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.

A157105 a(n) = 137842*n - 30996.

Original entry on oeis.org

106846, 244688, 382530, 520372, 658214, 796056, 933898, 1071740, 1209582, 1347424, 1485266, 1623108, 1760950, 1898792, 2036634, 2174476, 2312318, 2450160, 2588002, 2725844, 2863686, 3001528, 3139370, 3277212, 3415054
Offset: 1

Views

Author

Vincenzo Librandi, Feb 23 2009

Keywords

Comments

The identity (5651522*n^2 - 2541672*n + 285769)^2 - (1681*n^2 - 756*n + 85)*(137842*n - 30996)^2 = 1 can be written as (A157106(n))^2 - (A157010(n))*(a(n))^2 = 1.

Links

Crossrefs

Cf. A157010, A157106.

Programs

  • Magma
    I:=[106846, 244688, 382530]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];
  • Mathematica
    LinearRecurrence[{2,-1},{106846,244688},30] (* Harvey P. Dale, Mar 31 2013 *)
82*(1681*Range[30] -378) (* G. C. Greubel, Jan 11 2022 *)
  • PARI
    a(n) = 137842*n - 30996
  • Sage
    [82*(1681*n - 378) for n in (1..30)] # G. C. Greubel, Jan 11 2022

Formula

a(n) = 2*a(n-1) -a(n-2).
G.f: 82*x*(1303 + 378*x)/(1-x)^2.
E.g.f.: 82*(378 - (378 - 1681*x)*exp(x)). - G. C. Greubel, Jan 11 2022

A157106 a(n) = 5651522*n^2 - 2541672*n + 285769.

Original entry on oeis.org

3395619, 17808513, 43524451, 80543433, 128865459, 188490529, 259418643, 341649801, 435184003, 540021249, 656161539, 783604873, 922351251, 1072400673, 1233753139, 1406408649, 1590367203, 1785628801, 1992193443, 2210061129
Offset: 1

Views

Author

Vincenzo Librandi, Feb 23 2009

Keywords

Comments

The identity (5651522*n^2-2541672*n+285769)^2-(1681*n^2-756*n+85)*(137842*n-30996)^2=1 can be written as a(n)^2-A157010(n)*A157105(n)^2=1.

Links

Crossrefs

Cf. A157010, A157105.

Programs

  • Magma
    I:=[3395619, 17808513, 43524451]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];
  • Maple
    A157106:=n->5651522*n^2 - 2541672*n + 285769; seq(A157106(n), n=1..30); # Wesley Ivan Hurt, Jan 23 2014
  • Mathematica
    LinearRecurrence[{3,-3,1},{3395619,17808513,43524451},30]
  • PARI
    a(n) = 5651522*n^2 - 2541672*n + 285769

Formula

a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).

A157110 a(n) = 1681*n^2 - 2606*n + 1010.

Original entry on oeis.org

85, 2522, 8321, 17482, 30005, 45890, 65137, 87746, 113717, 143050, 175745, 211802, 251221, 294002, 340145, 389650, 442517, 498746, 558337, 621290, 687605, 757282, 830321, 906722, 986485, 1069610, 1156097, 1245946, 1339157, 1435730, 1535665
Offset: 1

Views

Author

Vincenzo Librandi, Feb 23 2009

Keywords

Comments

The identity (5651522*n^2 - 8761372*n + 3395619)^2 - (1681*n^2 - 2606*n + 1010)*(137842*n - 106846)^2 = 1 can be written as A157112(n)^2 - a(n)*A157111(n)^2 = 1. - Vincenzo Librandi, Jan 25 2012
The continued fraction expansion of sqrt(a(n)) is [41n-32; {4, 1, 1, 4, 82n-64}]. - Magus K. Chu, Oct 03 2022
From Klaus Purath, Apr 18 2025: (Start)
a(n)*41^2-1 is a square, and a(n) is the sum of two squares (see FORMULA). There are no squares in this sequence. The odd prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*(41*y)^2 = -1 for any integer n where a(1-n) = A157010(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*41^2 - 1), x(0) = 1, x(1) = 4*D*41^2 - 1, y(0) = 1, y(1) = 4*D*41^2 - 3. The two recurrences are of the form (4*D*41^2 - 2, -1).
It follows from the above that the terms of this sequence and of A157010 belong to A031396. (End)

Links

Crossrefs

Cf. A157111, A157112.

Programs

  • Magma
    I:=[85, 2522, 8321]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jan 25 2012
  • Mathematica
    LinearRecurrence[{3,-3,1},{85,2522,8321},40] (* Vincenzo Librandi, Jan 25 2012 *)
Table[1681*n^2-2606*n+1010,{n,40}] (* Harvey P. Dale, Nov 24 2024 *)
  • PARI
    for(n=1, 22, print1(1681*n^2 - 2606*n + 1010", ")); \\ Vincenzo Librandi, Jan 25 2012

Formula

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jan 25 2012
G.f.: x*(-85 - 2267*x - 1010*x^2)/(x-1)^3. - Vincenzo Librandi, Jan 25 2012
From Klaus Purath, Apr 18 2025: (Start)
a(n) = (9*n - 7)^2 + (40*n - 31)^2 for any integer n.
1681*a(n) - 1 = (1681*n - 1303)^2 for any integer n. (End)
Showing 1-3 of 3 results.

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