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A349792
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Numbers k such that k*(k+1) is the median of the primes between k^2 and (k+1)^2.
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2
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2, 3, 5, 6, 8, 25, 29, 38, 59, 101, 135, 217, 260, 295, 317, 455, 551, 686, 687, 720, 825, 912, 1193, 1233, 1300, 1879, 1967, 2200, 2576, 2719, 2857, 3303, 3512, 4215, 4241, 4448, 4658, 5825, 5932, 5952, 6155, 6750, 7275, 10305, 10323, 10962, 11279, 13495, 14104
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OFFSET
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1,1
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LINKS
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MATHEMATICA
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Select[Range@3000, Median@Select[Range[#^2, (#+1)^2], PrimeQ]==#(#+1)&] (* Giorgos Kalogeropoulos, Dec 05 2021 *)
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PROG
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(PARI) a349791(n) = {my(p1=nextprime(n^2), p2=precprime((n+1)^2), np1=primepi(p1), np2=primepi(p2), nm=(np1+np2)/2); if(denominator(nm)==1, prime(nm), (prime(nm-1/2)+prime(nm+1/2))/2)};
for(k=2, 5000, my(t=k*(k+1)); if(t==a349791(k), print1(k, ", ")))
(Python)
from sympy import primerange
from statistics import median
def ok(n): return n>1 and int(median(primerange(n**2, (n+1)**2)))==n*(n+1)
(Python)
from itertools import count, islice
from sympy import primepi, prime, nextprime
def A349792gen(): # generator of terms
p1 = 0
for n in count(1):
p2 = primepi((n+1)**2)
b = p1 + p2 + 1
if b % 2:
p = prime(b//2)
q = nextprime(p)
if p+q == 2*n*(n+1):
yield n
p1 = p2
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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