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.

Previous Showing 41-45 of 45 results.

A022109 Fibonacci sequence beginning 1, 19.

Original entry on oeis.org

1, 19, 20, 39, 59, 98, 157, 255, 412, 667, 1079, 1746, 2825, 4571, 7396, 11967, 19363, 31330, 50693, 82023, 132716, 214739, 347455, 562194, 909649, 1471843, 2381492, 3853335, 6234827, 10088162, 16322989, 26411151, 42734140, 69145291, 111879431, 181024722
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n-1) = Sum(P(19;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=18. These are the SW-NE diagonals in P(19;n,k), the (19,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Links

Crossrefs

a(n) = A109754(18, n+1) = A101220(18, 0, n+1).

Programs

  • Magma
    a0:=1; a1:=19; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
  • Mathematica
    LinearRecurrence[{1, 1}, {1, 19}, 35] (* Paolo Xausa, Feb 22 2024 *)

Formula

a(n) = a(n-1)+a(n-2), n >= 2, a(0) = 1, a(1) = 19.
G.f.: (1+18*x)/(1-x-x^2).

A022105 Fibonacci sequence beginning 1, 15.

Original entry on oeis.org

1, 15, 16, 31, 47, 78, 125, 203, 328, 531, 859, 1390, 2249, 3639, 5888, 9527, 15415, 24942, 40357, 65299, 105656, 170955, 276611, 447566, 724177, 1171743, 1895920, 3067663, 4963583, 8031246, 12994829
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n-1)=sum(P(15;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=14. These are the SW-NE diagonals in P(15;n,k), the (15,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Links

Crossrefs

a(n) = A109754(14, n+1).
a(k) = A118654(4, k).

Programs

  • Magma
    a0:=1; a1:=15; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
  • Mathematica
    a={};b=1;c=15;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,12,1}];a (* Vladimir Joseph Stephan Orlovsky, Jul 23 2008 *)
LinearRecurrence[{1,1},{1,15},40] (* Harvey P. Dale, Oct 11 2015 *)

Formula

a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=15. a(-1):=14.
G.f.: (1+14*x)/(1-x-x^2).
a(n) = A101220(14,0,n+1). - Ross La Haye, May 02 2006

A022107 Fibonacci sequence beginning 1, 17.

Original entry on oeis.org

1, 17, 18, 35, 53, 88, 141, 229, 370, 599, 969, 1568, 2537, 4105, 6642, 10747, 17389, 28136, 45525, 73661, 119186, 192847, 312033, 504880, 816913, 1321793, 2138706, 3460499, 5599205, 9059704, 14658909
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n-1)=sum(P(17;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=16. These are the SW-NE diagonals in P(17;n,k), the (17,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Links

Crossrefs

a(n) = A109754(16, n+1) = A101220(16, 0, n+1).

Programs

  • Magma
    a0:=1; a1:=17; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
  • Mathematica
    a={};b=1;c=17;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,12,1}];a (* Vladimir Joseph Stephan Orlovsky, Jul 23 2008 *)
LinearRecurrence[{1,1},{1,17},40] (* Harvey P. Dale, Aug 04 2017 *)

Formula

a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=17. a(-1):=16.
G.f.: (1+16*x)/(1-x-x^2).

A022108 Fibonacci sequence beginning 1, 18.

Original entry on oeis.org

1, 18, 19, 37, 56, 93, 149, 242, 391, 633, 1024, 1657, 2681, 4338, 7019, 11357, 18376, 29733, 48109, 77842, 125951, 203793, 329744, 533537, 863281, 1396818, 2260099, 3656917, 5917016, 9573933, 15490949
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n-1)=sum(P(18;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=17. These are the SW-NE diagonals in P(18;n,k), the (18,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Links

Crossrefs

a(n) = A109754(17, n+1) = A101220(17, 0, n+1).

Programs

  • Magma
    a0:=1; a1:=18; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
  • Mathematica
    a={};b=1;c=18;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,12,1}];a (* Vladimir Joseph Stephan Orlovsky, Jul 23 2008 *)
LinearRecurrence[{1,1},{1,18},40] (* Harvey P. Dale, Apr 15 2018 *)

Formula

a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=18. a(-1):=17.
G.f.: (1+17*x)/(1-x-x^2).

A119587 2^n + 1 - 2*Fibonacci(n+1).

Original entry on oeis.org

0, 1, 1, 3, 7, 17, 39, 87, 189, 403, 847, 1761, 3631, 7439, 15165, 30795, 62343, 125905, 253783, 510759, 1026685, 2061731, 4136991, 8295873, 16627167, 33311647, 66716029, 133582107, 267406999, 535206833, 1071049287
Offset: 0

Views

Author

Ross La Haye, May 31 2006, Jun 27 2007

Keywords

Links

Crossrefs

Cf. A000032, A000045, A000051, A000071, A006355, A008466, A027934, A101220, A118654.

Programs

  • Mathematica
    Table[2^n + 1 - 2 Fibonacci[n + 1], {n, 0, 30}]

Formula

a(n) = 2^n + 1 - 2*Fibonacci(n+1) = 2^n + 1 + Fibonacci(n) - Fibonacci(n+3) = 2^n + 1 - Fibonacci(n) - Lucas(n). a(n) = 2(2^(n-1) - Fibonacci(n+1)) + 1, for n > 0. a(n) = A000051(n) - A006355(n+2) = A000051(n) - A000045(n) - A000032(n). a(n) = A101220(2,2,n-1) - A101220(1,1,n-3), for n > 2. a(n) = A008466(n) - A000071(n-1), for n > 0. a(n) = 2*A008466(n-1) + 1, for n > 0.
a(n) = 2*A101220(2,2,n-2) + 1, for n > 1. a(n) = Sum[2^(n-k)Fibonacci(k) - Fibonacci(k-2),{k,0,n}] = antidiagonal sums of A118654. a(n+1) - a(n) = 2(2^(n-1) - Fibonacci(n)), for n > 0. a(n+1) - a(n) = 2*A027934(n-2), for n > 1. a(n+1) - a(n) = 2*A101220(1,2,n-1), for n > 0. a(0) = 0; a(1) = 1; a(n) = a(n-1) + a(n-2) + 2^(n-2) - 1, for n > 1. a(0) = 0; a(1) = 1; a(2) = 1; a(3) = 3; a(n) = 4*a(n-1) - 4*a(n-2) - a(n-3) + 2*a(n-4), for n > 3.
O.g.f. = x(1-3x+3x^2)/((1-x)(1-2x)(1-x-x^2)).
Previous Showing 41-45 of 45 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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