|
| |
|
|
A245624
|
|
Sequence of distinct least positive numbers such that the average of the first n terms is a cube.
|
|
2
|
|
|
|
1, 15, 8, 84, 27, 249, 64, 552, 125, 1035, 216, 1740, 343, 2709, 512, 3984, 729, 5607, 1000, 7620, 1331, 10065, 1728, 12984, 2197, 16419, 2744, 20412, 3375, 25005, 4096, 30240, 4913, 36159, 5832, 42804, 6859, 50217, 8000, 58440, 9261, 67515, 10648, 77484, 12167, 88389, 13824
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Colin Barker's formulas are true if the curve x^3 = 7*y^3 + 6*y^2 + 2*y has no positive integer solutions. This is a curve of genus 1 (equivalent to the elliptic curve s^3 + t^2 + 20), and does have some rational points, but no positive integer solutions at least for y <= 10^21. - Robert Israel, May 17 2015
Now confirmed: that curve has no positive integer solutions. See the Mathematics Stack Exchange link. - Robert Israel, May 18 2015
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(2*n-1) = n^3, a(2*n) = 7*n^3 + 6*n^2 + 2*n.
a(n) = 4*a(n-2) - 6*a(n-4) + 4*a(n-6) - a(n-8) for n > 8. - Colin Barker, Nov 05 2014
G.f.: x*(3*x^5 + x^4 + 24*x^3 + 4*x^2 + 15*x + 1) / ((x-1)^4*(x+1)^4). - Colin Barker, Nov 05 2014
|
|
|
MAPLE
|
seq(op([k^3, 7*k^3+6*k^2+2*k]), k=1..100); # Robert Israel, May 18 2015
|
|
|
MATHEMATICA
|
Flatten[Table[{n^3, 7 n^3 + 6 n^2 + 2 n}, {n, 25}]] (* Vincenzo Librandi, May 19 2015 *)
|
|
|
PROG
|
(PARI) v=[]; n=1; while(n<10^5, num=(vecsum(v)+n); if(num%(#v+1)==0&&vecsearch(vecsort(v), n)==0, for(i=1, n+2, if(i^3>(num/(#v+1)), break); if(i^3==(num/(#v+1)), print1(n, ", "); v=concat(v, n); n=1; break))); n++)
(Magma) &cat[[k^3, 7*k^3+6*k^2+2*k]: k in [1..25]]; // Vincenzo Librandi, May 19 2015
(PARI) Vec(x*(3*x^5+x^4+24*x^3+4*x^2+15*x+1)/((x-1)^4*(x+1)^4) + O(x^100)) \\ Colin Barker, May 19 2015
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
