Problems & Puzzles: Puzzles

Puzzle 466. phi(n-1)=phi(n)=phi(n+1) Farideh sent the one more nice puzzle: For n=5187 we have phi(n-1)=phi(n)=phi(n+1) Q. Does there exist another such number? Note: See A001274

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Farideh herself provided an interesting extension for this same puzzle:

If we define a(n) the smallest number m such that phi(m-n)=phi(m)=phi(m+n)
then the terms for n=1,2,...,12 are:

5187, 10, ?, 20, 25935, 78, ?, 40, ? , ?, 57057,156.

It seems that for each n, a(6n) exists!
Also we can show that if gcd(n, 34889017254)=1 then a(n) exists.

Q: Can you find some of the numbers a(n) for n=3, 7, 9 & 10 or can you
prove that at least one of them doesn't exist.

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Giovanni Resta wrote on Feb. 2014

For Puzzle 466, I have computed phi(n) for n < 10^13,
looking for values n such that phi(n-k) = phi(n) = phi(n+k),
in particular for k = 1, 3, 7, 9 and 10, as requested by
you and Farideh.

For n < 10^13, the only new result I obtained is

phi(30057431145 - 7) = phi(30057431145) = phi(30057431145 + 7).

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