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A276319
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Least number k such that phi(Sum_{j=0..n}{k-j}) = phi(Sum_{j=0..n}{k+j}).
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3
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1, 4, 9, 21, 10, 50, 27, 24, 36, 41, 25, 42, 54, 51, 147, 115, 34, 100, 61, 40, 133, 87, 55, 132, 121, 100, 102, 144, 46, 171, 167, 52, 89, 126, 82, 318, 122, 148, 121, 236, 85, 177, 144, 145, 216, 205, 115, 299, 216, 115, 206, 300, 94, 211, 246, 154, 192, 399
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OFFSET
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0,2
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LINKS
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FORMULA
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Solutions of the equation phi((n+1)*(2*k-n)/2) = phi((n+1)*(2*k+n)/2).
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EXAMPLE
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a(3) = 21 because phi(18+19+20+21) = phi(21+22+23+24) = 24.
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MAPLE
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with(numtheory): P:= proc(q) local k, n; print(1);
for n from 1 to q do for k from n to q do
if phi((n+1)*(2*k-n)/2)=phi((n+1)*(2*k+n)/2)
then print(k); break; fi; od; od; end: P(10^9);
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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