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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A316939 Triangle read by rows formed using Pascal's rule except that n-th row begins and ends with Fibonacci(n+2).

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 5, 7, 7, 5, 8, 12, 14, 12, 8, 13, 20, 26, 26, 20, 13, 21, 33, 46, 52, 46, 33, 21, 34, 54, 79, 98, 98, 79, 54, 34, 55, 88, 133, 177, 196, 177, 133, 88, 55, 89, 143, 221, 310, 373, 373, 310, 221, 143, 89, 144, 232, 364, 531, 683, 746, 683, 531, 364, 232, 144, 233, 376, 596, 895, 1214, 1429
Offset: 0

Views

Author

Vincenzo Librandi, Jul 28 2018

Keywords

Examples

			Triangle begins:
   1;
   2,  2;
   3,  4,   3;
   5,  7,   7,   5;
   8, 12,  14,  12,   8;
  13, 20,  26,  26,  20,  13;
  21, 33,  46,  52,  46,  33,  21;
  34, 54,  79,  98,  98,  79,  54, 34;
  55, 88, 133, 177, 196, 177, 133, 88, 55;
  ...
		

Crossrefs

Cf. A316528 (row sums).
Columns k=0..2: A000045, A000071, A001924.
Other Fibonacci borders: A074829, A108617, A316938.

Programs

  • Maple
    f:= proc(n,k) option remember;
      if k=0 or k=n then combinat:-fibonacci(n+2) else procname(n-1,k)+procname(n-1,k-1) fi
    end proc:
    for n from 0 to 10 do
      seq(f(n,k),k=0..n)
    od; # Robert Israel, Sep 20 2018
  • Mathematica
    t={}; Do[r={}; Do[If[k==0||k==n, m=Fibonacci[n + 2], m=t[[n, k]] + t[[n, k + 1]]]; r=AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t // Flatten

Extensions

Incorrect g.f. removed by Georg Fischer, Feb 18 2020

A316937 a(n) = 3*a(n-1) - a(n-2) - 2*a(n-3) for n > 2, a(0)=3, a(1)=10, a(2)=26.

Original entry on oeis.org

3, 10, 26, 62, 140, 306, 654, 1376, 2862, 5902, 12092, 24650, 50054, 101328, 204630, 412454, 830076, 1668514, 3350558, 6723008, 13481438, 27020190, 54133116, 108416282, 217075350, 434543536, 869722694, 1740473846, 3482611772, 6967916082, 13940188782, 27887426720
Offset: 0

Views

Author

Vincenzo Librandi, Jul 17 2018

Keywords

Comments

Row sums of triangle A316938.

Crossrefs

Programs

  • GAP
    List([0..35],n->13*2^n-2*Fibonacci(n+5)); # Muniru A Asiru, Jul 22 2018
    
  • Magma
    I:=[3, 10, 26]; [n le 3 select I[n] else 3*Self(n-1)-Self(n-2)-2*Self(n-3): n in [1..40]];
    
  • Magma
    [((2^(-n)*(65*4^n + (1-Sqrt(5))^n*(-25 + 11*Sqrt(5)) - (1 + Sqrt(5))^n*(25 + 11*Sqrt(5)))) / 5): n in [0..20]]; // Vincenzo Librandi, Aug 24 2018
  • Maple
    seq(coeff(series((3+x-x^2)/((1-2*x)*(1-x-x^2)), x,n+1),x,n),n=0..35); # Muniru A Asiru, Jul 22 2018
  • Mathematica
    CoefficientList[Series[(3 + x - x^2) / ((1 - 2 x) (1 - x - x^2)), {x, 0, 33}], x] (* or *) RecurrenceTable[{a[n]==3 a[n-1] - a[n-2] - 2 a[n-3], a[0]==3, a[1]==10, a[2]==26}, a, {n, 0, 40}]
    f[n_] := 13*2^n - 2 Fibonacci[n + 5]; Array[f, 32, 0] (* or *)
    LinearRecurrence[{3, -1, -2}, {3, 10, 26}, 32] (* Robert G. Wilson v, Jul 21 2018 *)
  • PARI
    Vec((3 + x - x^2) / ((1 - 2*x)*(1 - x - x^2)) + O(x^40)) \\ Colin Barker, Jul 22 2018
    

Formula

G.f.: (3 + x - x^2) / ((1 - 2*x)*(1 - x - x^2)).
a(n) = 13*2^n - 2*Fibonacci(n+5) for n>0.
a(n) = (2^(-n)*(65*4^n + (1-sqrt(5))^n*(-25+11*sqrt(5)) - (1+sqrt(5))^n*(25+11*sqrt(5)))) / 5. - Colin Barker, Jul 22 2018
Showing 1-2 of 2 results.