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 1-4 of 4 results.

A137878 Perfect squares among 17-gonal numbers A051869(k) = k*(15*k - 13)/2.

Original entry on oeis.org

1, 17689, 378225, 4109707449, 87870152041, 954775454112481, 20414169462254569, 221815343046210267025, 4742660677722035990769, 51532584126226886201833161, 1101824413949324675985344641, 11972153009151467313136073526409, 255978051492792346696545201859225
Offset: 1

Views

Author

Alexander Adamchuk, Feb 19 2008

Keywords

Comments

Corresponding square roots sqrt(a(n)) are listed in A137879.
Indices of perfect squares among the 17-gonal numbers A051869(k) = k*(15*k - 13)/2 are listed in A137880. Note that all such indices are also perfect squares, their square roots are listed in A137881(k) = sqrt(A137880(k)).

Crossrefs

Cf. A051869 (17-gonal numbers), A137879, A137880, A137881.

Programs

  • PARI
    Vec(x*(1+17688*x+128214*x^2+17688*x^3+x^4)/((1-x)*(1-482*x+x^2)*(1+482*x+x^2)) + O(x^20)) \\ Colin Barker, Jun 19 2016

Formula

a(n) = A137879(n)^2 = A051869( A137880(n) ) = A051869( A137881(n)^2 ).
From Colin Barker, Jun 19 2016: (Start)
a(n) = a(n-1) + 232322*a(n-2) - 232322*a(n-3) - a(n-4) + a(n-5) for n > 5.
G.f.: x*(1 + 17688*x + 128214*x^2 + 17688*x^3 + x^4) / ((1-x)*(1 - 482*x + x^2)*(1 + 482*x + x^2)).
(End)

Extensions

Edited and extended by Max Alekseyev, Oct 19 2008

A137879 Numbers k such that k^2 is a 17-gonal number.

Original entry on oeis.org

1, 133, 615, 64107, 296429, 30899441, 142878163, 14893466455, 68866978137, 7178619931869, 33193740583871, 3460079913694403, 15999314094447685, 1667751339780770377, 7711636199783200299, 803852685694417627311, 3716992648981408096433
Offset: 1

Views

Author

Alexander Adamchuk, Feb 19 2008

Keywords

Comments

Corresponding 17-gonal numbers equal k^2 are listed in A137878.
The 17-gonal numbers A051869(n) = n*(15n - 13)/2 are perfect squares for indices n listed in A137880. Note that all such indices are also perfect squares of numbers listed in A137881.

Crossrefs

Cf. A051869 (17-gonal numbers), A137878 (17-gonal numbers that are perfect squares), A137880, A137881.

Programs

Formula

a(n) = sqrt(A137878(n)) = sqrt(A051869(A137880(n))) = sqrt(A051869(A137881(n)^2)).
From Max Alekseyev, Oct 19 2008: (Start)
For n>=5, a(n) = 482*a(n-2) - a(n-4).
a(2n) = (-60 + 17*sqrt(30))/120 * (11 + 2*sqrt(30))^(2n) + (-60 - 17*sqrt(30))/120 * (11 - 2*sqrt(30))^(2n).
a(2n+1) = (60 + 17*sqrt(30))/120 * (11 + 2*sqrt(30))^(2n) + (60 - 17*sqrt(30))/120 * (11 - 2*sqrt(30))^(2n). (End)

Extensions

Extended by Max Alekseyev, Oct 19 2008

A137880 Indices k of perfect squares among 17-gonal numbers A051869(k) = k*(15*k - 13)/2.

Original entry on oeis.org

1, 49, 225, 23409, 108241, 11282881, 52171729, 5438325025, 25146664929, 2621261378961, 12120640323841, 1263442546333969, 5842123489426225, 608976686071593889, 2815891401263116401, 293525499243961920321, 1357253813285332678849, 141478681658903574000625
Offset: 1

Views

Author

Alexander Adamchuk, Feb 19 2008

Keywords

Comments

Corresponding perfect squares are listed in A137878.
Note that all a(n) are perfect squares themselves, their square roots are listed in A137881.

Crossrefs

Cf. A051869 (17-gonal numbers), A137878 (17-gonal numbers that are perfect squares), A137879, A137881.

Programs

  • Mathematica
    Rest@ CoefficientList[Series[x (1 + 48 x - 306 x^2 + 48 x^3 + x^4)/((1 - x) (1 - 22 x + x^2) (1 + 22 x + x^2)), {x, 0, 18}], x] (* Michael De Vlieger, Jun 18 2016 *)
  • PARI
    Vec(x*(1+48*x-306*x^2+48*x^3+x^4)/((1-x)*(1-22*x+x^2)*(1+22*x+x^2)) + O(x^20)) \\ Colin Barker, Jun 18 2016

Formula

A051869( a(n) ) = A137878(n); a(n) = A137881(n)^2.
From Max Alekseyev, Oct 19 2008: (Start)
a(n) = 482*a(n-2) - a(n-4) - 208.
a(2n) = ( (15 - sqrt(30))/30 * (11 + 2*sqrt(30))^n + (15 + sqrt(30))/30 * (11 - 2*sqrt(30))^n )^2.
a(2n+1) = ( (15 + sqrt(30))/30 * (11 + 2*sqrt(30))^n + (15 - sqrt(30))/30 * (11 - 2*sqrt(30))^n )^2. (End)
a(n) = a(n-1) + 482*a(n-2) - 482*a(n-3) - a(n-4) + a(n-5). - Matthew House, Jun 18 2016
G.f.: x*(1 + 48*x - 306*x^2 + 48*x^3 + x^4) / ((1-x)*(1 - 22*x + x^2)*(1 + 22*x + x^2)). - Colin Barker, Jun 18 2016

Extensions

Edited and extended by Max Alekseyev, Oct 19 2008

A238245 Positive integers n such that x^2 - 22xy + y^2 + n = 0 has integer solutions.

Original entry on oeis.org

20, 39, 56, 71, 80, 84, 95, 104, 111, 116, 119, 120, 156, 180, 191, 224, 239, 255, 284, 296, 311, 320, 336, 351, 359, 380, 399, 404, 416, 431, 444, 455, 464, 471, 476, 479, 480, 500, 504, 551, 596, 599, 624, 639, 680, 695, 696, 719, 720, 756, 764, 791, 824
Offset: 1

Views

Author

Colin Barker, Feb 20 2014

Keywords

Examples

			39 is in the sequence because x^2 - 22xy + y^2 + 39 = 0 has integer solutions, for example (x, y) = (2, 43).
		

Crossrefs

Cf. A157014 (n = 20), A137881 (n = 104), A077422 (n = 120), A133275 (n = 180).
Showing 1-4 of 4 results.