A085692 -id:A085692 - cp's OEIS frontend

A087399 Duplicate of A085692.

25, 121, 5041

Offset: 1

dead

A005563 a(n) = n*(n+2) = (n+1)^2 - 1.

0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499, 2600

Offset: 0

N. J. A. Sloane

Erdős conjectured that n^2 - 1 = k! has a solution if and only if n is 5, 11 or 71 (when k is 4, 5 or 7).
Second-order linear recurrences y(m) = 2y(m-1) + a(n)*y(m-2), y(0) = y(1) = 1, have closed form solutions involving only powers of integers. - Len Smiley, Dec 08 2001
Number of edges in the join of two cycle graphs, both of order n, C_n * C_n. - Roberto E. Martinez II, Jan 07 2002
Let k be a positive integer, M_n be the n X n matrix m_(i,j) = k^abs(i-j) then det(M_n) = (-1)^(n-1)*a(k-1)^(n-1). - Benoit Cloitre, May 28 2002
Also numbers k such that 4*k + 4 is a square. - Cino Hilliard, Dec 18 2003
For each term k, the function sqrt(x^2 + 1), starting with 1, produces an integer after k iterations. - Gerald McGarvey, Aug 19 2004
a(n) mod 3 = 0 if and only if n mod 3 > 0: a(A008585(n)) = 2; a(A001651(n)) = 0; a(n) mod 3 = 2*(1-A079978(n)). - Reinhard Zumkeller, Oct 16 2006
a(n) is the number of divisors of a(n+1) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007
Nonnegative X values of solutions to the equation X^3 + X^2 = Y^2. To find Y values: b(n) = n(n+1)(n+2). - Mohamed Bouhamida, Nov 06 2007
Sequence allows us to find X values of the equation: X + (X + 1)^2 + (X + 2)^3 = Y^2. To prove that X = n^2 + 2n: Y^2 = X + (X + 1)^2 + (X + 2)^3 = X^3 + 7*X^2 + 15X + 9 = (X + 1)(X^2 + 6X + 9) = (X + 1)*(X + 3)^2 it means: (X + 1) must be a perfect square, so X = k^2 - 1 with k>=1. we can put: k = n + 1, which gives: X = n^2 + 2n and Y = (n + 1)(n^2 + 2n + 3). - Mohamed Bouhamida, Nov 12 2007
From R. K. Guy, Feb 01 2008: (Start)
Toads and Frogs puzzle:
This is also the number of moves that it takes n frogs to swap places with n toads on a strip of 2n + 1 squares (or positions, or lily pads) where a move is a single slide or jump, illustrated for n = 2, a(n) = 8 by
   T T - F F
   T - T F F
   T F T - F
   T F T F -
   T F - F T
   - F T F T
   F - T F T
   F F T - T
   F F - T T
I was alerted to this by the Holton article, but on consulting Singmaster's sources, I find that the puzzle goes back at least to 1867.
Probably the first to publish the number of moves for n of each animal was Edouard Lucas in 1883. (End)
a(n+1) = terms of rank 0, 1, 3, 6, 10 = A000217 of A120072 (3, 8, 5, 15). - Paul Curtz, Oct 28 2008
Row 3 of array A163280, n >= 1. - Omar E. Pol, Aug 08 2009
Final digit belongs to a periodic sequence: 0, 3, 8, 5, 4, 5, 8, 3, 0, 9. - Mohamed Bouhamida, Sep 04 2009 [Comment edited by N. J. A. Sloane, Sep 24 2009]
Let f(x) be a polynomial in x. Then f(x + n*f(x)) is congruent to 0 (mod f(x)); here n belongs to N. There is nothing interesting in the quotients f(x + n*f(x))/f(x) when x belongs to Z. However, when x is irrational these quotients consist of two parts, a) rational integers and b) integer multiples of x. The present sequence represents the non-integer part when the polynomial is x^2 + x + 1 and x = sqrt(2), f(x+n*f(x))/f(x) = A056108(n) + a(n)*sqrt(2). - A.K. Devaraj, Sep 18 2009
For n >= 1, a(n) is the number for which 1/a(n) = 0.0101... (A000035) in base (n+1). - Rick L. Shepherd, Sep 27 2009
For n > 0, continued fraction [n, 1, n] = (n+1)/a(n); e.g., [6, 1, 6] = 7/48. - Gary W. Adamson, Jul 15 2010
Starting (3, 8, 15, ...) = binomial transform of [3, 5, 2, 0, 0, 0, ...]; e.g., a(3) = 15 = (1*3 + 2*5 +1*2) = (3 + 10 + 2). - Gary W. Adamson, Jul 30 2010
a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_0(n) = 2*n-n^2 and a(n) = -P_0(n+2). See also A067998 and for the case k=1 A080956. - Peter Luschny, Jul 08 2011
a(n) is the maximal determinant of a 2 X 2 matrix with integer elements from {1, ..., n+1}, so the maximum determinant of a 2x2 matrix with integer elements from {1, ..., 5} = 5^2 - 1 = a(4) = 24. - Aldo González Lorenzo, Oct 12 2011
Using four consecutive triangular numbers t1, t2, t3 and t4, plot the points (0, 0), (t1, t2), and (t3, t4) to create a triangle. Twice the area of this triangle are the numbers in this sequence beginning with n = 1 to give 8. - J. M. Bergot, May 03 2012
Given a particle with spin S = n/2 (always a half-integer value), the quantum-mechanical expectation value of the square of the magnitude of its spin vector evaluates to  = S(S+1) = n(n+2)/4, i.e., one quarter of a(n) with n = 2S. This plays an important role in the theory of magnetism and magnetic resonance. - Stanislav Sykora, May 26 2012
Twice the harmonic mean [H(x, y) = (2*x*y)/(x + y)] of consecutive triangular numbers A000217(n) and A000217(n+1). - Raphie Frank, Sep 28 2012
Number m such that floor(sqrt(m)) = floor(m/floor(sqrt(m))) - 2 for m > 0. - Takumi Sato, Oct 10 2012
The solutions of equation 1/(i - sqrt(j)) = i + sqrt(j), when i = (n+1), j = a(n). For n = 1, 2 + sqrt(3) = 3.732050.. = A019973. For n = 2, 3 + sqrt(8) = 5.828427... = A156035. - Kival Ngaokrajang, Sep 07 2013
The integers in the closed form solution of a(n) = 2*a(n-1) + a(m-2)*a(n-2), n >= 2, a(0) = 0, a(1) = 1 mentioned by Len Smiley, Dec 08 2001, are m and -m + 2 where m >= 3 is a positive integer. - Felix P. Muga II, Mar 18 2014
Let m >= 3 be a positive integer. If a(n) = 2*a(n-1) + a(m-2) * a(n-2), n >= 2, a(0) = 0, a(1) = 1, then lim_{n->oo} a(n+1)/a(n) = m. - Felix P. Muga II, Mar 18 2014
For n >= 4 the Szeged index of the wheel graph W_n (with n + 1 vertices). In the Sarma et al. reference, Theorem 2.7 is incorrect. - Emeric Deutsch, Aug 07 2014
If P_{k}(n) is the n-th k-gonal number, then a(n) = t*P_{s}(n+2) - s*P_{t}(n+2) for s=t+1. - Bruno Berselli, Sep 04 2014
For n >= 1, a(n) is the dimension of the simple Lie algebra A_n. - Wolfdieter Lang, Oct 21 2015
Finding all positive integers (n, k) such that n^2 - 1 = k! is known as Brocard's problem, (see A085692). - David Covert, Jan 15 2016
For n > 0, a(n) mod (n+1) = a(n) / (n+1) = n. - Torlach Rush, Apr 04 2016
Conjecture: When using the Sieve of Eratosthenes and sieving (n+1..a(n)), with divisors (1..n) and n>0, there will be no more than a(n-1) composite numbers. - Fred Daniel Kline, Apr 08 2016
a(n) mod 8 is periodic with period 4 repeating (0,3,0,7), that is a(n) mod 8 = 5/2 - (5/2) cos(n*Pi) - sin(n*Pi/2) + sin(3*n*Pi/2). - Andres Cicuttin, Jun 02 2016
Also for n > 0, a(n) is the number of times that n-1 occurs among the first (n+1)! terms of A055881. - R. J. Cano, Dec 21 2016
The second diagonal of composites (the only prime is number 3) from the right on the Klauber triangle (see Kival Ngaokrajang link), which is formed by taking the positive integers and taking the first 1, the next 3, the following 5, and so on, each centered below the last. - Charles Kusniec, Jul 03 2017
Also the number of independent vertex sets in the n-barbell graph. - Eric W. Weisstein, Aug 16 2017
Interleaving of A000466 and A033996. - Bruce J. Nicholson, Nov 08 2019
a(n) is the number of degrees of freedom in a triangular cell for a Raviart-Thomas or Nédélec first kind finite element space of order n. - Matthew Scroggs, Apr 22 2020
From Muge Olucoglu, Jan 19 2021: (Start)
For n > 1, a(n-2) is the maximum number of elements in the second stage of the Quine-McCluskey algorithm whose minterms are not covered by the functions of n bits. At n=3, we have a(3-2) = a(1) = 1*(1+2) = 3 and f(A,B,C) = sigma(0,1,2,5,6,7).
.
          0 1 2   5 6 7
        +---------------
  *(0,1)| X X
   (0,2)| X   X
   (1,5)|   X     X
  *(2,6)|     X     X
  *(5,7)|         X   X
   (6,7)|           X X
.
*: represents the elements that are covered. (End)
1/a(n) is the ratio of the sum of the first k odd numbers and the sum of the next n*k odd numbers. - Melvin Peralta, Jul 15 2021
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {1, 2n}]. - Magus K. Chu, Sep 09 2022
Number of diagonals parallel to an edge in a regular (2*n+4)-gon (cf. A367204). - Paolo Xausa, Nov 21 2023
For n >= 1, also the number of minimum cyclic edge cuts in the (n+2)-trapezohedron graph. - Eric W. Weisstein, Nov 21 2024
For n >= 1, a(n) is the sum of the interior angles of a polygon with n+2 sides, in radians, multiplied by (n+2)/Pi. - Stuart E Anderson, Aug 06 2025

			G.f. = 3*x + 8*x^2 + 15*x^3 + 24*x^4 + 35*x^5 + 48*x^6 + 63*x^7 + 80*x^8 + ...

E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see index under Toads and Frogs Puzzle.
Martin Gardner, Perplexing Puzzles and Tantalizing Teasers, p. 21 (for "The Dime and Penny Switcheroo").
R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.
Derek Holton, Math in School, 37 #1 (Jan 2008) 20-22.
Edouard Lucas, Récréations Mathématiques, Gauthier-Villars, Vol. 2 (1883) 141-143.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Spencer Daugherty, Pamela E. Harris, Ian Klein, and Matt McClinton, Metered Parking Functions, arXiv:2406.12941 [math.CO], 2024. See p. 22.
R. K. Guy, Letter to N. J. A. Sloane, May 1990.
R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
Kival Ngaokrajang, Klauber triangle
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
K. V. S. Sarma and I. V. N. Uma, On Szeged index of standard graphs, International J. of Math. Archive, 3(8), 2012, 3129-3135. - _Emeric Deutsch_, Aug 07 2014
Stanislav Sýkora, Magnetic Resonance on OEIS, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.
Leo Tavares, Illustration: Bounded Squares.
Leo Tavares, Illustration: Trapezoids.
Leo Tavares, Illustration: Trapagons.
Eric Weisstein's World of Mathematics, Barbell Graph.
Eric Weisstein's World of Mathematics, Cyclic Edge Cut.
Eric Weisstein's World of Mathematics, Independent Vertex Set.
Eric Weisstein's World of Mathematics, Near-Square Prime.
Eric Weisstein's World of Mathematics, Trapezohedral Graph.
Wikipedia, Quine-McCluskey Algorithm.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

A column of triangle A102537.
a(n+1), n>=2, first column of triangle A120070.
Cf. A000035, A000217, A000290, A000466, A001651, A002378, A005563.
Cf. A007531, A008585, A008833, A013468, A016742, A019973, A028560.
Cf. A033996, A046092, A053186, A054000, A055881, A056108, A062196.
Cf. A068310, A067725, A067728, A067998, A069817, A079978, A080956.
Cf. A085692, A120072, A123865, A123866, A123867, A123868, A134582.
Cf. A140091, A140681, A143053, A144396, A156035, A163280, A212331.
Cf. A253607, A367204.

a005563 n = n * (n + 2)
a005563_list = zipWith (*) [0..] [2..]  -- Reinhard Zumkeller, Dec 16 2012

[n*(n+2): n in [0..60]]; // G. C. Greubel, Mar 29 2024

Table[n^2 - 1, {n, 42}] (* Zerinvary Lajos, Mar 21 2007 *)
ListCorrelate[{1, 2}, Range[-1, 50], {1, -1}, 0, Plus, Times] (* Harvey P. Dale, Aug 29 2015 *)
Range[20]^2 - 1 (* Eric W. Weisstein, Aug 16 2017 *)
Table[n (n + 2), {n, 20}] (* Eric W. Weisstein, Nov 21 2024 *)
CoefficientList[Series[(-3 + x)/(-1 + x)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Nov 21 2024 *)
LinearRecurrence[{3, -3, 1}, {3, 8, 15}, 20] (* Eric W. Weisstein, Nov 21 2024 *)

makelist(n*(n+2), n, 0, 56); /* Martin Ettl, Oct 15 2012 */

a(n)=n*(n+2) \\ Charles R Greathouse IV, Dec 22 2011

concat(0, Vec(x*(3-x)/(1-x)^3 + O(x^90))) \\ Altug Alkan, Oct 22 2015

[n*(n+2) for n in range(61)] # G. C. Greubel, Mar 29 2024

G.f.: x*(3-x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = A000290(n+1) - 1.
A002378(a(n)) = A002378(n)*A002378(n+1); e.g., A002378(15)=240=12*20. - Charlie Marion, Dec 29 2003
a(n) = A067725(n)/3. - Zerinvary Lajos, Mar 06 2007
a(n) = Sum_{k=1..n} A144396(k). - Zerinvary Lajos, May 11 2007
a(n) = A134582(n+1)/4. - Zerinvary Lajos, Feb 01 2008
A143053(a(n)) = A000290(n+1), for n > 0. - Reinhard Zumkeller, Jul 20 2008
a(n) = Real((n+1+i)^2). - Gerald Hillier, Oct 12 2008
A053186(a(n)) = 2*n. - Reinhard Zumkeller, May 20 2009
a(n) = (n! + (n+1)!)/(n-1)!, n > 0. - Gary Detlefs, Aug 10 2009
a(n) = floor(n^5/(n^3+1)) with offset 1 (a(1)=0). - Gary Detlefs, Feb 11 2010
a(n) = a(n-1) + 2*n + 1 (with a(0)=0). - Vincenzo Librandi, Nov 18 2010
Sum_{n>=1} 1/a(n) = 3/4. - Mohammad K. Azarian, Dec 29 2010
a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^(n-1)*(cos(x))^3), for n > 0. - Francesco Daddi, Aug 02 2011
a(n) = A002378(n) + floor(sqrt(A002378(n))); pronic number + its root. - Fred Daniel Kline, Sep 16 2011
a(n-1) = A008833(n) * A068310(n) for n > 1. - Reinhard Zumkeller, Nov 26 2011
G.f.: U(0) where U(k) = -1 + (k+1)^2/(1 - x/(x + (k+1)^2/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 19 2012
a(n) = 15*C(n+4,3)*C(n+4,5)/(C(n+4,2)*C(n+4,4)). - Gary Detlefs, Aug 05 2013
a(n) = (n+2)!/((n-1)! + n!), n > 0. - Ivan N. Ianakiev, Nov 11 2013
a(n) = 3*C(n+1,2) - C(n,2) for n >= 0. - Felix P. Muga II, Mar 11 2014
a(n) = (A016742(n+1) - 4)/4 for n >= 0. - Felix P. Muga II, Mar 11 2014
a(-2 - n) = a(n) for all n in Z. - Michael Somos, Aug 07 2014
A253607(a(n)) = 1. - Reinhard Zumkeller, Jan 05 2015
E.g.f.: x*(x + 3)*exp(x). - Ilya Gutkovskiy, Jun 03 2016
For n >= 1, a(n^2 + n - 2) = a(n-1) * a(n). - Miko Labalan, Oct 15 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/4. - Amiram Eldar, Nov 04 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = 2.
Product_{n>=1} (1 - 1/a(n)) = -sqrt(2)*sin(sqrt(2)*Pi)/Pi. (End)
a(n) = A000290(n+2) - n*2. See Bounded Squares illustration. - Leo Tavares, Oct 05 2021
From Leo Tavares, Oct 10 2021: (Start)
a(n) = A008585(n) + 2*A000217(n-1). See Trapezoids illustration.
2*A005563 = A054000(n+1). See Trapagons illustration.
a(n) = 2*A000217(n) + n. (End)
a(n) = (n+2)!!/(n-2)!! for n > 1. - Jacob Szlachetka, Jan 02 2022

Partially edited by Joerg Arndt, Mar 11 2010

More terms from N. J. A. Sloane, Aug 01 2010

A146968 Brocard's problem: positive integers n such that n!+1 = m^2.

Original entry on oeis.org

4, 5, 7

Offset: 1

Marco Bellaccini (marcomurk(AT)tele2.it), Nov 03 2008

No other terms below 10^9.

See A085692 for more comments and references. - M. F. Hasler, Nov 20 2018

			7! + 1 = 5041 = 71^2, hence 7 is in the sequence. - _Klaus Brockhaus_, Nov 05 2008

Berndt, B. C. and Galway, W. F. On the Brocard-Ramanujan Diophantine Equation n!+1=m^2, The Ramanujan Journal, March 2000, Volume 4, Issue 1, pp 41-42.
Apoloniusz Tyszka, On sets X subset of N for which we know an algorithm that computes a threshold number t(X) in N such that X is infinite if and only if X contains an element greater than t(X), 2019.
Eric Weisstein's World of Mathematics, Brocard's Problem.

A085692, A146968, A216071 are all essentially the same sequence. - N. J. A. Sloane, Sep 01 2012

[ : n in [1..8047] | t where t,p:=IsSquare(Factorial(n)+1) ]; // Klaus Brockhaus, Nov 05 2008

Select[Range[10],IntegerQ[Sqrt[#!+1]]&] (* Harvey P. Dale, Jan 31 2015 *)

{ for (n=1, 60100, if(issquare(n!+1) == 1, print(n) ) ) } \\ Marco Bellaccini (marcomurk(AT)tele2.it), Nov 08 2008

#!/bin/sh n=0 while(true) do n=`echo $n + 1 | bc` calc "($n! + 1)" ^ "(1 / 2)" | grep -v \. done

Edited by Max Alekseyev, Feb 06 2010

A227644 Perfect powers equal to the sum of 2 factorial numbers.

Original entry on oeis.org

4, 8, 25, 121, 144, 5041

Offset: 1

Giovanni Resta, Jul 19 2013

a(7), if it exists, is greater than 10^100.
a(7), if it exists, is greater than 10000!. - Filip Zaludek, Jul 18 2017
a(7), if it exists, is greater than 11750!. - Filip Zaludek, Sep 07 2018
a(7), if it exists, is greater than 20000!. - Filip Zaludek, Nov 04 2020

			5041 = 71^2 = 1! + 7!.

Giovanni Resta, Decompositions of a(1)-a(6)

Cf. A114377, A227645, A227646, A227647, A227648, A227649, A227650, A227651, A065433, A085692.

/* To compile: gcc -Wall -O2 A227644.c -o A227644 -lgmp */
#include 
#include 
#include 
int main()
{
   int bsz=256, a=0;
   mpz_t *f, t;
   f = malloc(sizeof(mpz_t) * bsz);
   mpz_init(t); mpz_init(f[0]); mpz_set_ui(f[0], 1);
   while (1)
   {
       a += 1;
       if (a == bsz)
       {
           bsz *= 2;
           f = (mpz_t *) realloc(f, sizeof(mpz_t) * bsz);
       }
       mpz_init(f[a]);
       mpz_mul_ui(f[a], f[a-1], a);
       for (int i=1; i<=a; i++)
       {
           mpz_add(t, f[a], f[i]);
           if (mpz_perfect_power_p(t))
           {
               gmp_printf("%Zd, ", t);
               fflush(stdout);
           }
       }
   }
   return 0;
}

A064237 Numbers k such that k! + 1 is divisible by a square.

Original entry on oeis.org

4, 5, 7, 12, 23

Offset: 1

Vladeta Jovovic, Sep 22 2001

229 is another term because 613^2 divides 229!+1. See A115091 for primes whose square divides m!+1 for some m. An examination of the factorizations of m!+1 for m<=100 found no additional squares. - T. D. Noe, Mar 01 2006
562 is also a term because 562!+1 is divisible by 563^2. - Vladeta Jovovic, Mar 30 2004
A web search reveals that for 1 <= k <= 228 there are 82 values of k for which k! + 1 has not been completely factored (the smallest is k=103), so showing that 229 and 562 are indeed the next two terms will be a huge task. I checked that k!+1 is not divisible by p^2 for k <= 1000 and prime p < 10^8. - Francois Brunault, Nov 23 2008
It is very likely that 229 and 562 are the next two terms, but this has not yet been proved. - Francois Brunault, Nov 29 2008
Contains A007540(n)-1 for all n.  That sequence is conjectured to be infinite. - Robert Israel, Jul 04 2016
This sequence includes A146968 (solutions of Brocard's problem). - Salvador Cerdá, Mar 08 2016
If k > 562 and k! + 1 is divisible by p^2 where p is prime, then either k > 10000 or p > 2038074743 (the hundred millionth prime). - Jason Zimba, Oct 21 2021

			4 is in the sequence because 4! + 1 = 5^2.
5 is in the sequence because 5! + 1 = 11^2.
6 is not in the sequence because 6! + 1 = 721
7 is in the sequence because 7! + 1 = 71^2.
12 is in the sequence because 12! + 1 = 13^2 * 2834329.
23 is a term because 23!+1 = 47^2*79*148139754736864591.
From _Thomas Richard_, Aug 31 2021: (Start)
229 and 562 are terms because
229!+1 = 613^2 * 38669 * 1685231 * 3011917759 * (417-digit composite)
562!+1 = 563^2 * 64467346976659839517037 * 112870688711507255213769871 * 63753966393108716329397432599379239 * (1214-digit prime). (End)

Hisanori Mishima, Factorizations of m!+1

Cf. A007540 (Wilson primes), A115091, A146968, A038507, A085692.

remove(t -> numtheory:-issqrfree(t!+1), [$1..50]); # Robert Israel, Jul 04 2016

Flatten[Position[MoebiusMu[Range[30]!+1], 0]]; (* T. D. Noe, Mar 01 2006, Nov 21 2008 *)

lista(nn) = for(n=1, nn, if(!issquarefree(n!+1), print1(n, ", "))); \\ Altug Alkan, Mar 08 2016

Example corrected by T. D. Noe, Nov 26 2008

A216071 Brocard's problem: positive integers m such that m^2 = n! + 1 for some n.

Original entry on oeis.org

5, 11, 71

Offset: 1

V. Raman, Sep 01 2012

See A085692 and A146968 for links, references and comments. - M. F. Hasler, Nov 20 2018

Wikipedia, Brocard's problem

A085692, A146968, A216071 are all essentially the same sequence. - N. J. A. Sloane, Sep 01 2012

Sqrt[#!+1]&/@Select[Range[1000],IntegerQ[Sqrt[#!+1]]&] (* Harvey P. Dale, Sep 29 2012 *)

apply( sqrtint, A085692) \\ M. F. Hasler, Nov 20 2018

select( is_A216071(m)=m^2==A084558(m^2)!+1, [0..99]) \\ M. F. Hasler, Nov 20 2018

a(n) = A000196(A085692(n)) = A000196(A038507(A146968(n))) where A000196 = sqrt and A038507(n) = n! + 1. - M. F. Hasler, Nov 20 2018

A181764 Numbers n such that n!+1 is a product of two distinct prime numbers.

Original entry on oeis.org

6, 8, 10, 13, 14, 19, 20, 24, 25, 26, 28, 34, 38, 48, 54, 55, 59, 71, 75, 92, 109, 114, 115

Offset: 1

Vladimir Joseph Stephan Orlovsky, Nov 13 2010

n! + 1 must be the product of two distinct prime numbers and also the product of only two prime numbers counted with multiplicity. Thus, 12 is NOT a term of the sequence because 12! + 1 = 13*13*2834329. - Harvey P. Dale, Jul 22 2019

Other terms in this sequence: 392, 551, 601, 770, 772, 878, 1033, 1320, 1831, 2620, 2808, 3752, 4233, 4616, 4984, 7260. - Chai Wah Wu, Feb 28 2020

			6!+1=7*103; 8!+1=61*661; 10!+1=11*329891; 13!+1=83*75024347; 14!+1=23*3790360487; 19!+1=71*1713311273363831;..

Bruce C. Berndt and William F. Galway, On the Brocard-Ramanujan diophantine equation n!+1=m^2, The Ramanujan Journal, March 2000, Volume 4, Issue 1, pp 41-42.

Cf. A033312, A038507, A078781, A085692, A085747, A088332.

Subsequence of A078778.

fQ[n_]:=Last/@FactorInteger[n]=={1,1}; Select[Range[40], fQ[#!+1]&]

Extended by D. S. McNeil, Nov 13 2010
One more term (114) (factored by Womack et al.) from Sean A. Irvine, May 25 2015
One more term (115) (factored by Womack et al.) from Sean A. Irvine, Feb 08 2016

A085963 Denominators of Farey fractions with prime numerators and denominators.

Original entry on oeis.org

3, 5, 5, 3, 7, 5, 7, 5, 3, 7, 11, 11, 7, 5, 7, 11, 5, 11, 3, 7, 13, 11, 13, 11, 7, 13, 5, 7, 11, 13, 5, 11, 3, 7, 13, 17, 13, 17, 11, 13, 11, 7, 17, 13, 5, 17, 7, 11, 13, 5, 11, 17, 3, 7, 17, 13, 19, 17, 13, 19, 17, 11, 13, 19, 11, 7, 17, 19, 13, 5, 17, 7, 11, 13, 19, 5, 11, 17, 3, 19, 7

Offset: 3

Cino Hilliard, Aug 17 2003

Cf. A085692.

\ Farey sequence of order n fareyct(n) = { forprime(x=2,n, y = farey(x); \ print1(y","); ) } farey(n) = { c=0; m=n*(n-2)+2; a=vector(m); forprime(x=1,n, forprime(y=x,n, v = x/y; if(v<1, c++; a[c]=v; ) ) ); a = vecsort(a); c=0; for(x=2,m, if(a[x]<>a[x-1] & a[x]<>0, print1(numerator(a[x])","); c++; ) ); return(c) }

A232802 Number of solution pairs (x,y) for x <= 11 such that x! + n = y^2 (Brocard-Ramanujan Diophantine equation) is soluble over the integers.

Original entry on oeis.org

3, 1, 2, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 1, 1, 1, 0

Offset: 1

Frank M Jackson, Nov 30 2013

nonn

The Mathematica program will find the number of integer pairs (x,y) solving x!+n = y^2 for each n from 1 to 200 with x not exceeding 11. Dabrowski showed that the abc conjecture implies only finite solutions for each n. Berndt and Galway found that 11 was the highest value that x reached for a solution with n in the range 1 to 2500 and could find no further solution pairs (x,y) in that range even when x was increased to 10^5.

For n = 1 the number of solutions and arbitrary x is Brocard's problem, and it is conjectured - but verified only in the range x <= 10^12 - that there are 3 solution pairs (x,y): (4,5), (5,11), (7,71). - Georg Fischer, Nov 27 2020

Bruce Berndt and William Galway, On the Diophantine equation n! + 1 = m^2
Andrzej Dabrowski, On the Diophantine equation x! + A = y^2, Nieuw Archief voor Wiskunde, Vierde serie Deel 14 No. 3 (Nov. 1996) pp. 321-324.
Wikipedia, Brocard's problem
Wikipedia, abc conjecture - some consequences

Cf. A085692, A146968, A216071 (Brocard's problem; all essentially the same sequence).

Cf. A038202, A068869.

Table[Length@Select[Sqrt[Range[11]!+n], IntegerQ[#] &], {n, 1, 200}]

Definition narrowed by Georg Fischer, Nov 27 2020

A321883 Nonnegative integers n for which n! + 1 is not a square.

Original entry on oeis.org

0, 1, 2, 3, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69

Offset: 1

Felix Fröhlich, Nov 20 2018

nonn

Complement of A146968 = positive integers n such that n!+1 is a square (Brocard's problem, so far {4, 5, 7} are the only known terms).

A weak form of Szpiro's conjecture implies that there are only finitely many nonnegative integers that are not in the sequence (cf. Overholt, 1993).

B. C. Berndt and W. F. Galway, On the Brocard-Ramanujan Diophantine Equation n! + 1 = m^2, The Ramanujan Journal, Vol. 4, No. 1 (2000), 41-42.
M. Overholt, The Diophantine Equation n! + 1 = m^2, Bulletin of the London Mathematical Society, Vol. 25, No. 2 (1993), 104.
Wikipedia, Brocard's problem

Cf. A085692, A146968, A216071.

Select[Range[0,100], !IntegerQ[Sqrt[#!+1]] &] (* Amiram Eldar, Nov 21 2018 *)

select( is(n)=!issquare(n!+1), [0..99]) \\ M. F. Hasler, Nov 20 2018

cp's OEIS Frontend

A087399 Duplicate of A085692.

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A146968 Brocard's problem: positive integers n such that n!+1 = m^2.

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A227644 Perfect powers equal to the sum of 2 factorial numbers.

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A064237 Numbers k such that k! + 1 is divisible by a square.

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A216071 Brocard's problem: positive integers m such that m^2 = n! + 1 for some n.

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A181764 Numbers n such that n!+1 is a product of two distinct prime numbers.

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