A069232 -id:A069232 - cp's OEIS frontend

A069230 Number of primes p such that n < p < n + tau(n)^2 where tau(n) = A000005(n).

0, 2, 1, 3, 1, 5, 0, 5, 3, 5, 1, 10, 0, 4, 4, 6, 1, 9, 0, 8, 3, 4, 0, 14, 2, 4, 4, 9, 1, 14, 0, 8, 4, 4, 4, 19, 0, 4, 4, 15, 1, 14, 0, 8, 8, 4, 0, 19, 1, 8, 3, 8, 0, 14, 3, 14, 4, 5, 1, 29, 0, 3, 7, 11, 4, 13, 0, 8, 4, 13, 1, 27, 0, 3, 8, 8, 3, 13, 0, 19, 5, 3, 0, 26, 2, 3, 3, 13, 0, 27, 3, 7, 4, 5, 5

Offset: 1

Benoit Cloitre, Apr 13 2002

			a(12) = 10 as there are 10 primes between (exclusive) 12 and 12 + tau(12)^2 = 12 + 6^2 = 12 + 36 = 48 namely the 10 primes 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. - _David A. Corneth_, Sep 20 2020

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A000005, A069231, A069232, A069233.

Table[Count[Range[n+1,n+DivisorSigma[0,n]^2-1],?PrimeQ],{n,100}] (* _Harvey P. Dale, Oct 25 2021 *)

a(n) = #select(x->isprime(x), vector(numdiv(n)^2-1, k, k+n)); \\ Michel Marcus, Jun 18 2017

A069231 Numbers n such that there are exactly 3 primes p satisfying the inequality n < p < n + tau(n)^2 where tau(n) = A000005(n).

Original entry on oeis.org

4, 9, 21, 51, 55, 62, 74, 77, 82, 86, 87, 91, 106, 122, 123, 129, 134, 142, 143, 145, 146, 155, 158, 159, 161, 177, 183, 214, 215, 217, 237, 249, 254, 259, 265, 274, 278, 298, 299, 301, 309, 334, 335, 339, 341, 343, 358, 365, 371, 377, 382, 386, 394, 395, 407

Offset: 1

Benoit Cloitre, Apr 13 2002

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

Cf. A000005, A049591, A069230, A069232, A069233.

filter:= n -> nops(select(isprime, [$(n+1) .. (n+numtheory:-tau(n)^2-1)]))=3:
select(filter, [$1..1000]); # Robert Israel, Jan 05 2018

fQ[n_] := Block[{r = Range[n, n + DivisorSigma[0, n]^2]}, If[ PrimeQ@ n, r = Rest@ r]; If[ PrimeQ[ r[[-1]]], r = Most@ r]; Length@ Select[r, PrimeQ] == 3]; Select[Range@410, fQ] (* Robert G. Wilson v, Jan 05 2018 *)

isok(n) = #select(x->isprime(x), vector(numdiv(n)^2-1, k, k+n)) == 3; \\ Michel Marcus, Jun 18 2017

A069233 Numbers k such that there is exactly 1 prime p satisfying the inequality k < p < k + tau(k)^2 where tau(k) = A000005(k).

Original entry on oeis.org

3, 5, 11, 17, 29, 41, 49, 59, 71, 101, 107, 111, 115, 121, 137, 149, 169, 179, 191, 197, 201, 202, 203, 205, 206, 227, 239, 269, 281, 287, 289, 291, 295, 311, 314, 319, 321, 347, 361, 403, 419, 431, 461, 469, 471, 505, 521, 526, 527, 569, 599, 617, 622, 623

Offset: 1

Benoit Cloitre, Apr 13 2002

Numbers k such that A069230(k) = 1. - Amiram Eldar, Jan 29 2025

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

Cf. A000005, A049591, A069230, A069231, A069232.

q[k_] := Count[k + Range[1, DivisorSigma[0, k]^2 - 1], ?PrimeQ] == 1; Select[Range[640], q] (* _Amiram Eldar, Jan 29 2025 *)

isok(n) = #select(x->isprime(x), vector(numdiv(n)^2-1, k, k+n)) == 1; \\ Michel Marcus, Jun 18 2017

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A069230 Number of primes p such that n < p < n + tau(n)^2 where tau(n) = A000005(n).

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A069231 Numbers n such that there are exactly 3 primes p satisfying the inequality n < p < n + tau(n)^2 where tau(n) = A000005(n).

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A069233 Numbers k such that there is exactly 1 prime p satisfying the inequality k < p < k + tau(k)^2 where tau(k) = A000005(k).

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