Emirhan Üçok - cp's OEIS frontend

User: Emirhan Üçok

Emirhan Üçok has authored 2 sequences.

A385821 Numbers k such that ceiling((k^2 + 1)/2) is prime.

2, 3, 5, 6, 9, 11, 12, 15, 18, 19, 25, 29, 35, 39, 42, 45, 48, 49, 51, 54, 59, 60, 61, 65, 66, 69, 71, 72, 79, 84, 85, 90, 95, 101, 121, 131, 132, 139, 141, 144, 145, 150, 159, 165, 169, 171, 174, 175, 181, 186, 192, 195, 198, 199, 201, 204, 205, 209, 210, 219, 221, 231, 245, 246

Offset: 1

Emirhan Üçok, Jul 09 2025

			5 is a term since ceiling((5^2 + 1)/2) = 13, which is prime.
8 is not a term since ceiling((8^2 + 1)/2) = 33, which is not prime.

James C. McMahon, Table of n, a(n) for n = 1..10000

Positions of primes in A080827.

Select[Range[246],PrimeQ[Ceiling[(#^2+1)/2]]&] (* James C. McMahon, Jul 15 2025 *)

isok(k) = isprime(ceil((k^2+1)/2)); \\ Michel Marcus, Jul 10 2025

from sympy import isprime
def ok(k): return isprime((k*k + 2) // 2)
print([k for k in range(1, 247) if ok(k)])

A385466 Primes that are at the end of the local maxima in the sequence of consecutive prime gaps.

Original entry on oeis.org

11, 17, 29, 37, 67, 79, 97, 107, 127, 137, 149, 191, 197, 239, 251, 277, 307, 331, 347, 367, 397, 419, 431, 439, 457, 479, 499, 521, 541, 557, 587, 631, 673, 701, 719, 751, 769, 787, 809, 821, 827, 853, 877, 907, 929, 967, 991, 1009, 1019, 1031, 1049, 1061, 1087

Offset: 1

Emirhan Üçok, Jun 29 2025

This sequence lists the larger prime in each consecutive prime pair where the difference is a local maximum in the sequence of prime gaps.

A local maximum occurs when p(n)-p(n-1) < p(n+1)-p(n) > p(n+2)-p(n+1) where p(n) is the n-th prime.

			The primes 7 and 11 differ by 4, which is larger than the previous gap (2) and the next gap (2). So 11 is in the sequence.

Cf. A001223 (prime gaps), A198696.

Module[{primes = Prime[Range[1, 200]], diffs, res = {}}, diffs = Differences[primes];
Do[If[diffs[[i]] > diffs[[i - 1]] && diffs[[i]] > diffs[[i + 1]],
AppendTo[res, primes[[i + 1]]]], {i, 2, Length[diffs] - 1}]; res]

from sympy import primerange
primes = list(primerange(2, 2000))
diffs = [primes[i+1] - primes[i] for i in range(len(primes)-1)]
local_max_asals = []
for i in range(1, len(diffs)-1):
    if diffs[i] > diffs[i-1] and diffs[i] > diffs[i+1]:
        local_max_asals.append(primes[i+1])
print(local_max_asals[:70])

a(n) = prime(A198696(n)+1). - Michel Marcus, Jul 01 2025

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User: Emirhan Üçok

A385821 Numbers k such that ceiling((k^2 + 1)/2) is prime.

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