cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A018253 Divisors of 24.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 12, 24
Offset: 1

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Author

N. J. A. Sloane

Keywords

Comments

The divisors of 24 greater than 1 are the only positive integers n with the property m^2 == 1 (mod n) for all integer m coprime to n. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 10 2001
Numbers n for which all Dirichlet characters are real. - Benoit Cloitre, Apr 21 2002
These are the numbers n that are divisible by all numbers less than or equal to the square root of n. - Tanya Khovanova, Dec 10 2006 [For a proof, see the Tauvel paper in references. - Bernard Schott, Dec 20 2012]
Also, numbers n such that A160812(n) = 0. - Omar E. Pol, Jun 19 2009
It appears that these are the only positive integers n such that A160812(n) = 0. - Omar E. Pol, Nov 17 2009
24 is a highly composite number: A002182(6)=24. - Reinhard Zumkeller, Jun 21 2010
Chebolu points out that these are exactly the numbers for which the multiplication table of the integers mod n have 1s only on their diagonal, i.e., ab == 1 (mod n) implies a = b (mod n). - Charles R Greathouse IV, Jul 06 2011
It appears that 3, 4, 6, 8, 12, 24 (the divisors >= 3 of 24) are also the only numbers n whose proper non-divisors k are prime numbers if k = d-1 and d divides n. - Omar E. Pol, Sep 23 2011
About the last Pol's comment: I have searched to 10^7 and have found no other terms. - Robert G. Wilson v, Sep 23 2011
Sum_{i=1..8} A000005(a(i))^3 = (Sum_{i=1..8} A000005(a(i)))^2, see Kordemsky in References and Barbeau et al. in Links section. - Bruno Berselli, Dec 29 2014

Examples

			Square root of 12 = 3.46... and 1, 2 and 3 divide 12.
From the tenth comment: 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 6^3 + 8^3 = (1+2+2+3+4+4+6+8)^2 = 900. - _Bruno Berselli_, Dec 28 2014

References

  • Harvey Cohn, "Advanced Number Theory", Dover, chap.II, p. 38
  • Boris A. Kordemsky, The Moscow Puzzles: 359 Mathematical Recreations, C. Scribner's Sons (1972), Chapter XIII, Paragraph 349.
  • Patrick Tauvel, "Exercices d'algèbre générale et d'arithmétique", Dunod, 2004, exercice 70 page 368.

Links

Crossrefs

Cf. A174228, A018256, A018261, A018266, A018293, A018321, A018350, A018412, A018609, A018676, A178877, A178878, A165412, A178858-A178864.
Cf. A000005, A158649. - Bruno Berselli, Dec 29 2014
Cf. A303704 (with respect to Astudillo's 2001 comment above).

Programs

Formula

a(n) = A161710(n-1). - Reinhard Zumkeller, Jun 21 2009

A254328 Numbers k such that all x^2 mod k are squares (including 0 and 1).

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 12, 16
Offset: 1

Views

Author

Joerg Arndt, Jan 28 2015

Keywords

Comments

Are there any more terms > 16?
There are no more terms less than 10^12. Probably the sequence is finite. - Charles R Greathouse IV, Jan 29 2015
This is a subsequence of A303704, so it is full. - Jianing Song, Feb 14 2019

Examples

			Terms k <= 16 and the squares mod k:
1: [0]
2: [0, 1]
3: [0, 1, 1]
4: [0, 1, 0, 1]
5: [0, 1, 4, 4, 1]
8: [0, 1, 4, 1, 0, 1, 4, 1]
12: [0, 1, 4, 9, 4, 1, 0, 1, 4, 9, 4, 1]
16: [0, 1, 4, 9, 0, 9, 4, 1, 0, 1, 4, 9, 0, 9, 4, 1]
k = 10 is not a term: in the list of squares mod 10, [0, 1, 4, 9, 6, 5, 6, 9, 4, 1], the numbers 5 and 6 are not squares.

Links

Crossrefs

Cf. A065428, A254329, A096008, A303704.

Programs

  • Mathematica
    f[n_] := Mod[Range[n]^2, n]; Select[Range@ 10000, AllTrue[f@ #, IntegerQ[Sqrt[#]] &] &] (* AllTrue function introduced in version 10; Michael De Vlieger, Jan 29 2015 *)
  • PARI
    isok(n)=for(k=2,n-1,if(!issquare(lift(Mod(k,n)^2)),return(0)));return(1);
for(n=1,10^9,if(isok(n),print1(n,", ")));
  • PARI
    is(n)=for(k=sqrtint(n)+1,n\2, if(!issquare(k^2%n), return(0))); 1
for(m=10,10^6,for(k=0,sqrtint(2*m),if(is(t=m^2-k^2),print(t))))
\\ Charles R Greathouse IV, Jan 29 2015

Extensions

Keywords fini and full added by Jianing Song, Feb 14 2019

A330423 Least nonsquare k that is a quadratic residue modulo n and is coprime to n.

Original entry on oeis.org

2, 3, 7, 5, 6, 7, 2, 17, 7, 11, 3, 13, 3, 11, 19, 17, 2, 7, 5, 21, 22, 3, 2, 73, 6, 3, 7, 29, 5, 19, 2, 17, 31, 13, 11, 13, 3, 5, 10, 41, 2, 37, 6, 5, 19, 3, 2, 73, 2, 11, 13, 17, 6, 7, 14, 57, 7, 5, 3, 61, 3, 5, 22, 17, 14, 31, 6, 13, 13, 11, 2, 73, 2, 3, 19, 5, 15
Offset: 1

Views

Author

Jianing Song, Dec 14 2019

Keywords

Comments

a(n) > n if and only if n is in A303704.
It seems that lim_{n->oo} a(n)/n = 0. Conjectured last term m such that a(m)/m > 1/k, k = 1, 2, 3, ...: 840, 1680, 2640, 9240, 10920, 10920, 18480, 18480, 21840, 29640, ...

Examples

			k is a coprime quadratic residue modulo 16 if and only if k == 1 (mod 8). Since 1 and 9 are squares, a(16) = 17.
k is a coprime quadratic residue modulo 24 if and only if k == 1 (mod 24). Since 1, 25 and 49 are squares, a(24) = 73.
k is a coprime quadratic residue modulo 840 if and only if k == 1, 121, 169, 289, 361, 529 (mod 840). Since 1, 121, 169, 289, 361, 529, 841, 961 are squares, a(840) = 840+169 = 1009.

Links

Crossrefs

Cf. A303704, A330404.

Programs

  • PARI
    a(n) = my(k=1); while(!issquare(Mod(k,n)) || issquare(k) || gcd(k,n)>1, k++); k
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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