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.

A001091 a(n) = 8*a(n-1) - a(n-2); a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 31, 244, 1921, 15124, 119071, 937444, 7380481, 58106404, 457470751, 3601659604, 28355806081, 223244789044, 1757602506271, 13837575261124, 108942999582721, 857706421400644, 6752708371622431, 53163960551578804
Offset: 0

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Author

Keywords

Comments

a(15+30k)-1 and a(15+30k)+1 are consecutive odd powerful numbers. The first pair is 13837575261124 +- 1. See A076445. - T. D. Noe, May 04 2006
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 15*y^2 = 1. The corresponding y values are in A001090. - Vincenzo Librandi, Nov 12 2010 [edited by Jon E. Schoenfield, May 04 2014]
The square root of 15*(n^2-1) at those numbers = 5 * A136325. - Richard R. Forberg, Nov 22 2013
For the above Diophantine equation x^2-15*y^2=1, x + y = A105426. - Richard R. Forberg, Nov 22 2013
a(n) solves for x in the Diophantine equation floor(3*x^2/5)= y^2. The corresponding y solutions are provided by A136325(n). x + y = A070997(n). - Richard R. Forberg, Nov 22 2013
Except for the first term, values of x (or y) in the solutions to x^2 - 8xy + y^2 + 15 = 0. - Colin Barker, Feb 05 2014

References

  • Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A001090, A090965, A098269, A322836 (column 4).

Programs

  • GAP
    a:=[1,4];; for n in [3..20] do a[n]:=8*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 26 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1-4*x)/(1-8*x+x^2) )); // G. C. Greubel, Aug 26 2019
    
  • Mathematica
    LinearRecurrence[{8,-1},{1,4},20] (* Harvey P. Dale, May 01 2014 *)
  • PARI
    a(n)=subst(poltchebi(n),x,4)
    
  • PARI
    a(n)=n=abs(n); polcoeff((1-4*x)/(1-8*x+x^2)+x*O(x^n),n) /* Michael Somos, Jun 07 2005 */
    
  • Sage
    def A001091_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( (1-4*x)/(1-8*x+x^2) ).list()
    A001091_list(20) # G. C. Greubel, Aug 26 2019
    

Formula

G.f.: A(x) = (1-4*x)/(1-8*x+x^2). - Simon Plouffe in his 1992 dissertation
For all elements x of the sequence, 15*(x^2 -1) is a square. Limit_{n -> infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 11 2002
a(n) = sqrt(15*((A001090(n))^2)+1).
a(n) = ((4+sqrt(15))^n + (4-sqrt(15))^n)/2.
a(n) = 4*S(n-1, 8) - S(n-2, 8) = (S(n, 8) - S(n-2, 8))/2, n>=1; S(n, x) := U(n, x/2) with Chebyshev's polynomials of the 2nd kind, A049310, with S(-1, x) := 0 and S(-2, x) := -1.
a(n) = T(n, 4) with Chebyshev's polynomials of the first kind; see A053120.
a(n)=a(-n). - Ralf Stephan, Jun 06 2005
a(n)*a(n+3) - a(n+1)*a(n+2) = 120. - Ralf Stephan, Jun 06 2005
From Peter Bala, Feb 19 2022: (Start)
a(n) = Sum_{k = 0..floor(n/2)} 4^(n-2*k)*15^k*binomial(n,2*k).
a(n) = [x^n] (4*x + sqrt(1 + 15*x^2))^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 16*x + 4*x^2) is the g.f. of A098269.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k. (End)
From Peter Bala, Aug 17 2022: (Start)
Sum_{n >= 1} 1/(a(n) - (5/2)/a(n)) = 1/3.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (3/2)/a(n)) = 1/5.
Sum_{n >= 1} 1/(a(n)^2 - 5/2) = 1/3 - 1/sqrt(15). (End)
a(n) = A001090(n+1)-4*A001090(n). - R. J. Mathar, Dec 12 2023

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Aug 25 2000
Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A110124 A scaled Legendre triangle.

Original entry on oeis.org

1, 0, 1, -2, 2, 1, 0, 4, 4, 1, 6, 8, 22, 6, 1, 0, 16, 136, 52, 8, 1, -20, 32, 886, 504, 94, 10, 1, 0, 64, 5944, 5136, 1232, 148, 12, 1, 70, 128, 40636, 53856, 16966, 2440, 214, 14, 1, 0, 256, 281488, 575296, 240368, 42256, 4248, 292, 16, 1, -252, 512, 1968934, 6225792, 3468844, 752800, 88566, 6776, 382, 18, 1
Offset: 0

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Author

Paul Barry, Jul 13 2005

Keywords

Comments

Row sums are A110125. Diagonal sums are A110126. Columns include A000079, A069835, A084773, and A098269.

Examples

			Rows begin
1;
0,1;
-2,2,1;
0,4,4,1;
6,8,22,6,1;
0,16,136,62,8,1;
-20,32,886,504,94,10,1;
		

Programs

Formula

Number triangle T(n, k)=2^(n-k)*LegendreP(n-k, k); T(n, k)=sum{j=0..floor((n-k)/2), (-1)^j*C(n-k, j)C(2n-2k-2j, n-k)k^(n-k-2j)}.

A157077 Triangle read by rows, coefficients of the Legendre polynomials P(n, x) times 2^n: T(n, k) = 2^n * [x^k] P(n, x), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 2, -2, 0, 6, 0, -12, 0, 20, 6, 0, -60, 0, 70, 0, 60, 0, -280, 0, 252, -20, 0, 420, 0, -1260, 0, 924, 0, -280, 0, 2520, 0, -5544, 0, 3432, 70, 0, -2520, 0, 13860, 0, -24024, 0, 12870, 0, 1260, 0, -18480, 0, 72072, 0, -102960, 0, 48620, -252, 0, 13860, 0, -120120, 0, 360360, 0, -437580, 0, 184756
Offset: 0

Views

Author

Roger L. Bagula, Feb 22 2009

Keywords

Examples

			The term order is Q(x) = a_0 + a_1*x + ... + a_n*x^n. The coefficients of the first few polynomials in this order are:
{1},
{0, 2},
{-2, 0, 6},
{0, -12, 0, 20},
{6, 0, -60, 0, 70},
{0, 60, 0, -280, 0, 252},
{-20, 0, 420, 0, -1260, 0, 924},
{0, -280, 0, 2520, 0, -5544, 0, 3432},
{70, 0, -2520, 0, 13860, 0, -24024, 0, 12870},
{0, 1260, 0, -18480, 0, 72072, 0, -102960, 0, 48620},
{-252, 0, 13860, 0, -120120, 0, 360360, 0, -437580, 0, 184756}.
.
From _Jon E. Schoenfield_, Jul 04 2022: (Start)
As a right-aligned triangle:
                                                             1;
                                                     0,      2;
                                                 -2, 0,      6;
                                         0,     -12, 0,     20;
                                      6, 0,     -60, 0,     70;
                              0,     60, 0,    -280, 0,    252;
                         -20, 0,    420, 0,   -1260, 0,    924;
                  0,    -280, 0,   2520, 0,   -5544, 0,   3432;
              70, 0,   -2520, 0,  13860, 0,  -24024, 0,  12870;
        0,  1260, 0,  -18480, 0,  72072, 0, -102960, 0,  48620;
  -252, 0, 13860, 0, -120120, 0, 360360, 0, -437580, 0, 184756. (End)
		

Crossrefs

Programs

  • Maple
    with(orthopoly):with(PolynomialTools): seq(print(CoefficientList (2^n*P(n, x), x,termorder=forward)),n=0..10); # Peter Luschny, Dec 18 2014
  • Mathematica
    Table[CoefficientList[2^n*LegendreP[n, x], x], {n, 0, 10}]; Flatten[%]
  • PARI
    tabl(nn) = for (n=0, nn, print(Vecrev(2^n*pollegendre(n)))); \\ Michel Marcus, Dec 18 2014
    
  • Sage
    def A157077_row(n):
        if n==0: return [1]
        T = [c[0] for c in (2^n*gen_legendre_P(n, 0, x)).coefficients()]
        return [0 if is_odd(n+k) else T[k//2] for k in (0..n)]
    for n in range(9): print(A157077_row(n)) # Peter Luschny, Dec 19 2014

Formula

Row sums are 2^n.
From Peter Luschny, Dec 19 2014: (Start)
T(n,0) = A126869(n).
T(n,n) = A000984(n).
T(n,1) = (-1)^floor(n/2)*A005430(floor(n/2)+1) if n is odd else 0.
Let Q(n, x) = 2^n*P(n, x).
Q(n,0) = (-1)^floor(n/2)*A126869(floor(n/2)) if n is even else 0.
Q(n,1) = A000079(n).
Q(n,2) = A069835(n).
Q(n,3) = A084773(n).
Q(n,4) = A098269(n).
Q(n,5) = A098270(n). (End)
From Fabián Pereyra, Jun 30 2022: (Start)
n*T(n,k) = 2*(2*n-1)*T(n-1,k-1) - 4*(n-1)*T(n-2,k).
T(n,k) = (-1)^floor((n-k)/2)*binomial(n+k,k)*binomial(n,floor((n-k)/2))*(1+(-1)^(n-k))/2.
O.g.f.: A(x,t) = 1/sqrt(1-4*x*t+4*x^2) = 1 + (2*t)*x + (-2+6*t^2)*x^2 + (-12*t+20*t^3)*x^3 + (6-60*t^2+70*t^4)*x^4 + .... (End)

Extensions

Name clarified and edited by Peter Luschny, Dec 18 2014
Showing 1-3 of 3 results.