A166010 -id:A166010 - 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

A182200 a(n) = prime(n)^2-3.

Original entry on oeis.org

1, 6, 22, 46, 118, 166, 286, 358, 526, 838, 958, 1366, 1678, 1846, 2206, 2806, 3478, 3718, 4486, 5038, 5326, 6238, 6886, 7918, 9406, 10198, 10606, 11446, 11878, 12766, 16126, 17158, 18766, 19318, 22198, 22798, 24646, 26566, 27886, 29926, 32038, 32758, 36478

Offset: 1

Bruno Berselli, Apr 17 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A001248, A049001, A061725, A066872, A084920, A166010.

Magma
```
[NthPrime(n)^2-3: n in [1..43]];
```

A182200:=n->ithprime(n)^2-3; seq(A182200(k),k=1..50); # Wesley Ivan Hurt, Oct 11 2013

Mathematica
```
Table[Prime[n]^2 - 3, {n, 43}]
```

a(n) = A061725(n)-5 = A066872(n)-4 = A001248(n)-3 = A084920(n)-2 = A049001(n)-1 = A166010(n)+1. [Formulas revised and extended by Bruno Berselli, Oct 15 2012]

A166011 Least common multiple of prime(n)-3 and prime(n)+3.

Original entry on oeis.org

5, 0, 8, 20, 56, 80, 140, 176, 260, 416, 476, 680, 836, 920, 1100, 1400, 1736, 1856, 2240, 2516, 2660, 3116, 3440, 3956, 4700, 5096, 5300, 5720, 5936, 6380, 8060, 8576, 9380, 9656, 11096, 11396, 12320, 13280, 13940, 14960, 16016, 16376, 18236, 18620

Offset: 1

Vladimir Joseph Stephan Orlovsky, Oct 04 2009

From Altug Alkan, Apr 22 2016: (Start)
For n > 1, a(n) is (p-3)*(p+3)/2 where p is the n-th prime. The reason is that the greatest common divisor of p-3 and p+3 is always 2 where p is the n-th prime and n > 2.
Proof: Let us assume that q is the greatest common divisor of p-3 and p+3. Because of the fact that any divisor of a and b must divide a-b, we know that q must divide 6. Note that q cannot be a multiple of 3 because p is prime, that is, q must be 1 or 2. Since we know that p-3 and p+3 are always even numbers for odd prime p, q must be 2 because we define it as the greatest common divisor.
If the greatest common divisor of p-3 and p+3 is always 2 where p is the n-th prime and n > 2, then the least common multiple of p-3 and p+3 must be (p-3)*(p+3)/2 where p is the n-th prime and n > 2 because of the general identity lcm(a, b) * gcd(a, b) = a*b. Note that for p = 3, (p-3)*(p+3)/t always is equal to 0 for any nonzero integer t, so it can be said that a(n) is (p-3)*(p+3)/2 where p is the n-th prime and n > 1. (End)

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A066830, A084921, A166010.

A166011:=n->lcm(ithprime(n)+3,ithprime(n)-3): seq(A166011(n), n=1..100); # Wesley Ivan Hurt, Apr 22 2016

f[n_]:=LCM[n-3,n+3]; lst={};Do[p=Prime[n];AppendTo[lst,f[p]],{n,5!}]; lst
LCM[#+3,#-3]&/@Prime[Range[50]] (* Harvey P. Dale, Aug 09 2015 *)

a(n) = lcm(prime(n)-3, prime(n)+3); \\ Michel Marcus, Apr 22 2016

cp's OEIS Frontend

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

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

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A182200 a(n) = prime(n)^2-3.

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A166011 Least common multiple of prime(n)-3 and prime(n)+3.

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Author

Keywords

Comments

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Programs

Maple

Mathematica

PARI