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Least number k such that phi(Sum_{j=0..n}{k-j}) = phi(Sum_{j=0..n}{k+j}).
3

%I #12 Sep 04 2016 00:06:16

%S 1,4,9,21,10,50,27,24,36,41,25,42,54,51,147,115,34,100,61,40,133,87,

%T 55,132,121,100,102,144,46,171,167,52,89,126,82,318,122,148,121,236,

%U 85,177,144,145,216,205,115,299,216,115,206,300,94,211,246,154,192,399

%N Least number k such that phi(Sum_{j=0..n}{k-j}) = phi(Sum_{j=0..n}{k+j}).

%H Paolo P. Lava, <a href="/A276319/b276319.txt">Table of n, a(n) for n = 0..1000</a>

%F Solutions of the equation phi((n+1)*(2*k-n)/2) = phi((n+1)*(2*k+n)/2).

%e a(3) = 21 because phi(18+19+20+21) = phi(21+22+23+24) = 24.

%p with(numtheory): P:= proc(q) local k,n; print(1);

%p for n from 1 to q do for k from n to q do

%p if phi((n+1)*(2*k-n)/2)=phi((n+1)*(2*k+n)/2)

%p then print(k); break; fi; od; od; end: P(10^9);

%t Table[k = n; While[EulerPhi@ Sum[k - j, {j, 0, n}] != EulerPhi@ Sum[k + j, {j, 0, n}], k++]; k, {n, 0, 57}] (* _Michael De Vlieger_, Aug 30 2016 *)

%o (PARI) fa(n) = {if (n==0, k = 1, k = n); while (eulerphi((n+1)*(2*k-n)/2) != eulerphi((n+1)*(2*k+n)/2), k++); k;} \\ _Michel Marcus_, Aug 31 2016

%Y Cf. A000010, A276318, A276320.

%K nonn,easy

%O 0,2

%A _Paolo P. Lava_, Aug 30 2016