727272727

In both cases the pattern is obvious.

These are the only bases up to 10 where this happens.  In bases 2 and 
3, the sequence is only the number 1.  Base 4 gives us:

1,2,3,5,10 (1,2,3,11,22).

Base 5:

1,2,3,4,6,12,18,34 (1,2,3,4,11,22,33,114)

Base 6:

1,2,3,4,5,7,14,21,26 (1,2,3,4,5,11,22,33,42)

Base 9:

1,2,3,4,5,6,7,8,10,20,30,38,50,57,65,75,91,182,259,409,
546,587,656,984,1932

(1,2,3,4,5,6,7,8,
11,22,33,42,55,63,72,83,
111,222,317,504,666,722,808,
1313,2576)

And base 10 (as others have noted):

1,2,3,4,5,6,7,8,9,11,22,33,42,51,249,306,419,2587,
3413,5255,6064,7255,8844
,11156,22222,31778,
60444,72755,88344,111656,222088,333361,
422224,508887,622224,708887,922224,5138383

Franklin T. Adams-Watters

-----Original Message-----
From: Eric Angelini <Eric.Angelini at kntv.be>

Hello SeqFans,
we consider two consecutive integers ("a" and "b") from a
monotonically increasing seq S, and their sum "c"; we want
that:

"a" shares no digit with "b"
"b" shares no digit with "c"
"c" shares no digit with "a"

Starting with 1, do we have:

S = 1,2,3,4,5,6,7,8,9,11,22,33,42,51,249,302,449,... ?
(to prolong S, take the smallest integer not leading to a
 contradiction)

Does S stop at some point?
(is this old hat, BTW?)
Best,
É.



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