[seqfan] Integears
Eric Angelini
Eric.Angelini at kntv.be
Sun Jan 10 01:09:20 CET 2016
Hello SeqFans,
take an integer like 3126;
we call it "integear" because 32 is
divisible by 16.
The same with 1151946 because
1596 is divisible by 114. See the "gears":
1151946 --> 1 5 9 6
1151946 --> 1 1 4
We accept 1026 too as 12 is divisible
by (0)6.
And we accept also 1428 as 12 divides 48.
So, an "integear" G is such that:
- the concatenation "O" of the "odd-
ranked" digits of G either is divisible by,
or divides the concatenation "E" of
the "even-ranked" digits of G (with
"E" cleaned of all leading zeros,
should this happen).
I think the "Integear" seq starts:
G=11,12,13,14,15,16,17,18,19,21,22,24,
26,28,31,33,36,39,41,42,44,48,51,55,
61,62,63,66,71,77,81,82,84,88,91,93,
99,110,111,112,113,114,115,116,117,118,
119,120,122,124,126,128,132,135,138,
142,146,etc.
But what I like, of course, are "chains" like:
1,11,111,1021211,10103030504141211...
or
1,12,112,1031424,10105050109141828...
or
1,13,113,1041639,10107070005121037...
You get it:
-take the last "Integear" above and
compute I/O; this will give you the
term before it, 1041639;
- do the same with 1041639 and you'll
find 113.
- the last step (113/1=13) ends the
game.
Best,
Ã‰.
CatapultÃ© de mon aPhone
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