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A057635
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a(n) is the largest m such that phi(m) = n, where phi is Euler's totient function = A000010, or a(n) = 0 if no such m exists.
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19
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2, 6, 0, 12, 0, 18, 0, 30, 0, 22, 0, 42, 0, 0, 0, 60, 0, 54, 0, 66, 0, 46, 0, 90, 0, 0, 0, 58, 0, 62, 0, 120, 0, 0, 0, 126, 0, 0, 0, 150, 0, 98, 0, 138, 0, 94, 0, 210, 0, 0, 0, 106, 0, 162, 0, 174, 0, 118, 0, 198, 0, 0, 0, 240, 0, 134, 0, 0, 0, 142, 0, 270, 0, 0, 0, 0, 0, 158, 0, 330, 0
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OFFSET
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1,1
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COMMENTS
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To check that a property P holds for all EulerPhi(x) not exceeding n, for n with a(n) > 0, it suffices to check P for all EulerPhi(x) with x not exceeding a(n). - Joseph L. Pe, Jan 10 2002
Let f(n) = exp(gamma)*log(log(n)) + 2.5/log(log(n)), then a(n) < n*f(n^2) for all n > 1, where gamma = A001620.
Proof. Without loss of generality we suppose log(log(n)) > n_0 = sqrt(2.5/exp(gamma)) = 1.18475..., then f(n), n/f(n) and N(n) = ceiling(n*f(n^2)) are all monotonically increasing functions of n, and we have f(n) < 2*exp(gamma)*log(log(n)).
By the formula (3.41) in Theorem 15 by J. Barkley Rosser and Lowell Schoenfeld we have phi(k) > k/f(k) for k != 1, 2, 223092870. N(31802157) = 223092869 < 223092870, N(31802158) = 223092877 > 223092870, so N(n) != 223092870 (N(n) is increasing). So phi(N(n)) > N(n)/f(N(n)) > (n*f(n^2))/f(n*f(n^2)) (n/f(n) is increasing and log(log(n*f(n^2))) > n_0).
Note that f(n^2) < 2*exp(gamma)*log(log(n^2)) < 2*exp(gamma)*(log(n^2)/e) = 4*exp(gamma-1)*log(n) < 4*exp(gamma-2)*n < n, so n*f(n^2) < n^2, f(n*f(n^2)) < f(n^2) (f(n) is increasing and log(log(n*f(n^2))) > n_0), so phi(N(n)) > n. As a result, a(n) <= N(n) - 1 < n*f(n^2).
Conjecturally a(n) < n*f(n) for all n > 2. (End)
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LINKS
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FORMULA
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a(2n+1) = 0 for n > 0, and a(2n) = 0 iff 2n is in A005277.
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EXAMPLE
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m = 12 is the largest value of m such that phi(m) = 4, so a(4) = 12.
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MATHEMATICA
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a = Table[0, {100}]; Do[ t = EulerPhi[n]; If[t < 101, a[[t]] = n], {n, 1, 10^6}]; a
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PROG
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(PARI) a(n) = if(n%2, 2*(n==1), forstep(k=floor(exp(Euler)*n*log(log(n^2))+2.5*n/log(log(n^2))), n, -1, if(eulerphi(k)==n, return(k)); if(k==n, return(0)))) \\ Jianing Song, Feb 15 2019
(PARI) apply( {A057635(n, m=istotient(n))=if(!m, 0, n>1, m=log(log(n)*2); m=bitand(n*(exp(Euler)*m+2.5/m)\1, -2); while(eulerphi(m)!=n, m-=2); m, 2)}, [1..99]) \\ If n is known to be a totient, a nonzero 2nd arg can be given to avoid the check. - M. F. Hasler, Aug 13 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Edited and escape clause added to definition by M. F. Hasler, Aug 13 2021
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STATUS
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approved
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