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Numbers n satisfying phi(n+1) - phi(n-1) = 2.
2

%I #9 Sep 29 2019 02:37:26

%S 4,6,7,12,13,15,18,19,21,30,42,45,60,63,72,93,102,108,117,138,150,165,

%T 180,192,198,213,228,240,255,270,282,312,333,348,357,420,432,453,462,

%U 522,525,570,600,618,642,660,693,717,765,810,822,828,858,882,933,957

%N Numbers n satisfying phi(n+1) - phi(n-1) = 2.

%H Amiram Eldar, <a href="/A056853/b056853.txt">Table of n, a(n) for n = 1..10000</a>

%e phi(13+1)-phi(13-1) = 2, so 13 is a term of the sequence.

%t Select[Range[10^3], EulerPhi[ # + 1] - EulerPhi[ # - 1] == 2 &]

%t Flatten[Position[Partition[EulerPhi[Range[1000]],3,1],_?(Last[#]-First[#] == 2&),{1},Heads->False]]+1 (* _Harvey P. Dale_, Jul 14 2014 *)

%Y Cf. A000010. Essentially the same as A001838.

%K nonn

%O 1,1

%A _Joseph L. Pe_, Feb 12 2002