A353264 a(n) is the least number k such that A018804(k)/k = n.

1, 4, 15, 64, 48, 60, 144, 16384, 240, 1300, 1296, 960, 1008, 3564, 3840, 1073741824, 6000, 14580, 7056, 20800, 11520, 25500, 944784, 245760, 13104, 24948, 34560, 57024, 750000, 16380, 156816, 4611686018427387904, 102000, 364500, 46800, 233280, 134064, 174636

Offset

1,2

Comments

a(n) exist for all n>=1.

The solution k to A018804(k)/k = n is unique if and only if n is a power of 2: a(2^m) = 2^(2^(m+1)-2) = A051191(m+1).

The questions of existence and uniqueness are a part of a problem that was proposed during the Forty-Fifth International Mathematical Olympiad in Athens, Greece, July 7-19, 2004.

References

Dušan Djukić, Vladimir Janković, Ivan Matić, and Nikola Petrović, The IMO Compendium, A Collection of Problems Suggested for the International Mathematical Olympiads: 1959-2004, Springer, New York, 2006. See Problem 25, pp. 331, 726-727.

Links

Table of n, a(n) for n=1..38.

Dušan Djukić, Vladimir Janković, Ivan Matić, and Nikola Petrović, IMO Shortlist 2004, 2005, Problem 25, pp. 5-6, 18-19.

R. E. Woodrow, Problem N2, The Olympiad Corner, Crux Mathematicorum, Vol. 34, No. 7 (2008), pp. 419-421.

Index to sequences related to Olympiads.

Example

a(2) = 4 since A018804(4)/4 = 8/4 = 2 and 4 is the least number with this property.

Mathematica

f[p_, e_] := (e*(p - 1)/p + 1); r[n_] := Times @@ (f @@@ FactorInteger[n]); a[n_] := Module[{k = 1}, While[r[k] != n, k++]; k]; Array[a, 15]

Crossrefs

Cf. A018804, A051191.

Subsequence of A066862.

Sequence in context: A036728 A027216 A124541 * A117669 A323789 A341922

Adjacent sequences: A353261 A353262 A353263 * A353265 A353266 A353267

Keyword

nonn

Author

Amiram Eldar, Apr 09 2022

Status

approved