1,1

n such that bigomega(n)^(bigomega(n))=n, where bigomega=A001222. - Lekraj Beedassy (blekraj(AT)yahoo.com), Aug 21 2004

Positive n such that n' = n, where n' is the arithmetic derivative of n. - T. D. Noe (noe(AT)sspectra.com), Oct 12 2004

David Beckwith proposes (in the AMM reference): "Let n be a positive integer, and let p be a prime number. Prove that (p^p) | n! implies that (p^(p+1)) | n!" - Jonathan Vos Post (jvospost2(AT)yahoo.com), Feb 20 2006

Subsequence of A100716; A003415(m*a(n))=A129283(m)*a(n), especially A003415(a(n))=a(n). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Apr 07 2007

David Beckwith, Problem 11158, American Mathematical Monthly, Vol. 112, No. 5 (May 2005), p. 468.

J.-M. De Koninck & A. Mercier, 1001 Problemes en Theorie Classique Des Nombres, Problem 740 pp. 95; 312, Ellipses Paris 2004.

T. D. Noe, Table of n, a(n) for n=1..40

a(3) = 5^5 = 3125

Array[Prime[ # ]^Prime[ # ] &, 12] (from Vladimir Orlovsky (4vladimir(AT)gmail.com), May 01 2008)

Cf. A000040.

Cf. A003415 (arithmetic derivative of n).

Cf. A129150, A129151, A129152.

Sequence in context: A133032 A110763 A066352 this_sequence A132641 A008973 A132646

Adjacent sequences: A051671 A051672 A051673 this_sequence A051675 A051676 A051677

nonn

Asher Auel (asher.auel(AT)reed.edu)