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 1-4 of 4 results.

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.

Links

Crossrefs

Cf. A000045, A167821, A316528, A316938.

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

A347584 Triangle formed by Pascal's rule, except that the n-th row begins and ends with the n-th Lucas number.

Original entry on oeis.org

2, 1, 1, 3, 2, 3, 4, 5, 5, 4, 7, 9, 10, 9, 7, 11, 16, 19, 19, 16, 11, 18, 27, 35, 38, 35, 27, 18, 29, 45, 62, 73, 73, 62, 45, 29, 47, 74, 107, 135, 146, 135, 107, 74, 47, 76, 121, 181, 242, 281, 281, 242, 181, 121, 76, 123, 197, 302, 423, 523, 562, 523, 423, 302, 197, 123
Offset: 0

Views

Author

Noah Carey and Greg Dresden, Sep 07 2021

Keywords

Comments

Similar in spirit to the Fibonacci-Pascal triangle A074829, which uses Fibonacci numbers instead of Lucas numbers at the ends of each row.
If we consider the top of the triangle to be the 0th row, then the sum of terms in n-th row is 2*(2^(n+1) - Lucas(n+1)). This sum also equals 2*A027973(n-1) for n>0.

Examples

			The first two Lucas numbers (for n=0 and n=1) are 2 and 1, so the first two rows (again, for n=0 and n=1) of the triangle are 2 and 1, 1 respectively.
Triangle begins:
               2;
             1,  1;
           3,  2,  3;
         4,  5,  5,  4;
       7,  9, 10,  9,  7;
    11, 16, 19, 19, 16, 11;
  18, 27, 35, 38, 35, 27, 18;

Links

Crossrefs

Cf. A000032, A027973.
Cf. A227550, A228196 (general formula).
Fibonacci borders: A074829, A108617, A316938, A316939.

Programs

  • Mathematica
    T[n_, 0] := LucasL[n]; T[n_, n_] := LucasL[n];
T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten

Formula

a(n) = 2*A074829(n+1) - A108617(n).

A379837 Triangle read by rows formed using Pascal's rule except that n-th row begins and ends with Fibonacci(n+3).

Original entry on oeis.org

2, 3, 3, 5, 6, 5, 8, 11, 11, 8, 13, 19, 22, 19, 13, 21, 32, 41, 41, 32, 21, 34, 53, 73, 82, 73, 53, 34, 55, 87, 126, 155, 155, 126, 87, 55, 89, 142, 213, 281, 310, 281, 213, 142, 89, 144, 231, 355, 494, 591, 591, 494, 355, 231, 144
Offset: 0

Views

Author

Vincenzo Librandi, Jan 26 2025

Keywords

Examples

			Triangle begins:
      k= 0   1   2  3    4   5   6
  n=0:   2;
  n=1:   3,  3;
  n=2:   5,  6,  5;
  n=3:   8, 11, 11, 8;
  n=4:  13, 19, 22, 19, 13;
  n=5:  21, 32, 41, 41, 32, 21;
  n=6:  34, 53, 73, 82, 73, 53, 34;
  ...

Crossrefs

Cf. A000045, A074829, A316938, A316939.
Cf. A108617.

Programs

  • Mathematica
    // As triangle // t={};Do[r={};Do[If[k==0||k==n,m=Fibonacci[n+3],m=t[[n,k]]+t[[n,k+1]]];r=AppendTo[r,m],{k,0,n}];AppendTo[t,r],{n,0,11}];t
Showing 1-4 of 4 results.

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