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 1-6 of 6 results.

A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264
Offset: 0

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Author

Keywords

Comments

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Crossrefs

Programs

  • Magma
    I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011
    
  • Mathematica
    LinearRecurrence[{2,-10}, {0,1}, 50]
  • PARI
    a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016
    
  • SageMath
    [lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

Formula

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A102591 a(n) = Sum_{k=0..n} binomial(2n+1, 2k)*3^(n-k).

Original entry on oeis.org

1, 6, 44, 328, 2448, 18272, 136384, 1017984, 7598336, 56714752, 423324672, 3159738368, 23584608256, 176037912576, 1313964867584, 9807567290368, 73204678852608, 546407161659392, 4078438577864704, 30441879976280064
Offset: 0

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Author

Paul Barry, Jan 22 2005

Keywords

Comments

In general, Sum_{k=0..n} binomial(2n+1,2k)*r^(n-k) has g.f. (1-(r-1)x)/(1-2(r+1)+(r-1)^2x^2) and a(n) = ((sqrt(r)-1)^(2n+1) + (sqrt(r)+1)^(2n+1))/(2*sqrt(r)).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{8,-4},{1,6},20] (* Harvey P. Dale, Sep 28 2021 *)

Formula

G.f.: (1-2x)/(1-8x+4x^2);
a(n) = 8*a(n-1) - 4*a(n-2);
a(n) = sqrt(3)*(sqrt(3)-1)^(2n+1)/6 + sqrt(3)*(sqrt(3)+1)^(2n+1)/6.
a(n) = 2^n*A079935(n). - R. J. Mathar, Sep 20 2012
a(n) = 2^(2*n+1)*Sum_{k >= n} binomial(2*k,2*n)*(1/3)^(k+1). Cf. A099156. - Peter Bala, Nov 29 2021
3*a(n)^2 = A107903(n)^2 + 2^(2*n+1). - Philippe Deléham, Mar 21 2023

A107903 Generalized NSW numbers.

Original entry on oeis.org

1, 10, 76, 568, 4240, 31648, 236224, 1763200, 13160704, 98232832, 733219840, 5472827392, 40849739776, 304906608640, 2275853910016, 16987204845568, 126794223124480, 946404965613568, 7064062832410624, 52726882796830720
Offset: 0

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Author

Paul Barry, May 27 2005

Keywords

Comments

Counts total area under elevated Schroeder paths of length 2n+2, where horizontal steps can choose from three colors.
Case r=3 for family (1+(r-1)x)/(1-2(1+r)x+(1-r)^2*x^2). Case r=2 gives NSW numbers A002315 and case r=4 gives NSW numbers A096053.
Fifth binomial transform of (1+8x)/(1-16x^2), A107906.
If p is an odd prime, a((p-1)/2) == 1 mod p. - Altug Alkan, Mar 17 2016

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[2 n + 1, 2 k] 3^k, {k, 0, n}], {n, 0, 20}] (* or *) CoefficientList[Series[(1 + 2 x)/(1 - 8 x + 4 x^2), {x, 0, 20}], x] (* Michael De Vlieger, Mar 17 2016 *)
  • PARI
    Vec((1+2*x)/(1-8*x+4*x^2) + O(x^40)) \\ Michel Marcus, Mar 17 2016

Formula

G.f.: (1+2*x)/(1-8*x+4*x^2). [corrected by Ralf Stephan, Nov 30 2010]
a(n) = Sum_{k=0..n} binomial(2*n+1, 2*k)*3^k.
a(n) = ((1+sqrt(3))*(4+2*sqrt(3))^n+(1-sqrt(3))*(4-2*sqrt(3))^n)/2 = A099156(n+1)+2*A099156(n).
a(n) = 8*a(n-1) - 4*a(n-2); a(0) = 1, a(1) = 10. - Lekraj Beedassy, Apr 19 2020
a(n) = 2^n*A001834(n). - Philippe Deléham, Mar 18 2023

Extensions

Typo corrected and link added by Johannes W. Meijer, Aug 07 2010

A342133 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of g.f. 1/(1 - 2*k*x + k*x^2).

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 3, 0, 1, 6, 14, 4, 0, 1, 8, 33, 48, 5, 0, 1, 10, 60, 180, 164, 6, 0, 1, 12, 95, 448, 981, 560, 7, 0, 1, 14, 138, 900, 3344, 5346, 1912, 8, 0, 1, 16, 189, 1584, 8525, 24960, 29133, 6528, 9, 0, 1, 18, 248, 2548, 18180, 80750, 186304, 158760, 22288, 10, 0
Offset: 0

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Author

Seiichi Manyama, Mar 01 2021

Keywords

Examples

			Square array begins:
  1, 1,   1,    1,     1,     1, ...
  0, 2,   4,    6,     8,    10, ...
  0, 3,  14,   33,    60,    95, ...
  0, 4,  48,  180,   448,   900, ...
  0, 5, 164,  981,  3344,  8525, ...
  0, 6, 560, 5346, 24960, 80750, ...
		

Crossrefs

Columns 0..5 give A000007, A000027(n+1), A007070, A138395, A099156(n+1), A190987(n+1).
Rows 0..2 give A000012, A005843, A033991.
Main diagonal gives (-1) * A109520(n+1).

Programs

  • Maple
    T:= (n, k)-> (<<0|1>, <-k|2*k>>^(n+1))[1, 2]:
    seq(seq(T(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, Mar 01 2021
  • Mathematica
    T[n_, k_] := Sum[If[k == j == 0, 1, (2*k)^j] * (-2)^(j - n) * Binomial[j, n - j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Apr 27 2021 *)
  • PARI
    T(n, k) = sum(j=0, n\2, (2*k)^(n-j)*(-2)^(-j)*binomial(n-j, j));
    
  • PARI
    T(n, k) = sum(j=0, n, (2*k)^j*(-2)^(j-n)*binomial(j, n-j));
    
  • PARI
    T(n, k) = round(sqrt(k)^n*polchebyshev(n, 2, sqrt(k)));

Formula

T(0,k) = 1, T(1,k) = 2*k and T(n,k) = k*(2*T(n-1,k) - T(n-2,k)) for n > 1.
T(n,k) = Sum_{j=0..floor(n/2)} (2*k)^(n-j) * (-1/2)^j * binomial(n-j,j) = Sum_{j=0..n} (2*k)^j * (-1/2)^(n-j) * binomial(j,n-j).
T(n,k) = sqrt(k)^n * U(n, sqrt(k)) where U(n, x) is a Chebyshev polynomial of the second kind.

A110274 Expansion of (-16-7*x+6*x^2+28*x^3+8*x^4) / ((x-1)*(x^2+x+1)*(4*x^2-8*x+1)).

Original entry on oeis.org

16, 135, 1010, 7528, 56183, 419346, 3130024, 23362807, 174382354, 1301607592, 9715331319, 72516220178, 541268436136, 4040082608375, 30155587122450, 225084366546088, 1680052583878903, 12540083204846866, 93600455303259304
Offset: 0

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Author

Creighton Dement, Jul 18 2005

Keywords

Crossrefs

Cf. A110275.

Programs

  • Maple
    seriestolist(series((-16-7*x+6*x^2+28*x^3+8*x^4)/((x-1)*(x^2+x+1)*(4*x^2-8*x+1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: bisection of 4tessigcyczapsumseq[A*B] with A = - 'j + 'k - 'ii' - 'ij' - 'ik' and B = + .5'i + .5'j - .5'k + .5i' - .5j' + .5k' + .5'ij' + .5'ik' - .5'ji' - .5'ki'; Sumtype is set to: sum[(Y[0], Y[1], Y[2]),mod(3)
  • PARI
    Vec((16 + 7*x - 6*x^2 - 28*x^3 - 8*x^4) / ((1 - x)*(1 + x + x^2)*(1 - 8*x + 4*x^2)) + O(x^20)) \\ Colin Barker, May 12 2019

Formula

a(n) = 8*a(n-1) - 4*a(n-2) + a(n-3) - 8*a(n-4) + 4*a(n-5) for n>4. - Colin Barker, May 12 2019
117*a(n) = -47*A049347(n) -67*A049347(n-1) + 8*(209*A099156(n+1)+274*A099156(n)) +247. - R. J. Mathar, Sep 11 2019

A138240 Expansion of (1/4)(1-sqrt(1-12x)/sqrt(1-4x)).

Original entry on oeis.org

0, 1, 6, 40, 296, 2400, 20928, 192768, 1848960, 18277888, 184890368, 1904259072, 19898765312, 210424545280, 2247494172672, 24209586782208, 262696649785344, 2868744309571584, 31504024885002240, 347697247933169664
Offset: 0

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Author

Paul Barry, Mar 07 2008

Keywords

Comments

Hankel transform of a(n) is -4^comb(n,2)*A099156(n)=-4^comb(n,2)*[x^n](x/(1-8x+4x^2)).
Hankel transform of a(n+1) is 4^comb(n+1,2)=A053763(n+1).
Hankel transform of a(n+2) is 4^comb(n+1,2)*A102591(n+1)=4^comb(n+1,2)*[x^n](6-4x)/(1-8x+4x^2).

Crossrefs

Cf. A104498.

Programs

  • Mathematica
    CoefficientList[Series[1/4*(1-Sqrt[1-12*x]/Sqrt[1-4*x]), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

Formula

Recurrence: n*a(n) = 4*(4*n-5)*a(n-1) - 48*(n-2)*a(n-2) . - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(2*n-7/2)*3^(n+1/2)/(sqrt(Pi)*n^(3/2)) . - Vaclav Kotesovec, Oct 20 2012
Showing 1-6 of 6 results.