login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A115169 Integers b > 0 for which there exists a positive integer a <= b such that (a^2 + b^2)/(1 + ab) is an integer. 0
1, 8, 27, 30, 64, 112, 125, 216, 240, 343, 418, 512, 729, 1000, 1020, 1331, 1560, 1728, 2133, 2197, 2744, 3120, 3375, 4096, 4913, 5822, 5832, 6859, 7770, 8000, 9261, 10648, 12167, 13824, 15625, 16256, 16800, 17576, 18957, 19683 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

All positive cubes are in this sequence.

Indeed, if b = k^3, then for a = k, we have a^2 + b^2 = k^2 + k^6 = (1 + k^4)*k^2 = (a*b + 1)*a^2. More generally, if the ratio (a^2 + b^2)/(a*b + 1) is an integer, it is equal to gcd(a,b)^2, thus in particular a perfect square. (This was Question 6 in the 1988 IMO.) All solutions (a,b) are member of a sequence {(x(n), x(n+1)); n = 1,2,...} where x = (0, k, k^3, k^5 - k, ...) with x(n+1) = k^2*x(n) - x(n-1) and some k >= 2, cf. A052530 for k = 2, A065100 for k = 3. (One might consider >= 0 instead > 0 in the definition, but a = 0 yields a solution for any b.) - M. F. Hasler, Jun 12 2019

LINKS

Table of n, a(n) for n=1..40.

I. Lauko, G. Pinter and L. Pinter, Another Step Further... On a Problem of the 1988 IMO, Math. Mag. 79 (2006), 45-53.

Simon Pampena, The Legend of Question Six, Numberphile, YouTube, Aug 16 2016.

EXAMPLE

(2^2+8^2)/(1+2*8) = 68/17 = 4, an integer, so 8 is a term of the series.

From M. F. Hasler, Jun 12 2019: (Start)

The list of solutions starts:

     a      b     a^2+b^2   a*b+1   ratio

   ----------------------------------------

     1      1          2       2       1

     8      2         68      17       4

    27      3        738      82       9

    30      8        964     241       4

    64      4       4112     257      16

   112     30      13444    3361       4

   125      5      15650     626      25

   216      6      46692    1297      36

   240     27      58329    6481       9

   343      7     117698    2402      49

   418    112     187268   46817       4

   512      8     262208    4097      64

   729      9     531522    6562      81

  1000     10    1000100   10001     100

  1020     64    1044496   65281      16

(End)

PROG

(PARI) isok(n) = for(m=0, n, if (denominator((m^2+n^2)/(1+m*n))==1, return(1))); return (0); \\ Michel Marcus, Sep 18 2017

(PARI) is_A115169(n)=for(a=1, n\3+1, (a^2+n^2)%(1+a*n)||return(1)) \\ M. F. Hasler, Jun 12 2019

CROSSREFS

Cf. A000578 (cubes), A052530 (subsequence of terms for ratio 2^2, for n >= 2), A065100 (subsequence of terms for ratio 3^2).

Sequence in context: A070497 A070496 A304636 * A070495 A270421 A056729

Adjacent sequences:  A115166 A115167 A115168 * A115170 A115171 A115172

KEYWORD

nonn,changed

AUTHOR

John W. Layman, Mar 03 2006

EXTENSIONS

Edited by M. F. Hasler, Jun 12 2019

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified June 15 22:19 EDT 2019. Contains 324145 sequences. (Running on oeis4.)