This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
lst={}; s=0; Do[p=Prime[n]; s=s+p; r=s/p; If[r==IntegerPart[r], AppendTo[lst, r]], {n, 1, 10^6}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 07 2008 *)
2 divides 2; 5 divides 2 + 3 + 5; 71 divides 2 + 3 + 5 + 7 + ... + 61 + 67 + 71; etc.
sumOfPrimes = 0; Do[ sumOfPrimes += p; If[ Divisible[ sumOfPrimes, p], Print[p]], {p, Prime /@ Range[23000000]}] (* Jean-François Alcover, Oct 22 2012 *)
Transpose[Module[{nn=23000000,pr},pr=Prime[Range[nn]];Select[Thread[ {Accumulate[ pr], pr}], Divisible[#[[1]],#[[2]]]&]]][[2]] (* Harvey P. Dale, Feb 09 2013 *)
s=0;forprime(p=2,1e9,s+=p;if(s%p==0,print1(p", "))) \\ Charles R Greathouse IV, Jul 22 2013
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.
P:=Filtered([1..1000],IsPrime);
a:=List([1..70],i->Sum(P{[1..i]}) mod P[i]); # Muniru A Asiru, Mar 27 2018
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
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 *)
for(n=1,100,s=sum(i=1,n, prime(i)); print1(s-prime(n)*floor(s/prime(n)),","))
a(n) = vecsum(primes(n)) % prime(n); \\ Michel Marcus, Mar 27 2018
a(6) = 859 because 859 is the smallest odd prime that divides the sum of 6 + (sum of all primes smaller than itself). a(8) = 41 because 8+2+3+5+7+11+13+17+19+23+29+31+37+41 = 246 and 246/41 = 6.
f[n_] := Block[{p = 3, s = n +2}, While[ Mod[s, p] != 0, s = s + p; p = NextPrime@ p]; p]; Array[f, 47, 0] (* Robert G. Wilson v, Nov 12 2016 *)
# see link for an alternate that searches in parallel to a limit from sympy import nextprime def a(n): psum, p = 2, 3 while (n + psum)%p: psum, p = psum + p, nextprime(p) return p for n in range(12): print(a(n), end=", ") # Michael S. Branicky, May 03 2021
The first 4 composite numbers are: 4, 6, 8 and 9. As 4 + 6 + 8 + 9 = 3*9, 4 belongs to the sequence.
s=0; k=0; forcomposite (c=4, oo, k++; s+=c; if (s%c==0, print1 (k", ")))
a(2) = 129 = A007504(10), a(3) = 381 = A007504(16).
print1(s=0); forprime(p=2,1e4,s+=p; if(s%3==0, print1(", "s))) \\ Charles R Greathouse IV, Nov 03 2014
a(13)=3 because the 13th prime is 41 and the sum of primes up to 41 is 238, which has 3 distinct prime factors less than 41. a(20)=1 because the 20th prime is 71 and the sum of primes up to 71 is 639 = 7*71, which has only 1 distinct prime factor less than 71. - _Jon E. Schoenfield_, Apr 11 2018
s:= proc(n) option remember; `if`(n<1, 0, ithprime(n)+s(n-1)) end: a:= n-> nops(select(x-> x < ithprime(n), numtheory[factorset](s(n)))): seq(a(n), n=1..100); # Alois P. Heinz, Apr 11 2018
a[n_] := (S = Total[P = Prime[Range[n]]]; Count[P, p_ /; Divisible[S, p]]); Array[a, 100] (* Jean-François Alcover, Apr 30 2019 *)
a(n) = #select(x->(x < prime(n)), factor(sum(k=1, n, prime(k)))[,1]); \\ Michel Marcus, Apr 11 2018
a(4)=37 because prime(4)=7 divides the sum of 4 consecutive primes starting with 37 (37+41+43+47=168), but does not divide any earlier sum of 4 consecutive primes.
P:= [0,seq(ithprime(i),i=1..100000)]:
S:= ListTools:-PartialSums(P):
f:= proc(n) local p,k;
p:= ithprime(n);
for k from 1 to nops(S)-n do
if S[k+n]-S[k] mod p = 0 then
return P[k+1]
fi
od;
FAIL
end proc:
map(f, [$1..200]);
a[n_] := Block[{m=Prime@n, s=Sum[Prime@i, {i, n}], p=2, q}, q=m; While[Mod[s, m] > 0, s-=p; {p, q} = NextPrime@{p, q}; s+=q]; p]; Array[a, 70] (* Giovanni Resta, Oct 10 2019 *)
The subtotal for the primes up to 23 is 100. Between 23 and the next prime 29 there are the composites 24, 25, 26, 27 and 28 that are candidates for evenly dividing this subtotal of 100. We have 25 within this group and it does divide that subtotal of 100. Moving on to the subtotal up to 31, which is 160, we have the interval of composites 32, 33, 34, 35 and 36 as possible divisors of this current subtotal of 160. The number 32 divides this subtotal.
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