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.

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A075841 Numbers k such that 2*k^2 - 9 is a square.

Original entry on oeis.org

3, 15, 87, 507, 2955, 17223, 100383, 585075, 3410067, 19875327, 115841895, 675176043, 3935214363, 22936110135, 133681446447, 779152568547, 4541233964835, 26468251220463, 154268273357943, 899141388927195
Offset: 1

Views

Author

Gregory V. Richardson, Oct 14 2002

Keywords

Comments

Lim. n-> Inf. a(n)/a(n-1) = 3 + 2*sqrt(2).
Positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 36 = 0. - Colin Barker, Feb 08 2014
For each member t of the sequence there exists a nonnegative r such that t^2 = r^2 + (r+3)^2. The r values are in A241976. Example: 87^2 = 60^2 + 63^2. - Bruno Berselli, Jul 10 2017

References

  • A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
  • L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
  • Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Links

Programs

  • Mathematica
    CoefficientList[Series[3 (1 - x)/(1 - 6 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 11 2014 *)
LinearRecurrence[{6,-1},{3,15},20] (* Harvey P. Dale, Jun 05 2023 *)
  • PARI
    isok(n) = issquare(2*n^2-9); \\ Michel Marcus, Jul 10 2017

Formula

a(n) = 3*sqrt(2)/4*((1+sqrt(2))^(2*n-1)-(1-sqrt(2))^(2*n-1)) = 6*a(n-1) - a(n-2).
G.f.: 3*x*(1-x)/(1-6*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = 3*A001653(n). - R. J. Mathar, Sep 27 2014

A301383 Expansion of (1 + 3*x - 2*x^2)/(1 - 7*x + 7*x^2 - x^3).

Original entry on oeis.org

1, 10, 61, 358, 2089, 12178, 70981, 413710, 2411281, 14053978, 81912589, 477421558, 2782616761, 16218279010, 94527057301, 550944064798, 3211137331489, 18715879924138, 109084142213341, 635788973355910, 3705649697922121, 21598109214176818, 125883005587138789, 733699924308655918
Offset: 0

Views

Author

Bruno Berselli, Mar 20 2018

Keywords

Comments

y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+2). The corresponding x values are listed in A075841.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+1) are in A002315, and A075870 gives the x values.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y) are in A046090, and A001653 gives the x values.
Also, indices y for which 4*A000217(y) + 5 is a square. The next integers k such that k*A000217(y) + 5 is a square for infinitely many y values are 11, 20, 22, 29, 31, ...
First differences are in A106329.

Links

Crossrefs

Subsequence of A301451.
Cf. A000217, A000290, A001652, A002315, A033539, A046090, A053142, A075841, A077444, A106329, A241976.

Programs

  • Julia
    using Nemo
function A301383List(len)
    R, x = PowerSeriesRing(ZZ, len+2, "x")
    f = divexact(1+3*x-2*x^2, 1-7*x+7*x^2-x^3)
    [coeff(f, k) for k in 0:len]
end
A301383List(23) |> println # Peter Luschny, Mar 21 2018
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)));
  • Maple
    f:= gfun:-rectoproc({a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3), a(0)=1,a(1)=10,a(2)=61},a(n),remember):
map(f, [$0..50]); # Robert Israel, Mar 21 2018
  • Mathematica
    CoefficientList[Series[(1 + 3 x - 2 x^2)/(1 - 7 x + 7 x^2 - x^3), {x, 0, 30}], x]
  • Maxima
    makelist(coeff(taylor((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3), x, 0, n), x, n), n, 0, 30);
  • PARI
    Vec((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)+O(x^30))
  • Sage
    m=30; L. = PowerSeriesRing(ZZ, m); f=(1+3*x-2*x^2)/(1-7*x+7*x^2-x^3); print(f.coefficients())

Formula

O.g.f.: (1 + 3*x - 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) = 6*a(n-1) - a(n-2) + 2.
a(n) = (3/4)*((1 + sqrt(2))^(2*n + 1) + (1 - sqrt(2))^(2*n + 1)) - 1/2.
a(n) = A033539(2*n+2) = A241976(n+1) + 1 = 3*A001652(n) + 1 = 3*A046090(n) - 2.
a(n) = A053142(n+1) + 3*A053142(n) - 2*A053142(n-1), n>0.
2*a(n) = 3*A002315(n) - 1.
4*a(n) = 3*A077444(n+1) - 2.
E.g.f.: (3*exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - cosh(x) - sinh(x))/2. - Stefano Spezia, Mar 06 2020
Let T(n) be the n-th triangular number, A000217(n). Then T(a(n)-3) + 2*T(a(n)-2) + 3*T(a(n)-1) + 4*T(a(n)) + 3*T(a(n)+1) + 2*T(a(n)+2) + T(a(n)+3) = (A001653(n) + A001653(n+2))^2. - Charlie Marion, Mar 16 2021
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