A038202 -id:A038202 - cp's OEIS frontend

A068869 Smallest number k such that n! + k is a square.

0, 2, 3, 1, 1, 9, 1, 81, 729, 225, 324, 39169, 82944, 176400, 215296, 3444736, 26167684, 114349225, 255004929, 1158920361, 11638526761, 42128246889, 191052974116, 97216010329, 2430400258225, 1553580508516, 4666092737476, 565986718738441, 2137864362693921

Offset: 1

Amarnath Murthy, Mar 13 2002

nonn

Observation: for n < 2000, only for n = 1, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16 is a(n) a square (see A360210).
According to my conjecture that n! + n^2 != m^2 for n >= 1, m >= 0 (see A004664), for all terms: a(n) != n^2. - Alexander R. Povolotsky, Oct 06 2008
There are two cases: a(n) > sqrt(n!) in A182203 and a(n) < sqrt(n!) in A182204. - Artur Jasinski, Apr 13 2012

			a(6) = 9 as 6! + 9 = 729 is a square.

T. D. Noe, Table of n, a(n) for n=1..100

Cf. A038202, A048761, A055228, A066857, A360210.

Cf. A182203, A182204.

Table[ Ceiling[ Sqrt[n! ]]^2 - n!, {n, 1, 28}]

A068869(n)=(sqrtint(n!-1)+1)^2-n!  \\ M. F. Hasler, Apr 01 2012

from math import factorial, isqrt
def a(n): return (isqrt((f:=factorial(n))-1)+1)**2 - f
print([a(n) for n in range(1, 30)]) # Michael S. Branicky, Jan 30 2023

a(n) = A055228(n)^2 - n! = ceiling(sqrt(n!))^2 - n! = A048761(n!) - n!.

a(n) <= A038202(n)^2, with equality for the n listed in the first comment. - M. F. Hasler, Apr 01 2012

More terms from Vladeta Jovovic, Mar 21 2002

Edited by Robert G. Wilson v and N. J. A. Sloane, Mar 22 2002

A232176 Least positive k such that n^2 + triangular(k) is a square.

Original entry on oeis.org

1, 2, 6, 10, 14, 18, 7, 5, 8, 34, 6, 42, 46, 15, 54, 16, 14, 66, 70, 74, 23, 82, 9, 90, 17, 98, 102, 10, 110, 15, 25, 122, 126, 16, 39, 48, 40, 21, 150, 34, 158, 29, 54, 48, 30, 13, 182, 63, 55, 194, 56, 202, 14, 45, 214, 63, 222, 26, 41, 234, 31, 42, 39, 250, 32, 63

Offset: 0

Alex Ratushnyak, Nov 19 2013

nonn

Triangular(k) = A000217(k) = k*(k+1)/2.
a(n) <= 4*n - 2, because with k = 4*n-2: n^2 + k*(k+1)/2 = n^2 + (4*n-2)*(4*n-1)/2 = 9*n^2 - 6*n + 1 = (3*n-1)^2.
The sequence of numbers n such that a(n)=n begins: 8, 800, 7683200 ... - a subsequence of A220186.

Andrey Zabolotskiy, Table of n, a(n) for n = 0..5000

Cf. A000290, A000217, A038202, A055527, A220186, A232175.
Cf. A232179 (least k>=0 such that n^2 + triangular(k) is a triangular number).
Cf. A101157 (least k>0 such that triangular(n) + k^2 is a triangular number).
Cf. A232178 (least k>=0 such that triangular(n) + k^2 is a square).

lpk[n_]:=Module[{k=1},While[!IntegerQ[Sqrt[n^2+(k(k+1))/2]],k++];k]; Array[ lpk,70,0] (* Harvey P. Dale, May 04 2018 *)

a(n) = {k = 1; while (! issquare(n^2 + k*(k+1)/2), k++); k;} \\ Michel Marcus, Nov 20 2013

import math
for n in range(77):
  n2 = n*n
  y=1
  for k in range(1,10000001):
    sum = n2 + k*(k+1)//2
    r = int(math.sqrt(sum))
    if r*r == sum:
      print(str(k), end=',')
      y=0
      break
  if y: print('-', end=',')

A232175 Least positive k such that n^3 + k^2 is a square, or 0 if there is no such k.

Original entry on oeis.org

0, 1, 3, 6, 10, 3, 21, 8, 36, 15, 55, 6, 78, 35, 15, 48, 136, 27, 171, 10, 42, 99, 253, 10, 300, 143, 81, 42, 406, 15, 465, 64, 88, 255, 35, 63, 666, 323, 91, 3, 820, 21, 903, 55, 66, 483, 1081, 48, 1176, 125, 85, 39, 1378, 81, 165, 28, 76, 783, 1711, 15, 1830, 899, 63

Offset: 1

Alex Ratushnyak, Nov 19 2013

Numbers n such that a(n) = n*(n-1)/2 appear to be A000430.

n = 1 is the only number for which a(n) = 0. - T. D. Noe, Nov 21 2013

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..1000 from T. D. Noe).
StackExchange, The cube of integer can be written as the difference of two square.

Cf. A000290, A000430, A000578, A038202, A055527, A232176.

Join[{0}, Table[k = 1; While[! IntegerQ[Sqrt[n^3 + k^2]], k++]; k, {n, 2, 100}]] (* T. D. Noe, Nov 21 2013 *)

a(n) = {k = 1; while (!issquare(n^3+k^2), k++); k;} \\ Michel Marcus, Nov 20 2013

import math
for n in range(77):
   n3 = n*n*n
   y=1
   for k in range(1, 10000001):
     s = n3 + k*k
     r = int(math.sqrt(s))
     if r*r == s:
       print(k, end=', ')
       y=0
       break
   if y: print(end='-, ')

from _future_ import division
from sympy import divisors
def A232175(n):
    n3 = n**3
    ds = divisors(n3)
    for i in range(len(ds)//2-1,-1,-1):
        x = ds[i]
        y = n3//x
        a, b = divmod(y-x,2)
        if not b:
            return a
    return 0 # Chai Wah Wu, Sep 12 2017

A159082 Numbers whose squares added to 7! are prime.

Original entry on oeis.org

13, 23, 29, 59, 61, 73, 97, 101, 103, 109, 121, 127, 149, 169, 187, 191, 199, 221, 227, 251, 257, 263, 277, 299, 307, 317, 319, 331, 341, 367, 373, 383, 389, 397, 403, 407, 409, 433, 449, 451, 461, 463, 467, 491, 493, 499, 517, 527, 529, 533, 551, 563, 571

Offset: 1

Ulrich Krug (leuchtfeuer37(AT)gmx.de), Apr 05 2009

1) Necessarily a(n) is not divisible by 2, 3, 5, 7.
2) Sequence is conjectured to be infinite.
3) It is conjectured that an infinite number of terms are primes.
4) Note that sequence contains a(k), a(k+1) prime twin pairs, first are (59,61), (461,463), (827,829), (1319,1321).
5) It is conjectured that an infinite number of a(n) are squares, first are 121=11^2, 169=13^2, 529=23^2, 841=29^2, 961=31^2, 1681=41^2, ...
6) m!+k^2=n^2 are the generalized Brown number triples (m,k,n).

			1) 7!+1=71^2, (7, 71) is the largest (of three) Brown pairs; Erdos conjectured that there are no others.
2) 7!+3^2=5049= 3^3 * 11 * 17, 7!+5^2=5065 = 5 * 1013, 7!+7^2=5089 = 7 * 727, 7!+9^2=5121 = 3^2 * 569, 7!+11^2=5161 = 13 * 397.
3) 7!+13^2=5209 prime, so a(1)=13.

R. K. Guy, Unsolved Problems in Number Theory (2nd ed.) New York: Springer-Verlag, p. 193, 1994
I. Niven, H. S. Zuckerman and H. L. Montgomery: An Introduction to the Theory of Numbers (5th ed.). Wiley Text Books, 1991
David Wells, Prime Numbers: The Most Mysterious Figures in Math. John Wiley and Sons. 2005

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A038202, A158979.

With[{s = 7!}, Select[Range[600], PrimeQ[#^2 + s] &]] (* Harvey P. Dale, Jun 17 2015 *)

isok(n) = isprime(n^2+7!); \\ Michel Marcus, Jul 23 2013; corrected Jun 14 2022

7! + a(n)^2 = prime.

Edited by N. J. A. Sloane, Apr 05 2009

A181896 Least value x solving x^2 - y^2 = n!

Original entry on oeis.org

5, 11, 27, 71, 201, 603, 1905, 6318, 21888, 78912, 295260, 1143536, 4574144, 18859680, 80014848, 348776640, 1559776320, 7147792848, 33526120320, 160785625902, 787685472000, 3938427360000, 20082117976800, 104349745817240, 552166953609600, 2973510046027938

Offset: 4

Artur Jasinski, Mar 31 2012

nonn

Many of terms in this sequence are that same as A055228(n) but not all.

a(n) solves the Brocard-Ramanujan Problem, n! = a(n)^2 - 1, and thus (n, a(n)) are a pair of Brown Numbers, if and only if A038202(n) = 1. - Austin Hinkel, Dec 28 2022

Eric Weisstein's World of Mathematics, Brocard's Problem.

For least y value see A038202.

Cf. A055228.

cc = {}; Do[f = n!/4; x = Max[Select[Divisors[f], # <= Sqrt[f] &]]; kk = f/x - x; AppendTo[cc, Sqrt[n! + kk^2]], {n, 4, 30}]; cc

a(n)=my(N=n!,x=sqrtint(N));while(!issquare(x++^2-N),);x \\ Charles R Greathouse IV, Apr 10 2012

A269485 Least k > 0 such that n! + k^2 is prime.

Original entry on oeis.org

1, 1, 1, 1, 7, 11, 7, 13, 17, 31, 13, 1, 47, 17, 19, 19, 23, 73, 43, 29, 47, 31, 43, 29, 31, 37, 167, 1, 29, 43, 79, 229, 89, 71, 137, 37, 53, 1, 79, 131, 137, 1, 71, 83, 89, 89, 53, 97, 53, 101, 59, 173, 79, 71, 353, 191, 103, 523, 229, 191, 103, 401, 67, 257

Offset: 0

Jean-Marc Rebert, Feb 28 2016

nonn

a(n) = A033932(n) = 1 for n in A002981.

			a(4) = 7, because 4! + 7^2 = 73 is prime and for 0 < i < 7, 4! + i^2 is not prime.

Jean-Marc Rebert, Table of n, a(n) for n = 0..500

Cf. A002981, A033932, A038202.

Table[SelectFirst[Range@ 10000, PrimeQ[n! + #^2] &], {n, 120}]
(* Version 10, or *)
Table[k = 1; While[! PrimeQ[n! + k^2], k++]; k, {n, 120}] (* Michael De Vlieger, Feb 28 2016 *)

a(n) = {my(k=1); while (!isprime(n! + k^2), k++); k;} \\ Michel Marcus, Feb 29 2016

A181895 Difference between maximum and minimum positive value y in solutions to x^2 - y^2 = n!.

Original entry on oeis.org

4, 28, 176, 1258, 10070, 90692, 907184, 9979181, 119750111, 1556754911, 21794572379, 326918591535, 5230697470143, 88921857013919, 1600593426385151, 30411275101997759, 608225502043759679

Offset: 4

Artur Jasinski, Mar 31 2012

cc = {}; Do[f = n!/4; x = Max[Select[Divisors[f], # <= Sqrt[f] &]]; kk = f/x - x; kkk = ((n! - 4)/4); AppendTo[cc, kkk - kk], {n, 4, 20}]; cc

a(n) = A139174(n) - A038202(n).

A181899 Largest divisor of n!/4 which is less than sqrt(n!)/2.

Original entry on oeis.org

2, 5, 12, 35, 96, 288, 945, 3150, 10800, 39312, 147420, 571536, 2286144, 9424800, 39984000, 174283200, 779688000, 3573570000, 16761064320, 80379048750, 393826406400, 1969132032000, 10040487256800, 52174220175000, 276080056560000, 1486750296281250

Offset: 4

Artur Jasinski, Mar 31 2012

nonn

Comment from A038202: Let f=n!/4 and let a(n) be the largest divisor of f such that a(n) < sqrt(f). Then A038202(n) = f/a(n) - a(n). The greatest k such that n!+k^2 is a square is f-1. The number of k for which n!+k^2 is a square is A038548(f). - T. D. Noe, Nov 02 2004

Cf. A038202, A138196, A139151, A181892, A181893.

Table[f = n!/4; Select[Divisors[f], # <= Sqrt[f] &][[-1]], {n, 4, 20}]

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

cp's OEIS Frontend

A068869 Smallest number k such that n! + k is a square.

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A232176 Least positive k such that n^2 + triangular(k) is a square.

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A232175 Least positive k such that n^3 + k^2 is a square, or 0 if there is no such k.

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A159082 Numbers whose squares added to 7! are prime.

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A181896 Least value x solving x^2 - y^2 = n!

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A269485 Least k > 0 such that n! + k^2 is prime.

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A181895 Difference between maximum and minimum positive value y in solutions to x^2 - y^2 = n!.

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