A050482 Sum of remainders when n-th prime is divided by all preceding integers.

0, 1, 4, 8, 22, 28, 51, 64, 98, 151, 167, 233, 297, 325, 403, 505, 635, 645, 790, 904, 923, 1113, 1244, 1422, 1654, 1800, 1888, 2056, 2098, 2256, 2849, 3066, 3326, 3450, 3969, 4045, 4329, 4696, 5014, 5325, 5767, 5759, 6499, 6565, 6898

Offset: 1

Author

Brian Wallace (wallacebrianedward(AT)yahoo.co.uk), Dec 26 1999

Keywords

• nonn

Comments

a(n)/(n*log(n))^2 appears to approach a constant ~0.22... for large n. - Benedict W. J. Irwin, Dec 07 2016

Irwin's comment is incorrect. - Bill McEachen, Feb 04 2024. [Indeed, according to the first formula in A004125, a(n)/(n*log(n))^2 approaches a constant, which is not 0.22 but 1-Pi^2/12 = 0.1775... - Amiram Eldar, Feb 04 2024]

Examples

			a(4) = 8 because remainders when 7 is divided by 1..6 are 0,1,1,3,2,1, which add to 8.
a(2) = 3 mod (3-1) = 1.
a(3) = (5 mod (5-1)) + (5 mod (5-2)) + (5 mod (5-3)) = 2 + 1 + 1 = 4.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A004125, A034953.

Programs

Maple

A050482 := proc(n) local a,i; a := 0; for i from 1 to ithprime(n)-1 do a := a+(ithprime(n) mod i); od: end;

Mathematica

Table[Sum[Mod[Prime[n],k],{k,Prime[n]-1}],{n,45}] (* James C. McMahon, Feb 08 2024 *)

PARI

a(n)=my(p=prime(n));sum(k=2, p, p%k) \\ Charles R Greathouse IV, Jun 03 2013

Python

from math import isqrt
from sympy import prime
def A050482(n): return (p:=prime(n))**2+((s:=isqrt(p))**2*(s+1)-sum((q:=p//k)*((k<<1)+q+1) for k in range(1,s+1))>>1) # Chai Wah Wu, Nov 01 2023

Formula

a(n) = A004125(A000040(n)). - R. J. Mathar, Jun 12 2009