A050221 -id:A050221 - cp's OEIS frontend

A082661 Numbers n such that A082596(n) = A050221(n), sequence gives A060863(n).

2, 4, 52, 70, 95, 100, 124, 153, 169, 177, 201, 230, 252, 261, 272, 273, 275, 314, 316, 322, 377, 378, 384, 434, 450, 451, 507, 517, 527, 657, 702, 711, 720, 833, 884, 896, 904, 910, 930, 1005, 1025, 1047, 1081, 1137, 1189, 1202, 1228, 1351, 1358, 1415

Offset: 1

Naohiro Nomoto, May 18 2003

nonn

A122821 Number of ways n can be represented as the arithmetic mean of consecutive primes.

Original entry on oeis.org

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

Offset: 1

Ray Chandler, Sep 28 2006

nonn

Cf. A050221, A050237, A060863, A060864, A082370.

f[n_]:=Block[{i=1,j,c=0,m},While[Prime[i]<=n, j=1; While[m=Sum[Prime[k],{k,i,i+j-1}]/j; If[m==n,c++ ]; m

A082370 a(n) = number of sets of consecutive primes whose arithmetic mean is A000040(n).

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 2, 2, 1, 2, 2, 4, 1, 1, 1, 2, 1, 2, 1, 3, 2, 3, 1, 2, 1, 3, 3, 1, 2, 4, 3, 3, 5, 1, 1, 6, 2, 3, 1, 2, 1, 2, 3, 2, 3, 2, 1, 1, 2, 2, 2, 4, 2, 1, 2, 4, 3, 3, 3, 2, 2, 1, 2, 1, 4, 3, 5, 2, 1, 2, 1, 3, 1, 3, 1, 3, 3, 2, 3, 2, 3, 1, 1, 2, 1, 5, 2, 1, 2, 3, 1, 2, 1, 3, 3, 2, 1, 1, 5, 2, 2

Offset: 1

Naohiro Nomoto, May 11 2003

			For n=3; A000040(3) = 5. the two sets are 5/1 = 5, (3+5+7)/3 = 5. so a(3)=2.

Robert Israel, Table of n, a(n) for n = 1..4000

Cf. A050221, A060863, A082431, A122821.

N:= 300:
P:= [0,seq(ithprime(i),i=1..N)]:
S:= ListTools:-PartialSums(P):
mmax:= numtheory:-pi(floor(S[N]/N)):
V:= Vector(1..mmax,1):
for i from 1 to N+1 do
  for j from i+2 to N+1 do
    r:= (S[j]-S[i])/(j-i);
    if r::integer and isprime(r) then
      k:= numtheory:-pi(r);
      if k <= mmax then
        V[k]:= V[k]+1
      fi
    fi
od od:
convert(V,list); # Robert Israel, Mar 18 2018

a(n) = A122821(A000040(n)).

Extended by Ray Chandler, Oct 03 2006

A082431 a(n) = the smallest prime p such that there are exactly n sets of consecutive primes, each of which has an arithmetic mean of p.

Original entry on oeis.org

2, 5, 89, 53, 157, 173, 1597, 15233, 8803, 106753, 1570927, 5296771

Offset: 1

Naohiro Nomoto, May 11 2003

			a(4) = 53 because there are exactly four sets of consecutive primes which have means of 53: {53}, {47,53,59}, {41,...,67} and {31,...,73},

Cf. A050221, A050237, A060863, A082370.

{a(n)= m=2; starting_index=1; k=starting_index; sum_of_primes=0; prime_count=0; sets=0; until( (prime(starting_index)>m) && (sets==n), if( (prime(starting_index)>m) || (sets>n), m=nextprime(m+1); sets=0; starting_index=1; k=starting_index); sum_of_primes=sum_of_primes+prime(k); prime_count++; mean=sum_of_primes/prime_count; if(meanRick L. Shepherd, Jun 14 2004

Edited by Don Reble, Jun 17 2003

cp's OEIS Frontend

A082661 Numbers n such that A082596(n) = A050221(n), sequence gives A060863(n).

Views

Author

Keywords

A122821 Number of ways n can be represented as the arithmetic mean of consecutive primes.

Views

Author

Keywords

Crossrefs

Programs

Mathematica

A082370 a(n) = number of sets of consecutive primes whose arithmetic mean is A000040(n).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Formula

Extensions

A082431 a(n) = the smallest prime p such that there are exactly n sets of consecutive primes, each of which has an arithmetic mean of p.

Views

Author

Keywords

Examples

Crossrefs

Programs

PARI

Extensions