[seqfan] Magical sums: just glue the next digit
Eric Angelini
Eric.Angelini at kntv.be
Sat Apr 13 17:19:34 CEST 2013
Hello SeqFans,
Here is S(1) where the sum of any pair of consecutive terms T1 and T2
is given by T1 to which you concatenate the first digit of T2:
S(1)=1,10,99,899,8099,72898,656088,5916797,53251178,479260606,
4313345458,38820109125,349380982128,3144428839155,...
... we thus have 1+10=11 < 1+1 >
10+99=109 < 10+9 >
99+899=998 < 99+8 >
899+8099=8998 < 899+9 >
8099+72898=80997 < 8099+7 >
72898+656088=728986 < 72898+6 >...
So, if you want to impress your little kid, just ask him to pick
any integer of S(1) -- say T(n); then ask him again: "Do you want
me to add this number to the one just before it in S, or to the one
just after it?"
Whatever your beta-tester will select, you'll give him the answer
in a millisecond!
S(2), S(3) and S(4) start like this, I guess:
S(2)=2,20,181,1630,14671,132040,1188361,10695250,96257259,866315339,...
S(3)=3,30,272,2450,22052,198469,1786222,16075999,144683992,1302155929,...
S(4)=4,40,363,3270,29432,264890,2384012,21456110,193104991,1931049911,...
...
I've computed those seqs by hand (it is quite easy) -- I hope that I've
made no mistake and that each of them is the lexicofirst of its kind.
Best,
É.
P.-S.
To improve the little "addition trick" above, one could submit to
the poor guinea pig a list where the nine sequences S(1) to S(9) --
each of one having, say, 15 terms -- follow each other with no
visible interruption between them.
If the kid points to a single integer, you'll make the *real* addition,
left or right. If he doesn't, then the "gluing rule" (left or right)
will take place...
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