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Last Saturday [Decmeber 6] saw the annual Putnam competition. Since many on my PoW list do work on the Putnam problems, I am sending out this non-PoW mailing. The problems were nicely interesting, and B-5 especially caught my interest. This was the problem:

B-5:Leta_{j}denote a tower ofj2s: 2^{22.·'2}. Prove thata_{n}=a_{n-1}(modn).I was able to prove it using little more than Euler's formula that 2

^{Phi(n)}=1 (modn) when n is odd.My proof was constructive in that it allowed me to write some Mathematica code to get, given

jandm, the value of Mod[a_j, m]. It is always nice to see how to deal with such ridiculously large integers. And, naturally, computations lead to further questions, in particular, where does the congruence stop? Examples:There seems to be some pattern here. In particular, it looks like B-5 can be strengthened at least to

j= 2:a_{j}= 4 anda_{j-1}= 2. They are congruent mod 1 and 2, butnot3 (so B-5 as stated is best possible).j= 3:a_{j}= 16 anda_{j-1}= 4; they are congruent mod 1, 2, 3,and 4, but not 5.j= 4: First failure at modulus 11.j= 5: First failure at mod 23 (= 2*11 + 1)j= 6: First failure at mod 47 (= 2*47 + 1)j= 7: First failure at mod 283 (= 6*47 + 1)j= 8: First (I think) failure is at 1699 (= 6*1699+1)j= 9: First (I think) failure is at 1699*360 + 1IfNow, I have to admit it is possible my code is incorrect, so these observations should be taken with a grain of salt.n>3, thena_{n}=a_{n-1}(modk) for eachk= 1, 2, 3, ...,n,n+1.

Challenges:

- Prove the assertion about
n+1 above. (This might make a nice journal submission if true; on the other hand, it might follow from the solution of B-5; I will have to look at that carefully. Right now my solution proves:And the present question is whether this can be extended toa_{j}=a_{j-1}(modn) wheneverj>n.j>n- 1.)

- Say somethimg intelligent (either conjecture or proof) about the least
kfor whicha_{j}anda_{j-1}are not congruent modk.

More...I think I can define a simple TowerMod function. Recall that PowerMod[a,n,m] givesa^{n}modm. My TowerMod givesa^{aa.·'a}(nterms) modm.This is still work-in-progress. As usual, I have more questions than answers.

I would love some independent confirmation of these computations so that I can know my code is correct.

© Copyright 1997 Stan Wagon. Reproduced with permission.

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2 October 1998