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A276000
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Least k such that n! divides phi(k!) (k > 0).
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1
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1, 3, 6, 6, 10, 10, 14, 14, 14, 14, 22, 22, 26, 26, 26, 26, 34, 34, 38, 38, 38, 38, 46, 46, 46, 46, 46, 46, 58, 58, 62, 62, 62, 62, 62, 62, 74, 74, 74, 74, 82, 82, 86, 86, 86, 86, 94, 94, 94, 94, 94, 94, 106, 106, 106, 106, 106, 106, 118, 118, 122, 122, 122, 122
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OFFSET
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1,2
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LINKS
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FORMULA
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EXAMPLE
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a(2) = 3 because phi(3!) is divisible by 2!.
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MAPLE
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N:= 100: # for a(1)..a(N)
V:= Vector(N): n:= 0:
for k from 1 while n < N do
r:= numtheory:-phi(k!);
for i from n+1 to N while r mod (i!) = 0 do
V[i]:= k; n:= i;
od;
od:
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MATHEMATICA
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Array[Block[{k = 1}, While[Mod[EulerPhi[k!], #!] != 0, k++]; k] &, 64] (* Michael De Vlieger, Apr 24 2020 *)
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PROG
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(PARI) a(n) = {my(k = 1); while(eulerphi(k!) % n!, k++); k; }
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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