1,1

Unique numbers having 5 divisors (1, n-th prime, n-th prime^2=their square root, n-th prime^3, themselves). - Alexandre Wajnberg (alexandre.wajnberg(AT)skynet.be), Jan 15 2006

Subsequence of A036967. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Feb 05 2008

The n-th number with p divisors is equal to the n-th prime raised to power p-1, where p is prime. - Omar E. Pol (info(AT)polprimos.com), May 06 2008

The general product formula for even s is: product_{p=A000040} (p^s-1)/(p^s+1)= 2*Bernoulli(2s)/( binomial(2s,s)*Bernoulli^2(s)), where the infinite product is over all primes. Here, with s=4, product_{n=1,2,...} (a(n)-1)/(a(n)+1) = 6/7. In A030516, where s=6, the product of the ratios is 691/715. For s=8, the 8th row in A120458, the corresponding product of ratios is 7234/7293. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Feb 01 2009]

R. J. Mathar, Table of n, a(n) for n = 1..457

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

a(n)=A000040(n)^(5-1)=A000040(n)^4, where 5 is the number of divisors of a(n). - Omar E. Pol (info(AT)polprimos.com), May 06 2008

A000005(a(n))=5. Juri-Stepan Gerasimov (2stepan(AT)rambler.ru), Oct 10 2009

Array[Prime[ # ]^4&, 5! ] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Sep 01 2008]

(SAGE) BB = primes_first_n(36) list = [] for i in range(36): list.append(BB[i]^4) list - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 15 2007

Cf. A030078, A131991, A131992.

Cf. A000005, A000040.

Sequence in context: A153157 A113849 A046453 this_sequence A056571 A053909 A151502

Adjacent sequences: A030511 A030512 A030513 this_sequence A030515 A030516 A030517

nonn,easy

Jeff Burch (jmburch(AT)osprey.smcm.edu)

Description corrected by Eric Weisstein (eric(AT)weisstein.com)