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A362719 Number of numbers k, 1 <= k <= n, such that phi(k) = phi(n-k+1). 1

%I #16 May 02 2023 23:55:55

%S 1,2,1,0,1,2,1,2,3,0,1,2,1,2,3,2,3,0,3,2,3,2,1,2,1,2,1,0,1,2,3,2,3,2,

%T 3,0,1,4,3,2,1,0,3,2,5,2,1,6,3,2,1,0,5,2,1,6,3,0,1,0,3,2,3,4,3,0,3,4,

%U 3,0,3,2,5,2,1,4,3,0,5,2,3,2,3,0,1,4,3,0,1,6,7,2,7,2,3

%N Number of numbers k, 1 <= k <= n, such that phi(k) = phi(n-k+1).

%F a(n) = Sum_{k=1..n} [phi(k) = phi(n-k+1)], where phi is the Euler phi function (A000010) and [ ] is the Iverson bracket.

%e a(9) = 3, since phi(4) = phi(9-4+1), phi(5) = phi(9-5+1), and phi(6) = phi(9-6+1).

%t Table[Sum[KroneckerDelta[EulerPhi[n - k + 1], EulerPhi[k]], {k, n}], {n, 100}]

%o (PARI) a(n) = sum(k=1, n, eulerphi(k) == eulerphi(n-k+1)); \\ _Michel Marcus_, May 01 2023

%Y Cf. A000010 (phi).

%K nonn,easy

%O 1,2

%A _Wesley Ivan Hurt_, Apr 30 2023

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