A182174 -id:A182174 - cp's OEIS frontend

A084920 a(n) = (prime(n)-1)*(prime(n)+1).

3, 8, 24, 48, 120, 168, 288, 360, 528, 840, 960, 1368, 1680, 1848, 2208, 2808, 3480, 3720, 4488, 5040, 5328, 6240, 6888, 7920, 9408, 10200, 10608, 11448, 11880, 12768, 16128, 17160, 18768, 19320, 22200, 22800, 24648, 26568, 27888, 29928

Offset: 1

Reinhard Zumkeller, Jun 11 2003

Squares of primes minus 1. - Wesley Ivan Hurt, Oct 11 2013

Integers k for which there exist exactly two positive integers b such that (k+1)/(b+1) is an integer. - Benedict W. J. Irwin, Jul 26 2016

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Barry Brent, On the constant terms of certain meromorphic modular forms for Hecke groups, arXiv:2212.12515 [math.NT], 2022.
Barry Brent, On the Constant Terms of Certain Laurent Series, Preprints (2023) 2023061164.
Nik Lygeros and Olivier Rozier, A new solution to the equation tau(p) == 0 (mod p), J. Int. Seq. 13 (2010), Article 10.7.4.
Lực Ta, Enumeration of virtual quandles up to isomorphism, arXiv:2506.16536 [math.GT], 2025. See p. 6.

Cf. A000040, A005563, A049001, A154945, A166010, A182200, A182174.
Cf. A006093, A008864, A084921, A084922.
Cf. A066872, A001248, A024702, A013661, A065469.

a084920 n = (p - 1) * (p + 1) where p = a000040 n
-- Reinhard Zumkeller, Aug 27 2013

[p^2-1: p in PrimesUpTo(200)]; // Vincenzo Librandi, Mar 30 2015

A084920:=n->ithprime(n)^2-1; seq(A084920(k), k=1..50); # Wesley Ivan Hurt, Oct 11 2013

Table[Prime[n]^2 - 1, {n, 50}] (* Wesley Ivan Hurt, Oct 11 2013 *)
Prime[Range[50]]^2-1 (* Harvey P. Dale, Oct 02 2021 *)

a(n) = (prime(n)-1)*(prime(n)+1); \\ Michel Marcus, Jul 28 2016

[(p-1)*(p+1) for p in primes(200)] # Bruno Berselli, Mar 30 2015

a(n) = A006093(n) * A008864(n);
a(n) = A084921(n)*2, for n > 1; a(n) = A084922(n)*6, for n > 2.
Product_{n > 0} a(n)/A066872(n) = 2/5. a(n) = A001248(n) - 1. - R. J. Mathar, Feb 01 2009
a(n) = prime(n)^2 - 1 = A001248(n) - 1. - Vladimir Joseph Stephan Orlovsky, Oct 17 2009
a(n) ~ n^2*log(n)^2. - Ilya Gutkovskiy, Jul 28 2016
a(n) = (1/2) * Sum_{|k|<=2*sqrt(p)} k^2*H(4*p-k^2) where H() is the Hurwitz class number and p is n-th prime. - Seiichi Manyama, Dec 31 2017
a(n) = 24 * A024702(n) for n > 2. - Jianing Song, Apr 28 2019
Sum_{n>=1} 1/a(n) = A154945. - Amiram Eldar, Nov 09 2020
From Amiram Eldar, Nov 07 2022: (Start)
Product_{n>=1} (1 + 1/a(n)) = Pi^2/6 (A013661).
Product_{n>=1} (1 - 1/a(n)) = A065469. (End)

A049001 a(n) = prime(n)^2 - 2.

Original entry on oeis.org

2, 7, 23, 47, 119, 167, 287, 359, 527, 839, 959, 1367, 1679, 1847, 2207, 2807, 3479, 3719, 4487, 5039, 5327, 6239, 6887, 7919, 9407, 10199, 10607, 11447, 11879, 12767, 16127, 17159, 18767, 19319, 22199, 22799, 24647, 26567, 27887

Offset: 1

N. J. A. Sloane

Smallest numbers k such that k*prime(n)^2 + 1 is a square. - Bruno Berselli, Apr 19 2013

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Barry Brent, On the constant terms of certain meromorphic modular forms for Hecke groups, arXiv:2212.12515 [math.NT], 2022.
Barry Brent, On the Constant Terms of Certain Laurent Series, Preprints (2023) 2023061164.

Cf. A001248, A049002, A065481.

Cf. A084920, A166010, A182200; A182174.

a049001 = subtract 2 . a001248  -- Reinhard Zumkeller, Jul 30 2015

A049001:=n->ithprime(n)^2-2; seq(A049001(k), k=1..50); # Wesley Ivan Hurt, Oct 11 2013

Table[Prime[n]^2-2, {n, 50}] (* Wesley Ivan Hurt, Oct 11 2013 *)

a(n) = prime(n)^2 - 2; \\ Amiram Eldar, Nov 07 2022

a(n) = A001248(n) - 2.
a(n) = A182200(n) + 1. - Wesley Ivan Hurt, Oct 11 2013
Product_{n>=1} (1 - 1/a(n)) = A065481. - Amiram Eldar, Nov 07 2022

A166010 a(n) = prime(n)^2-4.

Original entry on oeis.org

0, 5, 21, 45, 117, 165, 285, 357, 525, 837, 957, 1365, 1677, 1845, 2205, 2805, 3477, 3717, 4485, 5037, 5325, 6237, 6885, 7917, 9405, 10197, 10605, 11445, 11877, 12765, 16125, 17157, 18765, 19317, 22197, 22797, 24645, 26565, 27885, 29925, 32037

Offset: 1

Vladimir Joseph Stephan Orlovsky, Oct 04 2009

Least common multiple of prime(n)-2 and prime(n)+2.

G. C. Greubel, Table of n, a(n) for n = 1..10000

Cf. A066830, A084921, A166011.

Cf. A049001, A084920, A182200; A182174.

[NthPrime(n)^2-4: n in [1..41]]; // Bruno Berselli, Apr 17 2012

f[n_]:=LCM[n-2,n+2]; lst={};Do[p=Prime[n];AppendTo[lst,f[p]],{n,5!}]; lst
Prime[Range[5!]]^2 - 4 (* Zak Seidov, Apr 17 2012 *)

a(n)=prime(n)^2-4 \\ Charles R Greathouse IV, Apr 17 2012

a(n) = A001248(n)-4 = A040976(n)*A052147(n). [Bruno Berselli, Apr 17 2012]

Definition rewritten by Bruno Berselli, Apr 17 2012

A182433 Smallest number such that the next n integers each have the square of one of the first n primes as a factor in order.

Original entry on oeis.org

7, 547, 29347, 1308247, 652312447, 180110691547, 65335225716547, 38733853511213647, 4368761145612023947, 1804216772228848838647, 14884872991210984993091647, 9816873967836575781598117447, 143397994078495393809327283088347

Offset: 2

Alonso del Arte, Apr 28 2012

nonn

These are found by an application of the Chinese remainder theorem. The remainders are the numbers prime(n)^2 - n (A182174), and the moduli are the squares of primes (A001248).
This guarantees a run of at least n nonsquarefree numbers. But just as n! + 1 guarantees a run of at least n - 1 composite numbers, this might not be the smallest run of n nonsquarefree numbers (for that, see A045882).
Marmet credits Erick Bryce Wong with the idea of applying the Chinese remainder theorem and a sieving process to obtain upper limits for squarefree gaps. From this it occurred to me to just apply the Chinese remainder theorem to find these squarefree gaps exhibiting the squares of primes in order.
Also, beyond a(4), that is n > 4, we will observe that some of the numbers in the run of nonsquarefree numbers are divisible by more than one prime power, e.g., a(n) + 5 is divisible both by 49 (the square of the fourth prime) and 4.

			a(3) = 547 as that is the solution to the simultaneous congruences x = 3 mod 4 = 7 mod 9 = 22 mod 25. We verify that the next 3 integers meet the requirement: 548 = 4 * 137, 549 = 9 * 61, 550 = 25 * 2 * 11.
a(4) = 29347 as that is the solution to the simultaneous congruences x = 3 mod 4 = 7 mod 9 = 22 mod 25 = 45 mod 49. We verify that the next 4 integers meet the requirement: 29348 = 4 * 11 * 23 * 29, 29349 = 9 * 3 * 1087, 29350 = 25 * 2 * 587, 29351 = 49 * 599.

A. Bogomolny, Chinese Remainder Theorem from Interactive Mathematics Miscellany and Puzzles
Louis Marmet, First occurrences of squarefree gaps...

Cf. A069021, A051681.

Table[ChineseRemainder[Prime[Range[n]]^2 - Range[n], Prime[Range[n]]^2], {n, 2, 14}]

cp's OEIS Frontend

A084920 a(n) = (prime(n)-1)*(prime(n)+1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

Sage

Formula

A049001 a(n) = prime(n)^2 - 2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Formula

A166010 a(n) = prime(n)^2-4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A182433 Smallest number such that the next n integers each have the square of one of the first n primes as a factor in order.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica