cp's OEIS Frontend

A056852 a(n) = (p^p + 1)/(p + 1), where p = prime(n).

%I A056852 #31 Dec 27 2022 16:54:06
%S A056852 7,521,102943,23775972551,21633936185161,45957792327018709121,
%T A056852 98920982783015679456199,870019499993663001431459704607,
%U A056852 85589538438707037818727607157700537549449,533411691585101123706582594658103586126397951,277766709362573247738903423315679814371773581141321037961
%N A056852 a(n) = (p^p + 1)/(p + 1), where p = prime(n).
%C A056852 From _Lorenzo Sauras Altuzarra_, Nov 27 2022: (Start)
%C A056852 Are all terms pairwise coprime? If so, they induce a permutation of the natural numbers, as Fermat numbers do (see A343767).
%C A056852 Are all terms squarefree?
%C A056852 A342173(n) <= length(a(n)) = A055642(a(n)) (the proof is due to _Jinyuan Wang_). (End)
%H A056852 T. D. Noe, <a href="/A056852/b056852.txt">Table of n, a(n) for n = 2..26</a>
%H A056852 Lorenzo Sauras-Altuzarra, <a href="https://doi.org/10.26493/2590-9770.1473.ec5">Some properties of the factors of Fermat numbers</a>, Art Discrete Appl. Math. (2022).
%F A056852 From _Lorenzo Sauras Altuzarra_, Nov 27 2022: (Start)
%F A056852 a(n) = Sum_{k=0..prime(n)-1} (-prime(n))^k.
%F A056852 a(n) = F(prime(n), 1)/F(prime(n), 0), where F(b, n) = b^b^n + 1 (i.e., F(b, n) is the n-th base-b Fermat number, see A129290). (End)
%p A056852 a := n -> (ithprime(n)^ithprime(n)+1)/(ithprime(n)+1): # _Lorenzo Sauras Altuzarra_, Nov 27 2022
%t A056852 Table[ (Prime[ n ]^Prime[ n ] + 1)/(Prime[ n ] + 1), {n, 2, 11} ]
%t A056852 (#^#+1)/(#+1)&/@Prime[Range[2,20]] (* _Harvey P. Dale_, Apr 23 2015 *)
%Y A056852 Cf. A001039, A055642, A129290, A342173, A343767.
%K A056852 nonn
%O A056852 2,1
%A A056852 _Robert G. Wilson v_, Aug 30 2000