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A115169 Integers b > 0 for which there exists a positive integer a <= b such that (a^2 + b^2)/(1 + ab) is an integer. 2
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
Charles R Greathouse IV, Table of n, a(n) for n = 1..500
Steve Chow, You, Me and The Legend of Question Six, BlackPenRedPen, YouTube, Apr 20 2020.
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
(PARI) is(n)=my(s=sqrtnint(n, 3), n2=n^2); for(b=1, s, if((n2+b^2)%(n*b+1)==0, return(1))); for(K=2, sqrtint((n2+(s+1)^2)\(n*s+n+1)), my(k=K^2); if(issquare(k^2*n2-4*n2+4*k), return(1))); 0 \\ Charles R Greathouse IV, Nov 08 2021
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
KEYWORD
nonn
AUTHOR
John W. Layman, Mar 03 2006
EXTENSIONS
Edited by M. F. Hasler, Jun 12 2019
STATUS
approved

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Last modified April 16 05:35 EDT 2024. Contains 371697 sequences. (Running on oeis4.)