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.

A077241 Combined Diophantine Chebyshev sequences A054488 and A077413.

Original entry on oeis.org

1, 2, 8, 13, 47, 76, 274, 443, 1597, 2582, 9308, 15049, 54251, 87712, 316198, 511223, 1842937, 2979626, 10741424, 17366533, 62605607, 101219572, 364892218, 589950899, 2126747701, 3438485822, 12395593988, 20040964033, 72246816227, 116807298376
Offset: 0

Views

Author

Wolfdieter Lang, Nov 08 2002

Keywords

Comments

-8*a(n)^2 + b(n)^2 = 17, with the companion sequence b(n)= A077242(n).
The number a > 0 belongs to the sequence A077241, if a^2 belongs to the sequence A034856. - Alzhekeyev Ascar M, Apr 27 2012
Numbers k such that k^2 + 2 is a triangular number (see A214838). - Alex Ratushnyak, Mar 07 2013

Examples

			8*a(2)^2 + 17 = 8*8^2+17 = 529 = 23^2 = A077242(2)^2.
		

Programs

  • Magma
    I:=[1,2,8,13]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Feb 18 2014
  • Mathematica
    LinearRecurrence[{0, 6, 0, -1}, {1, 2, 8, 13}, 30] (* Bruno Berselli, Mar 10 2013 *)
    CoefficientList[Series[(1 + x) (1 + x + x^2)/(1 - 6 x^2 + x^4), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 18 2014 *)
  • Maxima
    makelist(expand((-1)^n*((4-5*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor((n+1)/2))+(4+5*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor((n+1)/2)))/8), n, 0, 30); /* Bruno Berselli, Mar 10 2013 */
    

Formula

a(2k) = A054488(k) and a(2k+1)= A077413(k) for k>=0.
G.f.: (1+x)*(1+x+x^2)/(1-6*x^2+x^4).
a(n) = (-1)^n*((4-5*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor((n+1)/2))+(4+5*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor((n+1)/2)))/8. [Bruno Berselli, Mar 10 2013]

A077239 Bisection (odd part) of Chebyshev sequence with Diophantine property.

Original entry on oeis.org

7, 37, 215, 1253, 7303, 42565, 248087, 1445957, 8427655, 49119973, 286292183, 1668633125, 9725506567, 56684406277, 330380931095, 1925601180293, 11223226150663, 65413755723685, 381259308191447, 2222142093424997, 12951593252358535, 75487417420726213
Offset: 0

Views

Author

Wolfdieter Lang, Nov 08 2002

Keywords

Comments

a(n)^2 - 8*b(n)^2 = 17, with the companion sequence b(n)= A077413(n).
The even part is A077240(n) with Diophantine companion A054488(n).

Examples

			37 = a(1) = sqrt(8*A077413(1)^2 +17) = sqrt(8*13^2 + 17)= sqrt(1369) = 37.
		

Crossrefs

Cf. A077242 (even and odd parts).

Programs

  • Mathematica
    Table[2*ChebyshevT[n+1, 3] + ChebyshevT[n, 3], {n, 0, 19}]  (* Jean-François Alcover, Dec 19 2013 *)
  • PARI
    Vec((7-5*x)/(1-6*x+x^2) + O(x^40)) \\ Colin Barker, Oct 12 2015

Formula

a(n) = 6*a(n-1) - a(n-2), a(-1) := 5, a(0)=7.
a(n) = 2*T(n+1, 3)+T(n, 3), with T(n, x) Chebyshev's polynomials of the first kind, A053120. T(n, 3)= A001541(n).
G.f.: (7-5*x)/(1-6*x+x^2).
a(n) = (((3-2*sqrt(2))^n*(-8+7*sqrt(2))+(3+2*sqrt(2))^n*(8+7*sqrt(2))))/(2*sqrt(2)). - Colin Barker, Oct 12 2015

A077240 Bisection (even part) of Chebyshev sequence with Diophantine property.

Original entry on oeis.org

5, 23, 133, 775, 4517, 26327, 153445, 894343, 5212613, 30381335, 177075397, 1032071047, 6015350885, 35060034263, 204344854693, 1191009093895, 6941709708677, 40459249158167, 235813785240325, 1374423462283783, 8010726988462373, 46689938468490455
Offset: 0

Views

Author

Wolfdieter Lang, Nov 08 2002

Keywords

Comments

a(n)^2 - 8*b(n)^2 = 17, with the companion sequence b(n)= A054488(n).
The odd part is A077239(n) with Diophantine companion A077413(n).

Examples

			23 = a(1) = sqrt(8*A054488(1)^2 + 17) = sqrt(8*8^2 + 17)= sqrt(529) = 23.
		

Crossrefs

Cf. A077242 (even and odd parts).

Programs

  • Mathematica
    Table[ChebyshevT[n+1, 3] + 2*ChebyshevT[n, 3], {n, 0, 19}]  (* Jean-François Alcover, Dec 19 2013 *)
    LinearRecurrence[{6,-1},{5,23},30] (* Harvey P. Dale, Mar 29 2017 *)
  • PARI
    Vec((5-7*x)/(1-6*x+x^2) + O(x^40)) \\ Colin Barker, Oct 12 2015

Formula

a(n) = 6*a(n-1) - a(n-2), a(-1) = 7, a(0) = 5.
a(n) = T(n+1, 3)+2*T(n, 3), with T(n, x) Chebyshev's polynomials of the first kind, A053120. T(n, 3)= A001541(n).
G.f.: (5-7*x)/(1-6*x+x^2).
a(n) = (((3-2*sqrt(2))^n*(-4+5*sqrt(2))+(3+2*sqrt(2))^n*(4+5*sqrt(2))))/(2*sqrt(2)). - Colin Barker, Oct 12 2015

A144796 Expansion of x(5+215x-1253x^2-23x^3)/((1+34x+x^2)(1-34x+x^2)).

Original entry on oeis.org

5, 215, 4517, 248087, 5212613, 286292183, 6015350885, 330380931095, 6941709708677, 381259308191447, 8010726988462373, 439972911271998743, 9244372002975869765, 507728358348578357975, 10667997280707165246437
Offset: 1

Views

Author

Richard Choulet, Sep 21 2008

Keywords

Comments

The original definition "Numbers n such that (n^2-17)/2 is a square" describes A077242. - R. J. Mathar, Nov 27 2011

Programs

  • Mathematica
    Rest[CoefficientList[Series[x (5+215x-1253x^2-23x^3)/((1+34x+x^2)(1-34x+x^2)),{x,0,30}],x]] (* or *) LinearRecurrence[{0,1154,0,-1},{5,215,4517,248087},30] (* Harvey P. Dale, Apr 04 2014 *)
  • PARI
    Vec(x(5+215*x-1253*x^2-23*x^3)/((1+34*x+x^2)*(1-34*x+x^2))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

Formula

a(n) = 1154*a(n-2) - a(n-4).

A359438 For n >= 0, let S be the sequence of numbers m such that (m^2 - 2*n^2 + 1)/2 is a square. Then a(n) is the number k such that S(j) = 6*S(j-k) - S(j-2k) for all j for which S(j-2k) is defined.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 4, 2, 4, 2, 4, 2, 2, 4, 4, 2, 2, 4, 2, 2, 4, 4, 2, 3, 4, 4, 4, 4, 2, 4, 2, 8, 2, 2, 4, 2, 2, 2, 6, 2, 2, 4, 4, 2, 2, 4, 2, 4, 8, 4, 2, 4, 6, 2, 4, 4, 2, 2, 2, 8, 4, 4, 4, 2, 4, 4, 4, 2, 4, 4, 2, 4, 4, 4, 2, 2, 8, 4, 4, 2, 4
Offset: 0

Views

Author

Jon E. Schoenfield, Dec 31 2022

Keywords

Examples

			For n = 0, {S(j)} = A002315 (the NSW numbers), which satisfies S(j) = 6*S(j-1) - S(j-2), so a(0) = 1.
For n = 1, {S(j)} = A001541, which also satisfies S(j) = 6*S(j-1) - S(j-2), so a(1) = 1.
For n = 2, {S(j)} = A077443, which satisfies S(j) = 6*S(j-2) - S(j-4), so a(2) = 2.
For n = 5, {S(j)} = A106525, which satisfies S(j) = 6*S(j-3) - S(j-6), so a(5) = 3.
		

Crossrefs

Formula

a(0) = 1; for n >= 1, a(n) = A000005(2*n^2 - 1).
Showing 1-5 of 5 results.