A225195 -id:A225195 - cp's OEIS frontend

A242380 Lesser of consecutive primes whose average is a perfect power.

3, 7, 61, 79, 139, 223, 317, 439, 619, 1087, 1669, 1723, 2593, 3593, 4093, 5179, 6079, 8461, 12541, 13687, 16633, 17573, 19037, 19597, 21943, 25261, 27211, 28219, 29581, 36857, 38011, 39199, 45361, 46649, 47521, 51977, 56167, 74527, 87013, 88801, 91807, 92413, 95479

Offset: 1

Antonio Roldán, May 12 2014

nonn

			4093 is in the sequence because 4093 and 4099 are consecutive primes and (4093 + 4099)/2 = 4096 = 2^12.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Supersequence of A225195 and A242382.

Cf. A091624.

Select[Partition[Prime[Range[2, 10^4]], 2, 1], GCD @@ FactorInteger[(First[#] + Last[#])/2][[;; , 2]] > 1 &][[;; , 1]] (* Amiram Eldar, Jul 04 2022 *)

for(i=3, 10^5, if(isprime(i), k=(i+nextprime(i+1))/2; if(ispower(k), print1(i, ", "))))

A225077 Smaller of the two consecutive primes whose sum is a triangular number.

Original entry on oeis.org

17, 37, 59, 103, 137, 149, 313, 467, 491, 883, 911, 1277, 1423, 1619, 1783, 2137, 2473, 2729, 4127, 4933, 5437, 5507, 6043, 6359, 10039, 10453, 11717, 13397, 15809, 17489, 20807, 21821, 23027, 27631, 28307, 28813, 29669, 33029, 36947, 39103, 44203, 48281

Offset: 1

Alex Ratushnyak, May 28 2013

nonn

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A175132 (numbers n such that sum of two consecutive primes is triangular(n)).
Cf. A181902 and A154634 (average of two consecutive primes is a triangular number).
Cf. A075190 and A225195 (average of two consecutive primes is a square).
Cf. A074924 and A061275 (sum of two consecutive primes is a square).

f:= proc(n) local m,p,q;
  m:= n*(n+1)/2;
  p:= prevprime(ceil(m/2));
  q:= nextprime(p);
  if p+q=m then p fi
end proc:
map(f, [$3..500]); # Robert Israel, May 04 2020

tri[n_] := IntegerQ[Sqrt[1 + 8 n]]; t = {}; p1 = 2; While[Length[t] < 50, p2 = NextPrime[p1]; If[tri[p1 + p2], AppendTo[t, p1]]; p1 = p2]; t (* T. D. Noe, May 28 2013 *)

a(n) + nextprime(a(n)) = A000217(A175132(n)).

A242382 Lesser of consecutive primes whose average is a perfect cube.

Original entry on oeis.org

61, 1723, 4093, 17573, 21943, 46649, 110587, 195103, 287491, 314423, 405221, 474547, 1061189, 1191013, 1404919, 1601609, 1906621, 2000371, 2146687, 2196979, 3241783, 3511799, 4912991, 5268017, 6229501, 6751267, 6858997, 7077883, 11239421, 20346407, 21951997, 26198063

Offset: 1

Antonio Roldán, May 12 2014

nonn

			1723 is in the sequence because it is prime, nextprime(1723) = 1733, and average(1723,1733) = 1728 = 12^3.

Amiram Eldar, Table of n, a(n) for n = 1..128

Subsequence of A077037 and A242380.

Cf. A225195, A091624.

Select[Partition[Prime[Range[2, 10^5]], 2, 1], IntegerQ[Surd[(First[#] + Last[#])/2, 3]] &][[;; , 1]] (* Amiram Eldar, Jul 04 2022 *)

{for(i=3,3*10^7,if(isprime(i),k=(i+nextprime(i+1))/2;if(ispower(k,3),print1(i,", "))))}

A365706 For n >= 1, a(n) is the least prime p such that the arithmetic mean of (n + 1) consecutive primes starting with p is a perfect square, or a(n) = -1 if no such p exists.

Original entry on oeis.org

3, 2393, 5, 827, 53, 271, 1063, 23993, 197, 29, 193, 2143, 359, 6829, 397, 17, 433, 661, 2837, 25171, 13597, 563, 10301, 1814233, 51427, 6781, 316817, 7477, 71, 238919, 11491, 3109, 42293, 38653, 6263, 13043, 474497, 21433, 13, 21419, 16963, 5119, 705209, 183761

Offset: 1

Ctibor O. Zizka, Oct 15 2023

nonn

Does a(n) exists for all n >= 1 ?
From David A. Corneth, Oct 18 2023: (Start)
Let s(n) be the sum of n + 1 consecutive primes starting with a(n). Then s(n)/(n+1) = m^2 for some positive integer m.
This means s(n) = (n + 1) * m^2. If n is even then m is odd if a(n) > 2.
As s(n) >= A007504(n) we have m^2 >= s(n)/(n+1) >= A007504(n)/(n+1) i.e. m >= sqrt(A007504(n)/(n+1)). So for some m we can see if m^2 * (n+1) is the sum of n+1 consecutive primes and if so a(n) is the smallest prime of these n+1 primes after testing all candidates up to m. (End)
s(n) = (n + 1)* a(n) + Sum_{i=0..(n-1)} (n-i)*g(i+1), thus we have Sum_{i=0..(n-1)} (n-i)*g(i+1) = (m^2 - a(n)) * (n + 1), g(j) are the n gaps between n + 1 consecutive primes. - Ctibor O. Zizka, Oct 18 2023

			n = 2: we search for the least prime(i) such that (prime(i) + prime(i + 1) + prime(i + 2))/3 = m^2, m an integer. This is valid for (2393 + 2399 + 2411)/3 = 49^2 thus a(2) = 2393.

David A. Corneth, Table of n, a(n) for n = 1..1000

Cf. A000040, A007504, A053001, A225195.

isok(x) = (denominator(x)==1) && issquare(x);
a(n) = my(k=1); while (!isok((vecsum(primes(k+n))-vecsum(primes(k-1)))/(n+1)), k++); prime(k); \\ Michel Marcus, Oct 16 2023

a(n) = {my(m = n + 1, ps = vector(m, i, prime(i)), s); forprime(p = ps[m] + 1, , s = vecsum(ps); if(!(s % m) && issquare(s/m), return(ps[1])); ps = concat(vecextract(ps, "^1"), p));} \\ Amiram Eldar, Oct 18 2023

More terms from Amiram Eldar, Oct 18 2023

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A242380 Lesser of consecutive primes whose average is a perfect power.

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A225077 Smaller of the two consecutive primes whose sum is a triangular number.

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A242382 Lesser of consecutive primes whose average is a perfect cube.

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A365706 For n >= 1, a(n) is the least prime p such that the arithmetic mean of (n + 1) consecutive primes starting with p is a perfect square, or a(n) = -1 if no such p exists.

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