# [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...

```