A102655 -id:A102655 - cp's OEIS frontend

A075540 Integers that are the average of three successive primes.

5, 53, 157, 173, 211, 257, 263, 373, 511, 537, 563, 593, 607, 653, 733, 947, 977, 999, 1073, 1103, 1123, 1187, 1223, 1239, 1367, 1461, 1501, 1511, 1541, 1747, 1753, 1763, 1773, 1899, 1907, 1917, 2071, 2181, 2287, 2401, 2409, 2417, 2449, 2677, 2903, 2963

Offset: 1

Zak Seidov, Sep 21 2002

nonn

Not every three successive primes have an integer average. The integer averages are in the sequence.

Not all of these 3-averages are prime: the prime 3-averages are in A006562 (balanced primes). There are surprisingly many prime 3-averages: among the first 10000 terms of the sequence there are 2417 primes. Indices i(n) of first prime in sequence of three primes with integer average are in A075541, for prime 3-averages i(n) are in A064113. Interprimes (s-averages with s=2) are all composite, see A024675. (Edited by Zak Seidov, Sep 01 2015 )

			a(1) = 5 = (1/3)(3+5+7), first integer average of three successive primes; next is: a(2) = 53 = (1/3)(47 + 53 + 59); up to n=8 all terms are prime; while a(9) = 511 = (1/3)( 503 + 509 + 521) is the first nonprime 3-average: 511=7*73.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A006562, A024675, A075541, A064113.

Cf. A102655.

a075540 n = a075540_list !! (n-1)
a075540_list = map fst $ filter ((== 0) . snd) $
   zipWith3 (\x y z -> divMod (x + y + z) 3)
            a000040_list (tail a000040_list) (drop 2 a000040_list)
-- Reinhard Zumkeller, Jan 20 2012

N:= 10^4: # to get all terms using primes <= N
Primes:= select(isprime,[2,seq(2*i+1, i=1..(N-1)/2)]):
select(type,(Primes[1..-3] + Primes[2..-2] + Primes[3..-1])/3,integer); # Robert Israel, Sep 01 2015

Select[MovingAverage[Prime[Range[500]],3],IntegerQ] (* Harvey P. Dale, Aug 10 2012 *)

a(n) = (1/3) (p(i)+p(i+1)+p(i+2)), for some i(n).

Comment and example edited, inefficient Mma removed by Zak Seidov, Sep 01 2015

A123086 Numbers which are the arithmetic mean of 1000000 consecutive primes.

Original entry on oeis.org

11860710, 19524458, 30466003, 57980974, 63924288, 90876871, 98124660, 100711080, 107813124, 130902871, 130920140, 133345096, 137765645, 149928192, 187600902, 214934436, 223349020, 235566127, 238768532, 239934902, 264189816, 270196572, 278851320, 285021997

Offset: 1

Zak Seidov, Sep 27 2006

nonn

Corresponding indices of the first primes are: 275775, 740092, 1383476, 2948575, 3280201, 4764532, 5159226, 5299723, 5684491, 6926061, 6926985, 7056669, 7292768, 7940227, 9929283, 11358606, 11796712, 12431386, 12597486, 12657959, 13911879, 14221421, 14666768, 14983910, 15100050.

			11860710 is in the sequence since (p(275775) + p(275776) + ... + p(275775+999998) + p(275775+999999)) / 1000000 = 11860710 where p(n) is the n-th prime.

Zak Seidov, Table of n, a(n) for n = 1..467

Cf. A102655 (arithmetic mean of four successive primes).
Cf. A122040 (arithmetic mean of six successive primes).
Cf. A122480 (arithmetic mean of k consecutive primes).
Cf. A122808 (arithmetic mean of n, but no fewer, consecutive primes).

Timing[s = 7472966967499 ; a = 2; b = 15485863; Do[s = s - a + (b = NextPrime[b]); a = NextPrime[a]; If[Mod[s, 10^6] < 1, Print[s/10^6]], {10^8}]]

{s = 7472966967499 ; a = 2; b = 15485863; for (k = 1, 10^9,
if(s%10^6 < 1, print( s/10^6)); b = nextprime (b + 2);
s = s - a + b; a = nextprime (a + 1))}

A226153 Numbers n such that triangular(n) is an average of 4 consecutive primes.

Original entry on oeis.org

5, 10, 14, 15, 22, 34, 49, 54, 64, 66, 81, 93, 104, 116, 121, 122, 146, 154, 156, 180, 194, 221, 222, 236, 270, 299, 320, 332, 334, 337, 346, 360, 369, 371, 374, 387, 416, 417, 429, 435, 444, 472, 492, 498, 511, 520, 551, 556, 617, 622, 637, 654, 657, 670, 674, 677, 680

Offset: 1

Alex Ratushnyak, May 28 2013

nonn

Zak Seidov, Table of n, a(n) for n = 1..3000

Cf. A102655, A226151.

#include 
#include 
#include 
#define TOP (1ULL<<30)
int main() {
  unsigned long long i, j, p1, p2, p3, r, s;
  unsigned char *c = (unsigned char *)malloc(TOP/8);
  memset(c, 0, TOP/8);
  for (i=3; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0 /*&& i<(1ULL<<32)*/)
        for (j=i*i>>1; j>3] |= 1 << (j&7);
  for (p3=2, p2=3, p1=5, i=7; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
      s = p3 + p2 + p1 + i;
      if (s%4==0) {
        s/=4;
        r = sqrt(s*2);
        if (r*(r+1)==s*2) printf("%llu, ", r);
      }
      p3 = p2, p2 = p1, p1 = i;
    }
  return 0;
}

A034963 := proc(n)
    add(ithprime(i), i=n..n+3) ;
end proc:
istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return t1 ; else return -1; end if; end;
for n from 1 to 90000 do
    s := A034963(n)/4 ;
    if type(s,'integer') then
    tr := istriangular(s) ;
    if tr >= 0  then
        printf("%d, ", tr) ;
    end if;
    end if;
end do: # R. J. Mathar, Jun 06 2013

Module[{nn=30000,ntrs,m},ntrs=Table[{n,(n(n+1))/2},{n,nn}];m=Mean/@Partition[Prime[ Range[ nn]],4,1];Select[ntrs,MemberQ[m,#[[2]]]&]][[;;,1]] (* Harvey P. Dale, Jun 08 2023 *)
(Sqrt[8#+1]-1)/2&/@Select[Mean/@Partition[Prime[Range[25000]],4,1],OddQ[Sqrt[8#+1]]&] (* Harvey P. Dale, Sep 17 2024 *)

A123078 Integers that are the arithmetic mean of 1000 consecutive primes.

Original entry on oeis.org

10820, 15682, 19672, 23340, 25067, 33873, 37387, 39377, 47782, 59635, 75249, 82619, 86378, 87303, 89198, 92217, 97495, 111989, 131987, 133441, 162286, 164644, 178666, 181532, 182196, 200088, 203450, 215280, 215612, 218527, 221526, 229733

Offset: 1

Zak Seidov, Sep 27 2006

nonn

Corresponding indices of the first primes are: 812, 1323, 1731, 2097, 2268, 3121, 3456, 3645, 4428, 5518, 6914, 7569, 7899, 7980, 8146, 8412, 8875, ... .

			(p(812)+p(813)+...+p(812+998)+p(812+999))/1000=10820, p(n)=n-th prime.

Zak Seidov, Table of n, a(n) for n = 1..1000

Cf. A102655, A122040, A122480, A122808.

Select[Table[Sum[Prime[k], {k, n, n + 999}]/1000, {n, 50000}], IntegerQ] (* Elizabeth A. Blickley (Elizabeth.Blickley(AT)gmail.com), Oct 10 2006 *)
Select[Mean/@Partition[Prime[Range[25000]],1000,1],IntegerQ] (* Harvey P. Dale, Jul 27 2022 *)

Conjecture: a(n) ~ 1000 n log n. - Charles R Greathouse IV, Oct 23 2012

More terms from Elizabeth A. Blickley (Elizabeth.Blickley(AT)gmail.com), Oct 10 2006

A126096 Primes that are the arithmetic mean of four successive primes.

Original entry on oeis.org

127, 139, 149, 181, 241, 431, 967, 1021, 1031, 1061, 1597, 1759, 1913, 2113, 2437, 2593, 2833, 2953, 3769, 3793, 3947, 4219, 4261, 4463, 4603, 5011, 5869, 5923, 6449, 6701, 6959, 7103, 7489, 7549, 7727, 8273, 8803, 8839, 9137, 9241, 9421, 9931, 10069

Offset: 1

Zak Seidov, Mar 03 2007

nonn

			181 is in the sequence because it is prime and is the arithmetic mean of the consecutive primes 173, 179, 181 and 191.

Zak Seidov, Table of n, a(n) for n = 1..3000

Cf. A102655 (numbers that are the arithmetic mean of four successive primes).

a:=proc(n) local nn: nn:=(ithprime(n)+ithprime(n+1)+ithprime(n+2)+ithprime(n+3))/4: if type(nn,integer)=true and isprime(nn)=true then nn else fi end: seq(a(n),n=1..1300); # Emeric Deutsch, Mar 07 2007

lst={};Do[If[PrimeQ[p=(Prime[n]+Prime[n+1]+Prime[n+2]+Prime[n+3])/4],AppendTo[lst,p]],{n,8!}];lst (* Vladimir Joseph Stephan Orlovsky, Jul 28 2009 *)
pr=Prime[Range[114000(* for first 3015 terms *)]];
Select[Mean/@Partition[pr,4,1],PrimeQ] (* Zak Seidov, Apr 22 2011 *)

More terms from Emeric Deutsch, Mar 07 2007

A122037 Largest prime factor of number equal to the arithmetic mean of four successive primes.

Original entry on oeis.org

3, 3, 5, 3, 11, 5, 19, 7, 23, 11, 5, 17, 3, 29, 17, 7, 3, 19, 5, 127, 139, 3, 149, 31, 11, 7, 181, 31, 13, 5, 43, 19, 29, 241, 19, 23, 13, 53, 5, 139, 97, 17, 13, 53, 37, 31, 11, 17, 11, 127, 43, 19, 7, 71, 431, 7, 149, 11, 79, 37, 163, 11, 127, 257, 41, 271, 23, 43, 113, 19, 3

Offset: 1

Giovanni Teofilatto, Sep 14 2006

nonn

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A006530, A102655.

FactorInteger[ # ][[ -1, 1]] & /@ Select[Table[Sum[Prime[k], {k, n, n + 3}]/4, {n, 120}], IntegerQ] (* Ray Chandler, Sep 25 2006 *)
FactorInteger[#][[-1,1]]&/@Select[Mean/@Partition[Prime[Range[ 200]], 4,1], IntegerQ] (* Harvey P. Dale, May 21 2014 *)

a(n) = A006530(A102655(n)).

Extended by Ray Chandler, Sep 25 2006

A226152 Numbers n such that n^2 is an average of 4 consecutive primes.

Original entry on oeis.org

3, 9, 12, 21, 24, 35, 72, 126, 129, 179, 189, 194, 198, 214, 243, 253, 255, 279, 304, 322, 432, 443, 480, 487, 511, 523, 663, 681, 696, 699, 711, 717, 721, 734, 738, 796, 802, 838, 910, 975, 1008, 1034, 1070, 1144, 1215, 1230, 1237, 1265, 1276, 1370, 1375, 1386, 1469

Offset: 1

Alex Ratushnyak, May 28 2013

nonn

Integers of the form sqrt(A102655(k)) for any k. - R. J. Mathar, Jun 06 2013

Cf. A102655, A051395, A206280.

#include 
#include 
#include 
#define TOP (1ULL<<30)
int main() {
  unsigned long long i, j, p1, p2, p3, r, s;
  unsigned char *c = (unsigned char *)malloc(TOP/8);
  memset(c, 0, TOP/8);
  for (i=3; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0 /*&& i<(1ULL<<32)*/)
        for (j=i*i>>1; j>3] |= 1 << (j&7);
  for (p3=2, p2=3, p1=5, i=7; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
      s = p3 + p2 + p1 + i;
      if (s%4==0) {
        s/=4;
        r = sqrt(s);
        if (r*r==s) printf("%llu, ", r);
      }
      p3 = p2, p2 = p1, p1 = i;
    }
  return 0;
}

A034963 := proc(n)
    add(ithprime(i), i=n..n+3) ;
end proc:
for n from 1 to 90000 do
    s := A034963(n)/4 ;
    if type(s,'integer') then
    if issqr(s) then
        printf("%d, ", sqrt(s)) ;
    end if;
    end if;
end do: # R. J. Mathar, Jun 06 2013

a(n) = A051395(n)/2.

A226155 Smallest of four consecutive primes whose average is a triangular number.

Original entry on oeis.org

11, 47, 101, 109, 241, 587, 1217, 1481, 2069, 2203, 3313, 4357, 5443, 6779, 7351, 7489, 10723, 11927, 12239, 16267, 18911, 24517, 24733, 27953, 36571, 44839, 51347, 55249, 55931, 56941, 60017, 64951, 68239, 68993, 70117, 75041, 86719, 87133, 92227, 94819, 98773, 111611

Offset: 1

Alex Ratushnyak, May 28 2013

nonn

Cf. A102655, A226151 A226153, A226154.

#include 
#include 
#include 
#define TOP (1ULL<<30)
int main() {
  unsigned long long i, j, p1, p2, p3, r, s;
  unsigned char *c = (unsigned char *)malloc(TOP/8);
  memset(c, 0, TOP/8);
  for (i=3; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0 /*&& i<(1ULL<<32)*/)
        for (j=i*i>>1; j>3] |= 1 << (j&7);
  for (p3=2, p2=3, p1=5, i=7; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
      s = p3 + p2 + p1 + i;
      if (s%4==0) {
        s/=4;
        r = sqrt(s*2);
        if (r*(r+1)==s*2) printf("%llu, ", p3);
      }
      p3 = p2, p2 = p1, p1 = i;
    }
  return 0;
}

A000217inv:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return t1 ; else return -1; end if; end;
isA226155 := proc(n)
    local p1,p2,p3,a102655 ;
    if isprime(n) then
        p1 := nextprime(n) ;
        p2 := nextprime(p1) ;
        p3 := nextprime(p2) ;
        a102655 := (n+p1+p2+p3)/4 ;
        if type(a102655,'integer') then
            if A000217inv(a102655) >= 0 then
                return true;
            else
                return false;
            end if;
        else
            return false;
        end if;
    else
        false;
    end if;
end proc:
for n from 1 do
    p := ithprime(n) ;
    if isA226155(p) then
        printf("%d,\n",p) ;
    end if;
end do: # R. J. Mathar, Jun 06 2013

cp's OEIS Frontend

A075540 Integers that are the average of three successive primes.

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A123086 Numbers which are the arithmetic mean of 1000000 consecutive primes.

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A226153 Numbers n such that triangular(n) is an average of 4 consecutive primes.

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A123078 Integers that are the arithmetic mean of 1000 consecutive primes.

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A126096 Primes that are the arithmetic mean of four successive primes.

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A122037 Largest prime factor of number equal to the arithmetic mean of four successive primes.

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A226152 Numbers n such that n^2 is an average of 4 consecutive primes.

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A226155 Smallest of four consecutive primes whose average is a triangular number.