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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A159683 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 3*n(j) + 1 = a(j)*a(j) and 5*n(j) + 1 = b(j)*b(j) with positive integer numbers.

Original entry on oeis.org

0, 16, 1008, 62496, 3873760, 240110640, 14882985936, 922505017408, 57180428093376, 3544264036771920, 219687189851765680, 13617061506772700256, 844038126230055650208, 52316746764756677612656, 3242794261288683956334480, 201000927453133648615125120
Offset: 1

Views

Author

Paul Weisenhorn, Apr 19 2009

Keywords

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients(R!(16*x^2/((1-x)*(1-62*x+x^2)))); // G. C. Greubel, Jun 02 2018
    
  • Maple
    for a from 1 by 2 to 100000 do b:=sqrt((5*a*a-2)/3): if (trunc(b)=b) then
    n:=(a*a-1)/3: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:
    # Second program
    seq((4/15)*(simplify(ChebyshevU(n, 31) - 61*ChebyshevU(n-1, 31)) -1), n=1..30); # G. C. Greubel, Sep 27 2022
  • Mathematica
    CoefficientList[Series[16*x/((1-x)*(1-62*x+x^2)), {x, 0, 30}], x] (* G. C. Greubel, Jun 02 2018 *)
    LinearRecurrence[{63,-63,1},{0,16,1008},30] (* Harvey P. Dale, May 07 2022 *)
  • PARI
    concat(0, Vec(16*x^2/((1-x)*(1-62*x+x^2)) + O(x^30))) \\ Colin Barker, Sep 25 2015
    
  • SageMath
    [(4/15)*(-1 + chebyshev_U(n, 31) - 61*chebyshev_U(n-1, 31)) for n in range(1,30)] # G. C. Greubel, Sep 27 2022

Formula

The a(j) recurrence is a(1)=1, a(2)=7, a(t+2) = 8*a(t+1) - a(t) resulting in terms 1, 7, 55. 433, 3409, ... (A070997).
The b(j) recurrence is b(1)=1, b(2)=9, b(t+2) = 8*b(t+1) - b(t) resulting in terms 1, 9, 71, 559, 4401, ... (A057080).
The n(j) recurrence is n(0) = n(1) = 0, n(2)=16, n(t+3) = 63*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 16, 1008, 62496, ... (this sequence).
From Colin Barker, Sep 25 2015: (Start)
a(n) = 63*a(n-1) - 63*a(n-2) + a(n-3) for n>3.
G.f.: 16*x^2 / ((1-x)*(1-62*x+x^2)). (End)
a(n) = (-8+(4+sqrt(15))*(31+8*sqrt(15))^(-n) -(-4+sqrt(15))*(31+8*sqrt(15))^n)/30. - Colin Barker, Mar 03 2016
a(n) = (4/15)*(-1 + ChebyshevU(n, 31) - 61*ChebyshevU(n-1, 31)). - G. C. Greubel, Sep 27 2022

A200999 Triangular numbers, T(m), that are four-thirds of another triangular number; T(m) such that 3*T(m) = 4*T(k) for some k.

Original entry on oeis.org

0, 28, 5460, 1059240, 205487128, 39863443620, 7733302575180, 1500220836141328, 291035108908842480, 56459310907479299820, 10952815280942075322628, 2124789705191855133290040, 412198249991938953782945160, 79964335708730965178758071028
Offset: 0

Views

Author

Charlie Marion, Feb 15 2012

Keywords

Comments

Numbers h such that 6*h+1 and 8*h+1 are both squares. [Bruno Berselli, Jul 07 2014]

Examples

			3*0 = 4*0.
3*28 = 4*21.
3*5640 = 4*4095.
3*1059240 = 4*794430.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{195, -195, 1}, {0, 28, 5460}, 20] (* T. D. Noe, Feb 15 2012 *)
  • PARI
    concat(0, Vec(28*x/((1-x)*(1-194*x+x^2)) + O(x^15))) \\ Colin Barker, Mar 02 2016

Formula

For n>1, a(n) = 194*a(n-1) - a (n-2) + 28. See A200998 for generalization.
From Colin Barker, Mar 02 2016: (Start)
a(n) = ((97+56*sqrt(3))^(-n)*(-1+(97+56*sqrt(3))^n)*(-7+4*sqrt(3)+(7+4*sqrt(3))*(97+56*sqrt(3))^n))/96.
a(n) = 195*a(n-1)-195*a(n-2)+a(n-3) for n>2.
G.f.: 28*x / ((1-x)*(1-194*x+x^2)).
(End)
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