A071089 Remainder when sum of first n primes is divided by n-th prime.

0, 2, 0, 3, 6, 2, 7, 1, 8, 13, 5, 12, 33, 23, 46, 10, 27, 13, 32, 0, 55, 1, 44, 73, 90, 50, 28, 87, 63, 11, 69, 17, 70, 42, 41, 11, 72, 139, 75, 146, 44, 8, 9, 164, 88, 48, 7, 201, 121, 79, 224, 92, 46, 57, 170, 26, 145, 95, 216, 112, 58, 71, 293, 185, 129, 13, 255, 81, 128

Offset: 1

Author

Randy L. Ekl, May 26 2002

Keywords

• easy
• nonn

Comments

Conjecture: Every nonnegative integer can appear in the sequence at most finitely many times. - Thomas Ordowski, Jul 22 2013

I conjecture the opposite. Heuristically a given number should appear log log x times below x. - Charles R Greathouse IV, Jul 22 2013

In the first 10000 terms, one sees a(n) = n for n=2,7,12. Does this ever happen again? - J. M. Bergot, Mar 26 2018

Yes, it happens for n = 83408, too. - Michel Marcus, Mar 27 2018

Examples

			a[5] = 6 because s[5] = 2+3+5+7+11 = 28, p[5]=11 and q[5]= floor(28/11)=2, so a[5] = 28-11*2 = 6.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007506, A024011.

Programs

GAP

P:=Filtered([1..1000],IsPrime);
a:=List([1..70],i->Sum(P{[1..i]}) mod P[i]); # Muniru A Asiru, Mar 27 2018

Maple

s:= proc(n) option remember; `if`(n=0, 0, ithprime(n)+s(n-1)) end:
a:= n-> irem(s(n), ithprime(n)):
seq(a(n), n=1..100);  # Alois P. Heinz, Mar 27 2018

Mathematica

f[n_] := Mod[ Sum[ Prime[i], {i, 1, n - 1}], Prime[n]]; Table[ f[n], {n, 1, 70}] or
a[1] = 0; a[n_] := Block[{s = Sum[Prime[i], {i, 1, n}]}, s - Prime[n]*Floor[s/Prime[n]]]; Table[ f[n], {n, 1, 70}]
f[n_] := Mod[Plus @@ Prime@ Range@ n, Prime@ n]; Array[f, 70] (* Robert G. Wilson v, Nov 12 2016 *)
Module[{nn=70,t},t=Accumulate[Prime[Range[nn]]];Mod[#[[1]],#[[2]]]&/@ Thread[ {t,Prime[Range[nn]]}]] (* Harvey P. Dale, Sep 19 2019 *)

PARI

for(n=1,100,s=sum(i=1,n, prime(i)); print1(s-prime(n)*floor(s/prime(n)),","))

PARI

a(n) = vecsum(primes(n)) % prime(n); \\ Michel Marcus, Mar 27 2018

Formula

a(n) = s[n] - p[n]*q[n], where s[n] = sum of first n primes, p[n] is n-th prime and q[n] is floor(s[n]/p[n]).

a(A024011(n)) = 0. - Michel Marcus, Jan 22 2015

Extensions

Edited by Robert G. Wilson v and Benoit Cloitre, May 30 2002