A103208 -id:A103208 - cp's OEIS frontend

A111287 a(n) = smallest k such that prime(n) divides Sum_{i=1..k} prime(i).

1, 10, 2, 5, 8, 49, 4, 23, 23, 7, 39, 29, 6, 10, 39, 25, 30, 151, 38, 19, 139, 27, 174, 21, 287, 422, 240, 24, 94, 22, 16, 173, 861, 231, 143, 140, 213, 902, 18, 134, 143, 310, 70, 58, 12, 550, 237, 210, 229, 57, 221, 271, 194, 540, 145, 718, 116, 184, 90, 14, 168, 455, 61, 454

Offset: 1

N. J. A. Sloane, Nov 03 2005

nonn

It follows from a theorem of Daniel Shiu that k always exists. Shiu has proved that if (a,b) = 1 then the arithmetic progression a, a + b, ..., a + k*b, ... contains arbitrarily long sequences of consecutive primes. Since, for any positive integer b, there are thus arbitrarily long sequences of consecutive primes congruent to 1 mod b, there must be infinitely many a(n) that are divisible by b.

To clarify the previous comment: If the sum of the primes up to some point is s (mod b), then we need exactly b-s consecutive primes equal to 1 (mod b) to produce a sum divisible by b. Hence when there are b-1 consecutive primes congruent to 1 (mod b), then the sum of primes up to one of those primes will be divisible by b. - T. D. Noe, Dec 02 2009

			A007504 begins 2,5,10,17,28,41,58,77,100,129,... and the k=10th term is the first one that is divisible by prime(2) = 3, so a(2) = 10 (see also A103208).

T. D. Noe, Table of n, a(n) for n = 1..10000
D. K. L. Shiu, Strings of Congruent Primes, J. Lond. Math. Soc. 61 (2) (2000) 359-373 [MR1760689]

Cf. A000041, A007504, A053050, A111267, A111272, A103208, A168678.

read transforms; M:=1000; p0:=[seq(ithprime(i),i=1..M)]; q0:=PSUM(p0); w:=[]; for n from 1 to M do p:=p0[n]; hit := 0; for i from 1 to M do if q0[i] mod p = 0 then w:=[op(w),i]; hit:=1; break; fi; od: if hit = 0 then break; fi; od: w;

Table[p=Prime[n]; s=0; k=0; While[k++; s=Mod[s+Prime[k],p]; s>0]; k, {n,10}] (* T. D. Noe, Dec 02 2009 *)

A111287(n)= n=Mod(0,prime(n)); for(k=1,1e9, (n+=prime(k)) || return(k)) \\ M. F. Hasler, Nov 29 2009

The comments are based on correspondence with Paul Pollack and a posting to sci.math by Fred Helenius.

Typo in reference fixed by David Applegate, Dec 18 2009

A327449 Number of ways the first n primes can be partitioned into three sets with equal sums.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 423, 0, 2624, 0, 13474, 0, 0, 0, 611736, 0, 4169165, 0, 30926812, 0, 214975174, 0, 1590432628, 0, 11431365932, 0, 83946004461, 0, 0, 0, 4615654888831, 0, 35144700468737, 0, 271133285220726, 0, 2103716957561013, 0, 0, 0, 0, 0, 990170108748552983, 0, 7855344215856348141

Offset: 1

N. J. A. Sloane, Sep 19 2019

nonn

It is not true that a(2k+1) is always 0.

			One of the three solutions for n = 10: 3 + 17 + 23 = 2 + 5 + 7 + 29 = 11 + 13 + 19.

Keith F. Lynch, Posting to Math Fun Mailing List, Sep 17 2019.

Alois P. Heinz, Table of n, a(n) for n = 1..102

Cf. A000040, A007504, A022894, A112972, A275714, A103208, A111320, A113263, A327448, A327450.

s:= proc(n) option remember; `if`(n<2, 0, ithprime(n)+s(n-1)) end:
b:= proc(n, x, y) option remember; `if`(n=1, 1, (p-> (l->
      add(`if`(p>l[i], 0, b(n-1, sort(subsop(i=l[i]-p, l))
      [1..2][])), i=1..3))([x, y, s(n)-x-y]))(ithprime(n)))
    end:
a:= n-> `if`(irem(2+s(n), 3, 'q')=0, b(n, q-2, q)/2, 0):
seq(a(n), n=1..40);  # Alois P. Heinz, Sep 19 2019

s[n_] := s[n] = If[n < 2, 0, Prime[n] + s[n - 1]];
b[n_, x_, y_] := b[n, x, y] = If[n == 1, 1, Function[p, Function[l, Sum[If[ p > l[[i]], 0, b[n - 1, Sequence @@ Sort[ReplacePart[l, i -> l[[i]] - p]][[1;; 2]]]], {i, 1, 3}]][{x, y, s[n] - x - y}]][Prime[n]]];
a[n_] := If[Mod[2+s[n], 3]==0, q = Quotient[2+s[n], 3]; b[n, q-2, q]/2, 0];
Array[a, 40] (* Jean-François Alcover, Apr 09 2020, after Alois P. Heinz *)

EqSumThreeParts(v)={ my(n=#v, vs=vector(n), m=vecsum(v)/3, brk=0);
  for(i=1, n-1, vs[i+1]=vs[i]+v[i]; if(vs[i]<=min(1000,m), brk=i));
  my(q=Vecrev(prod(i=1, brk, 1+x^v[i]+y^v[i])));
  my(recurse(k,s,p)=if(k==brk, if(s<#q, polcoef(p*q[s+1],m,y)), if(s<=vs[k], self()(k-1, s, p*(1 + y^v[k]))) + if(s>=v[k], self()(k-1, s-v[k], p)) ));
  if(frac(m), 0, recurse(n-1, m, 1 + O(y*y^m))/2);
}
a(n)={EqSumThreeParts(primes(n))} \\ Andrew Howroyd, Sep 19 2019

a(n) > 0 <=> n in { A103208 }, with odd n in { A111320 }. - Alois P. Heinz, Sep 19 2019

Corrected and a(30)-a(52) added by Andrew Howroyd, Sep 19 2019

a(53) and beyond from Alois P. Heinz, Sep 19 2019

A053095 Number of primes having exactly the same digits as appear in first n primes.

Original entry on oeis.org

1, 1, 1, 8, 62, 568, 5370, 114546, 2898058, 0, 1119268465, 26053674813

Offset: 1

Enoch Haga, Mar 19 2000

a(n) = 0 if sum of digits is a multiple of three (see A103208). - Ray Chandler, Apr 07 2010

			a(4) = 8 because the first four prime n, concatenated, are 2+3+5+7 or 2357. There are 8 prime arrangements: 2357, 2753, 3257, 3527, 5237, 5273, 7253, 7523.

Cf. A054260, A053160, A053161, A103208.

from sympy import isprime, prime
from sympy.utilities.iterables import multiset_permutations
def A053095(n):
    return sum(1 for d in multiset_permutations(''.join(str(prime(m+1)) for m in range(n))) if isprime(int(''.join(d)))) # Chai Wah Wu, Mar 17 2019

a(7)-a(9) from Ray Chandler, Apr 07 2010
a(10) from Ray Chandler, Apr 08 2010
a(11) from Michael S. Branicky, Mar 17 2025
a(12) from Michael S. Branicky, Apr 07 2025

A102066 Sum of the first n primes, mod 6.

Original entry on oeis.org

2, 5, 4, 5, 4, 5, 4, 5, 4, 3, 4, 5, 4, 5, 4, 3, 2, 3, 4, 3, 4, 5, 4, 3, 4, 3, 4, 3, 4, 3, 4, 3, 2, 3, 2, 3, 4, 5, 4, 3, 2, 3, 2, 3, 2, 3, 4, 5, 4, 5, 4, 3, 4, 3, 2, 1, 0, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 3, 4, 3, 2, 3, 2, 3, 2, 3, 2, 3, 4, 3

Offset: 1

Alan Sutcliffe (alansut(AT)ntlworld.com), Dec 28 2004

Curiosity, it takes almost 60 terms before 1 and 0 appear. Stability arises from near equal (often alternating) occurrences of p mod 6 = 1 and p mod 6 = 5.

			a(5) = 4 since (2+3+5+7+11) mod 6 = 4.

H. L. Nelson, "Prime Sums", J. Rec. Math., 14 (1981), 205-206.

Cf. A007504, A103208.

Mod[Accumulate[Prime[Range[120]]],6] (* Harvey P. Dale, Jun 19 2014 *)

a(n) = sum(i=1, n, prime(i)) % 6 \\ Michel Marcus, Jul 12 2013

sum(p(n)) mod 6.

More terms from Harvey P. Dale, Jun 19 2014

cp's OEIS Frontend

A111287 a(n) = smallest k such that prime(n) divides Sum_{i=1..k} prime(i).

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A327449 Number of ways the first n primes can be partitioned into three sets with equal sums.

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Maple

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A053095 Number of primes having exactly the same digits as appear in first n primes.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Python

Extensions

A102066 Sum of the first n primes, mod 6.

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Author

Keywords

Comments

Examples

References

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions