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.

A144928 Values of k arising in A144927.

Original entry on oeis.org

49, 5341, 587461, 64615369, 7107103129, 781716728821, 85981733067181, 9457208920661089, 1040206999539652609, 114413312740441125901, 12584424194448984196501, 1384172248076647820489209, 152246362864236811269616489, 16745715742817972591837324581
Offset: 1

Views

Author

Richard Choulet, Sep 25 2008

Keywords

Comments

Numbers k such that there exists x in N : (x+7)^3-x^3=k^2. - Richard Choulet, Oct 16 2008

Examples

			a(1) = 49 because (7+7)^3 - 7^3 = 2041 = 49^2. - _Richard Choulet_, Oct 16 2008
a(2) = 5341 because (1162+7)^3 - 1162^3 = 28526281 = 5341^2. - _Colin Barker_, Jul 16 2016
		

Crossrefs

Cf. A144927.

Programs

  • Mathematica
    Sqrt[(# + 7)^3 - #^3] & /@ Rest@ CoefficientList[Series[7 x (-1 - 55 x + 2 x^2)/((x - 1) (x^2 - 110 x + 1)), {x, 0, 14}], x] (* or *)
    Rest@ CoefficientList[Series[49 x (1 - x)/(1 - 110 x + x^2), {x, 0, 14}], x] (* Michael De Vlieger, Jul 17 2016 *)
  • PARI
    Vec(49*x*(1-x)/(1-110*x+x^2) + O(x^20)) \\ Colin Barker, Oct 17 2014, corrected Jul 16 2016

Formula

a(n+2) = 110*a(n+1)-a(n). - Richard Choulet, Oct 16 2008
G.f.: 49*x*(1-x) / (1-110*x+x^2). - Colin Barker, Oct 17 2014, corrected Jul 16 2016

Extensions

More terms from Colin Barker, Oct 17 2014
All terms except the first corrected by Colin Barker, Jul 16 2016

A144929 Numbers n such that there exists an integer k with (n+1)^3 - n^3 = 7*k^2.

Original entry on oeis.org

1, 166, 18313, 2014318, 221556721, 24369225046, 2680393198393, 294818882598238, 32427396692607841, 3566718817304264326, 392306642506776468073, 43150163956928107223758, 4746125728619585018145361, 522030679984197423888766006, 57418628672533097042746115353, 6315527123298656477278183922878
Offset: 1

Views

Author

Richard Choulet, Sep 25 2008

Keywords

Examples

			a(1) = 1 because 2^3-1^3 = 7 = 7*1^2.
		

References

  • E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185.
  • P.-F. Teilhet, Query 2228, L'Intermédiaire des Mathématiciens, 11 (1904), 44-45. - N. J. A. Sloane, Mar 08 2022

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[1]==1,a[2]==166,a[n]==54+110a[n-1]-a[n-2]},a[n],{n,20}] (* or *) LinearRecurrence[{111,-111,1},{1,166,18313},20] (* Harvey P. Dale, Jun 11 2011 *)
    Rest@ CoefficientList[Series[x (-1 - 55 x + 2 x^2)/((x - 1) (x^2 - 110 x + 1)), {x, 0, 16}], x] (* or *) Last /@ Table[n /. {ToRules[Reduce[n > 0 && k >= 0 && (n + 1)^3 - n^3 == 7 k^2, n, Integers] /. C[1] -> c]} // Simplify, {c, 1, 16}] // Union (* Michael De Vlieger, Jul 14 2016 *)
  • PARI
    Vec(x*(-1-55*x+2*x^2)/((x-1)*(x^2-110*x+1)) + O(x^20)) \\ Colin Barker, Jul 14 2016

Formula

a(n+2) = 110*a(n+1)-a(n)+54.
a(n) = 111*a(n-1)-111*a(n-2)+a(n-3), with n>3. - Harvey P. Dale, Jun 11 2011
G.f.: x*(-1-55*x+2*x^2) / ((x-1)*(x^2-110*x+1)). - Harvey P. Dale, Jun 11 2011
a(n) = (A350967(n)-3)/6. - N. J. A. Sloane, Mar 03 2022

A145693 Numbers X such that there exists Y in N with X^2=21*Y^2+7.

Original entry on oeis.org

14, 1526, 167846, 18461534, 2030600894, 223347636806, 24566209447766, 2702059691617454, 297201999868472174, 32689517925840321686, 3595549769842566913286, 395477785164756520139774, 43498960818353374648461854, 4784490212233706454810664166
Offset: 1

Views

Author

Richard Choulet, Oct 16 2008

Keywords

Examples

			a(1)=14 because the first relation is 14^2=21*3^2+7.
		

Crossrefs

Programs

  • Magma
    I:=[14,1526]; [n le 2 select I[n] else 110*Self(n-1)-Self(n-2): n in [1..15]]; // Vincenzo Librandi, Oct 21 2014
  • Mathematica
    CoefficientList[Series[14 (1 - x)/(x^2 - 110 x + 1), {x, 0, 20}], x] (* Vincenzo Librandi, Oct 21 2014 *)
    LinearRecurrence[{110,-1},{14,1526},20] (* Harvey P. Dale, Sep 19 2024 *)
  • PARI
    Vec(-14*x*(x-1)/(x^2-110*x+1) + O(x^20)) \\ Colin Barker, Oct 21 2014
    

Formula

a(n+2) = 110*a(n+1)-a(n).
G.f.: -14*x*(x-1) / (x^2-110*x+1). - Colin Barker, Oct 21 2014

Extensions

Editing and more terms from Colin Barker, Oct 21 2014
Showing 1-3 of 3 results.