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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A015609 a(n) = 11*a(n-1) + 12*a(n-2).

Original entry on oeis.org

0, 1, 11, 133, 1595, 19141, 229691, 2756293, 33075515, 396906181, 4762874171, 57154490053, 685853880635, 8230246567621, 98762958811451, 1185155505737413, 14221866068848955, 170662392826187461
Offset: 0

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Author

Keywords

Comments

Number of walks of length n between any two distinct nodes of the complete graph K_13. Example: a(2)=11 because the walks of length 2 between the nodes A and B of the complete graph ABCDEFGHIJKLM are ACB, ADB, AEB, AFB, AGB, AHB, AIB, AJB, AKB, ALB and AMB. - Emeric Deutsch, Apr 01 2004

Crossrefs

Programs

  • Magma
    [(1/13)*(12^n-(-1)^n): n in [0..20]]; // Vincenzo Librandi, Oct 11 2011
    
  • Mathematica
    CoefficientList[Series[x/(1-11*x-12*x^2), {x, 0, 50}], x] (* or *) LinearRecurrence[{11,12}, {0,1}, 30] (* G. C. Greubel, Dec 30 2017 *)
  • PARI
    x='x+O('x^30); concat([0], Vec(x/(1-11*x-12*x^2))) \\ G. C. Greubel, Dec 30 2017
  • Sage
    [lucas_number1(n,11,-12) for n in range(0, 18)] # Zerinvary Lajos, Apr 27 2009
    
  • Sage
    [abs(gaussian_binomial(n,1,-12)) for n in range(0,18)] # Zerinvary Lajos, May 28 2009
    

Formula

From Emeric Deutsch, Apr 01 2004: (Start)
a(n) = 12^(n-1) - a(n-1).
G.f.: x/(1 - 11*x - 12*x^2). (End)
E.g.f.: exp(-x)*(exp(13*x) - 1)/13. - Stefano Spezia, Mar 11 2020

A271427 a(n) = 7^n - a(n-1) for n>0, a(0)=0.

Original entry on oeis.org

0, 7, 42, 301, 2100, 14707, 102942, 720601, 5044200, 35309407, 247165842, 1730160901, 12111126300, 84777884107, 593445188742, 4154116321201, 29078814248400, 203551699738807, 1424861898171642, 9974033287201501, 69818233010410500, 488727631072873507, 3421093417510114542
Offset: 0

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Author

Ilya Gutkovskiy, Apr 13 2016

Keywords

Comments

In general, the ordinary generating function for the recurrence b(n) = k^n - b(n-1), where n>0 and b(0)=0, is k*x/((1 + x)*(1 - k*x)). This recurrence gives the closed form b(n) = k*(k^n - (-1)^n)/(k + 1).

Examples

			a(2) = 7^2 - a(2-1) = 49 - 7 = 42.
a(4) = 7^4 - a(4-1) = 2401 - 301 = 2100.
		

Crossrefs

Cf. similar sequences with the recurrence b(n) = k^n - b(n-1): A125122 (k=1), A078008 (k=2), A054878 (k=3), A109499 (k=4), A109500 (k=5), A109501 (k=6), this sequence (k=7), A093134 (k=8), A001099 (k=n).

Programs

  • Mathematica
    LinearRecurrence[{6, 7}, {0, 7}, 30]
    Table[7 (7^n - (-1)^n)/8, {n, 0, 30}]
  • PARI
    vector(50, n, n--; 7*(7^n-(-1)^n)/8) \\ Altug Alkan, Apr 13 2016
    
  • Python
    for n in range(0,10**2):print((int)((7*(7**n-(-1)**n))/8))
    # Soumil Mandal, Apr 14 2016

Formula

O.g.f.: 7*x/(1 - 6*x - 7*x^2).
E.g.f.: (7/8)*(exp(7*x) - exp(-x)).
a(n) = 6*a(n-1) + 7*a(n-2).
a(n) = 7*(7^n - (-1)^n)/8.
a(n) = 7*A015552(n).
Sum_{n>0} 1/(a(n) + a(n-1)) = 1/6 = A020793.
Limit_{n->oo} a(n-1)/a(n) = 1/7 = A020806.
Previous Showing 11-12 of 12 results.