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A050473
Smallest k such that phi(k+n) = 2*phi(k).
6
2, 1, 1, 2, 1, 4, 3, 4, 3, 5, 5, 8, 26, 7, 5, 8, 9, 12, 5, 10, 7, 8, 46, 16, 5, 13, 9, 14, 7, 25, 21, 13, 9, 17, 7, 24, 62, 19, 11, 20, 76, 28, 13, 16, 15, 23, 17, 32, 21, 25, 17, 26, 52, 36, 11, 28, 13, 26, 13, 45, 74, 28, 17, 26, 13, 39, 33, 31, 21, 32, 13, 48, 39, 37, 25, 38
OFFSET
1,1
COMMENTS
Makowski proved that the sequence is well-defined.
It appears that k <= 2n, with equality for the n in A110196 only. Computations for n < 10^6 appear to show that k < n for all but a finite number of n. - T. D. Noe, Jul 15 2005
REFERENCES
Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B36, p. 138.
LINKS
Andrzej Makowski, On the equation phi(n+k)=2*phi(n), Elem. Math., Vol. 29, No. 1 (1974), p. 13.
EXAMPLE
phi(13+26) = 24 = 2*phi(13), so a(13) = 26.
MATHEMATICA
Table[k=1; While[EulerPhi[n+k] != 2*EulerPhi[k], k++ ]; k, {n, 100}] (Noe)
PROG
(PARI) f(n) = apply(x -> x - n, select(x -> x > n, invphi(2*eulerphi(n)))); \\ using Max Alekseyev's invphi.gp
lista(len) = {my(v = vector(len), c = 0, k = 1, s); while(c < len, s = f(k); for(i = 1, #s, if(s[i] <= len && v[s[i]] == 0, c++; v[s[i]] = k)); k++); v; } \\ Amiram Eldar, Nov 05 2024
CROSSREFS
Cf. A110179 (least k such that phi(n+k)=2*phi(n)).
Sequence in context: A002070 A326376 A106052 * A057593 A117008 A337366
KEYWORD
nonn,look
AUTHOR
Jud McCranie, Dec 24 1999
STATUS
approved