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 1-5 of 5 results.

A322780 First differences of A237262.

Original entry on oeis.org

1, 4, 9, 32, 71, 252, 559, 1984, 4401, 15620, 34649, 122976, 272791, 968188, 2147679, 7622528, 16908641, 60012036, 133121449, 472473760, 1048062951, 3719778044, 8251382159, 29285750592, 64962994321, 230566226692, 511452572409, 1815244062944
Offset: 1

Views

Author

N. J. A. Sloane, Jan 05 2019, following a suggestion from Stan Wagon

Keywords

Links

Crossrefs

Cf. A237262, A199336.

Programs

  • PARI
    Vec(x*(1 + 4*x + x^2) / (1 - 8*x^2 + x^4) + O(x^40)) \\ Colin Barker, Jan 06 2019

Formula

From Colin Barker, Jan 06 2019: (Start)
G.f.: x*(1 + 4*x + x^2) / (1 - 8*x^2 + x^4).
a(n) = 8*a(n-2) - a(n-4) for n>4.
(End)

A199336 x-values in the solution to 15*x^2 - 14 = y^2.

Original entry on oeis.org

1, 3, 5, 23, 39, 181, 307, 1425, 2417, 11219, 19029, 88327, 149815, 695397, 1179491, 5474849, 9286113, 43103395, 73109413, 339352311, 575589191, 2671715093, 4531604115, 21034368433, 35677243729, 165603232371, 280886345717, 1303791490535, 2211413522007
Offset: 1

Views

Author

Sture Sjöstedt, Nov 05 2011

Keywords

Comments

When are both n+1 and 15*n+1 perfect squares? This problem gives the equation 15*x^2-14 = y^2.
Values of x (or y) in the solutions to x^2 - 8xy + y^2 + 14 = 0. - Colin Barker, Feb 05 2014

Links

Crossrefs

Cf. A198947, A199338.
Essentially the second differences of A237262. Cf. also A322780.

Programs

  • Magma
    m:=29; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-x)*(1+4*x+x^2)/(1-8*x^2+x^4))); // Bruno Berselli, Nov 08 2011
  • Mathematica
    LinearRecurrence[{0, 8, 0, -1}, {1, 3, 5, 23}, 50] (* T. D. Noe, Nov 07 2011 *)

Formula

a(n+4) = 8*a(n+2) - a(n), a(1)=1, a(2)=3, a(3)=5, a(4)=23.
G.f.: x*(1-x)*(1+4*x+x^2)/(1-8*x^2+x^4). - Bruno Berselli, Nov 08 2011

Extensions

More terms from T. D. Noe, Nov 07 2011

A281584 Solutions x to the negative Pell equation x^2 - 15*y^2 = -11 with x, y > 0.

Original entry on oeis.org

2, 7, 23, 58, 182, 457, 1433, 3598, 11282, 28327, 88823, 223018, 699302, 1755817, 5505593, 13823518, 43345442, 108832327, 341257943, 856835098, 2686718102, 6745848457, 21152486873, 53109952558, 166533176882, 418133772007, 1311112928183, 3291960223498
Offset: 1

Views

Author

Colin Barker, Jan 25 2017

Keywords

Comments

The corresponding values of y are in A237262.

Examples

			7 is in the sequence because (x, y) = (7, 2) is a solution to x^2 - 15*y^2 = -11.

Links

Crossrefs

Cf. A237262.

Programs

  • Magma
    [n le 2 select 5*n-3 else IsOdd(n) select (11*Self(n-1)-4*Self(n-2))/3 else (11*Self(n-1)-3*Self(n-2))/4: n in [1..30]]; // Bruno Berselli, Jan 25 2017
  • Mathematica
    LinearRecurrence[{0, 8, 0, -1}, {2, 7, 23, 58}, 30] (* Bruno Berselli, Jan 25 2017 *)
  • PARI
    Vec(x*(1 + x)*(2 + x)*(1 + 2*x) / (1 - 8*x^2 + x^4) + O(x^30))

Formula

G.f.: x*(1 + x)*(2 + x)*(1 + 2*x) / (1 - 8*x^2 + x^4).
a(n) = 8*a(n-2) - a(n-4) for n>4.
From Bruno Berselli, Jan 25 2017: (Start)
a(n) = (11*a(n-1) - 4*a(n-2))/3 if n is odd, a(n) = (11*a(n-1) - 3*a(n-2))/4 if n is even (see also R. J. Mathar in A237262).
a(n)*a(n-3) - a(n-1)*a(n-2) = -15*(7-(-1)^n)/2, with n>3. Example: for n=8, a(8)*a(5) - a(7)*a(6) = 3598*182 - 1433*457 = -15*3. (End)

A341927 Bisection of the numerators of the convergents of cf(1,4,1,6,1,6,...,6,1).

Original entry on oeis.org

1, 6, 47, 370, 2913, 22934, 180559, 1421538, 11191745, 88112422, 693707631, 5461548626, 42998681377, 338527902390, 2665224537743, 20983268399554, 165200922658689, 1300624112869958, 10239791980300975, 80617711729537842, 634701901856001761, 4996997503118476246, 39341278123091808207
Offset: 0

Views

Author

John O. Oladokun, Feb 23 2021

Keywords

Comments

15*a(n)^2 - 11 is a square for all terms.
x = a(n) and y = a(n+1) satisfy x^2 + y^2 - 8*x*y = -11.
x = a(n) and y = a(n+2) satisfy x^2 + y^2 - 62*x*y = -704.

Examples

			a(3) = 8*6 - 1 = 47.

Links

Crossrefs

Bisection of A237262.
Cf. A341929.

Programs

  • Mathematica
    LinearRecurrence[{8, -1}, {1,6},15]
  • PARI
    my(p=Mod('x,'x^2-8*'x+1)); a(n) = subst(lift(p^n),'x,6); \\ Kevin Ryde, Mar 01 2021

Formula

a(0) = 1; a(1) = 6; a(n) = 8*a(n-1) - a(n-2).
G.f.: (1 - 2*x)/(1 - 8*x + x^2). - Stefano Spezia, Feb 26 2021
a(n) = A237262(2*n + 1).

A341929 Bisection of the numerators of the convergents of cf (1,1,6,1,6,1,...,6,1).

Original entry on oeis.org

1, 2, 15, 118, 929, 7314, 57583, 453350, 3569217, 28100386, 221233871, 1741770582, 13712930785, 107961675698, 849980474799, 6691882122694, 52685076506753, 414788729931330, 3265624762943887, 25710209373619766, 202416050226014241, 1593618192434494162, 12546529489249939055, 98778617721565018278
Offset: 0

Views

Author

John O. Oladokun, Feb 23 2021

Keywords

Comments

15*a(n)^2 - 11 is a square for all terms.
x = a(n) and y = a(n+1) satisfy the equation x^2 + y^2 - 8*x*y = -11.
x = a(n) and y = a(n+2) satisfy x^2 + y^2 - 62*x*y = -704.

Examples

			a(3) = 8*15 - 2 = 118.

Links

Crossrefs

After a(0), bisection of A237262.
Cf. A341927.

Programs

  • Mathematica
    LinearRecurrence [{8, -1}, {1,2}, 15]
  • PARI
    my(p=Mod('x,'x^2-8*'x+1)); a(n) = subst(lift(p^n),'x,2); \\ Kevin Ryde, Feb 27 2021

Formula

a(n) = 8*a(n-1) - a(n-2) for n >= 2.
a(n) = A237262(2*n) for n >= 1.
G.f.: (1 - 6*x)/(1 - 8*x + x^2). - Stefano Spezia, Mar 01 2021
Showing 1-5 of 5 results.

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