login
a(n) = Sum_{k=1..prime(n)-1} floor(k^5/prime(n)).
0

%I #6 Mar 15 2023 12:15:09

%S 0,10,258,1740,20070,48510,196920,350370,937860,3075030,4322160,

%T 10641330,17925180,22825110,35827560,65816010,113180910,133937670,

%U 215070570,288148140,331474860,493573080,633015810,899599140,1387338960,1700082450,1876303260,2272556790,2494333710

%N a(n) = Sum_{k=1..prime(n)-1} floor(k^5/prime(n)).

%H Jean-Christophe Pain, <a href="https://arxiv.org/abs/2303.07934">A prime sum involving Bernoulli numbers</a>, arXiv:2303.07934 [math.HO], 2023. See (23) p. 4.

%F a(n) = (p-2)*(p-1)*(p+1)*(2*p^2-2*p+3)/12 where p=prime(n).

%p a:= n-> (p-> (p-2)*(p-1)*(p+1)*(2*p^2-2*p+3)/12)(ithprime(n)):

%p seq(a(n), n=1..29); # _Alois P. Heinz_, Mar 15 2023

%o (PARI) a(n) = my(p=prime(n)); sum(k=1, p-1, k^5\p);

%Y Cf. A078837 (for k^3).

%K nonn

%O 1,2

%A _Michel Marcus_, Mar 15 2023