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.

A251703 4-step Fibonacci sequence starting with 1,1,0,0.

Original entry on oeis.org

1, 1, 0, 0, 2, 3, 5, 10, 20, 38, 73, 141, 272, 524, 1010, 1947, 3753, 7234, 13944, 26878, 51809, 99865, 192496, 371048, 715218, 1378627, 2657389, 5122282, 9873516, 19031814, 36685001, 70712613, 136302944, 262732372, 506432930, 976180859
Offset: 0

Views

Author

Arie Bos, Dec 07 2014

Keywords

Links

Crossrefs

Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251654, A251655, A251656, A251704, A251705.

Programs

  • J
    NB. see A251655 for the program and apply it to 1,1,0,0.
  • Mathematica
    LinearRecurrence[Table[1, {4}], {1, 1, 0, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)

Formula

a(n+4) = a(n) + a(n+1) + a(n+2) + a(n+3).
G.f.: (-1+2*x^2+2*x^3)/(-1+x+x^2+x^3+x^4) . - R. J. Mathar, Mar 28 2025
a(n) = A000078(n+3)-2*A000078(n+1)-2*A000078(n). - R. J. Mathar, Mar 28 2025

A000336 a(n) = a(n-1)*a(n-2)*a(n-3)*a(n-4); for n < 5, a(n) = n.

Original entry on oeis.org

1, 2, 3, 4, 24, 576, 165888, 9172942848, 21035720123168587776, 18437563379178327736384102280592359424, 590180110002114158896983994712576414865667267958188575935810179040280576
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

The next term has 139 digits. - Harvey P. Dale, Jan 21 2019

Links

Crossrefs

Cf. A000301, A000308, A001631, A003586, A251656.

Programs

  • Maple
    A000336 := proc(n) option remember; if n <=4 then n else A000336(n-1)*A000336(n-2)*A000336(n-3)*A000336(n-4); fi; end;
  • Mathematica
    t = {1, 2, 3, 4}; Do[AppendTo[t, t[[-1]]*t[[-2]]*t[[-3]]*t[[-4]]], {n, 5, 15}] (* T. D. Noe, Jun 19 2012 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,a b c d}; NestList[nxt,{1,2,3,4},10][[All,1]] (* Harvey P. Dale, Jan 21 2019 *)
  • PARI
    a(n,a=[24,1,2,3,4])={for(n=6,n,a[n%5+1]=a[(n-1)%5+1]^2\a[n%5+1]);a[n%5+1]} \\ M. F. Hasler, Apr 22 2018
  • PARI
    first(n) = n = max(n, 5); my(res = vector(n)); for(i=1, 4, res[i] = i); res[5]=24; for(i = 6, n, res[i] = res[i-1]^2 / res[i - 5]); res \\ David A. Corneth, Apr 22 2018

Formula

a(n) = 2^A251656(n) * 3^A001631(n-1). - Vaclav Kotesovec, Feb 02 2016
a(n) = a(n-1)^2 / a(n-5), for n > 5. - M. F. Hasler, Apr 22 2018

A251654 4-step Fibonacci sequence starting with 0, 1, 1, 0.

Original entry on oeis.org

0, 1, 1, 0, 2, 4, 7, 13, 26, 50, 96, 185, 357, 688, 1326, 2556, 4927, 9497, 18306, 35286, 68016, 131105, 252713, 487120, 938954, 1809892, 3488679, 6724645, 12962170, 24985386, 48160880, 92833081, 178941517, 344920864, 664856342, 1281551804
Offset: 0

Views

Author

Arie Bos, Dec 06 2014

Keywords

Links

Crossrefs

Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251655, A251656, A251672, A251703, A251704, A251705.

Programs

  • J
    NB. see A251655 for the program and apply it to 0,1,1,0.
  • Mathematica
    LinearRecurrence[Table[1, {4}], {0, 1, 1, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)

Formula

a(n+4) = a(n) + a(n+1) + a(n+2) + a(n+3).
G.f.: x*(-1+2*x^2)/(-1+x+x^2+x^3+x^4). - R. J. Mathar, Mar 28 2025
a(n) = A000078(n+2)-2*A000078(n). - R. J. Mathar, Mar 28 2025

A251655 4-step Fibonacci sequence starting with 0, 1, 1, 1.

Original entry on oeis.org

0, 1, 1, 1, 3, 6, 11, 21, 41, 79, 152, 293, 565, 1089, 2099, 4046, 7799, 15033, 28977, 55855, 107664, 207529, 400025, 771073, 1486291, 2864918, 5522307, 10644589, 20518105, 39549919, 76234920, 146947533, 283250477, 545982849, 1052415779, 2028596638
Offset: 0

Views

Author

Arie Bos, Dec 06 2014

Keywords

Links

Crossrefs

Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251654, A251656, A251672, A251703, A251704, A251705.

Programs

  • J
    (see www.jsoftware.com) First construct the generating matrix
   [M=: (#.@}: + {:)\"1&.|: <:/~i.4
1 1 1 1
1 2 2 2
2 3 4 4
4 6 7 8
Given that matrix, one can produce the first 4*250 numbers with
, M(+/ . *)^:(i.250) 0 1 1 1x
  • Mathematica
    LinearRecurrence[Table[1, {4}], {0, 1, 1, 1}, 36] (* Michael De Vlieger, Dec 09 2014 *)

Formula

a(n+4) = a(n) + a(n+1) + a(n+2) + a(n+3).
G.f.: x*(x-1)*(1+x)/(-1+x+x^2+x^3+x^4) . - R. J. Mathar, Mar 28 2025
a(n) = A000078(n+2)-A000078(n). - R. J. Mathar, Mar 28 2025

A251704 4-step Fibonacci sequence starting with 1, 1, 0, 1.

Original entry on oeis.org

1, 1, 0, 1, 3, 5, 9, 18, 35, 67, 129, 249, 480, 925, 1783, 3437, 6625, 12770, 24615, 47447, 91457, 176289, 339808, 655001, 1262555, 2433653, 4691017, 9042226, 17429451, 33596347, 64759041, 124827065, 240611904, 463794357, 893992367, 1723225693
Offset: 0

Views

Author

Arie Bos, Dec 07 2014

Keywords

Links

Crossrefs

Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251654, A251655, A251656, A251703, A251705.

Programs

  • J
    NB. see A251655 for the program and apply it to 1,1,0,1.
  • Mathematica
    LinearRecurrence[Table[1, {4}], {1, 1, 0, 1}, 36] (* Michael De Vlieger, Dec 09 2014 *)

Formula

a(n+4) = a(n) + a(n+1) + a(n+2) + a(n+3).
G.f.: (1+x)*(x^2+x-1)/(-1+x+x^2+x^3+x^4) . - R. J. Mathar, Mar 28 2025
a(n) = A001630(n-2)+A001630(n-1), n>2. - R. J. Mathar, Mar 28 2025

A251705 4-step Fibonacci sequence starting with 1, 1, 1, 0.

Original entry on oeis.org

1, 1, 1, 0, 3, 5, 9, 17, 34, 65, 125, 241, 465, 896, 1727, 3329, 6417, 12369, 23842, 45957, 88585, 170753, 329137, 634432, 1222907, 2357229, 4543705, 8758273, 16882114, 32541321, 62725413, 120907121, 233055969, 449229824, 865918327, 1669111241
Offset: 0

Views

Author

Arie Bos, Dec 07 2014

Keywords

Links

Crossrefs

Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251654, A251655, A251656, A251703, A251704.

Programs

  • J
    NB. see A251655 for the program and apply it to 1,1,1,0.
  • Mathematica
    LinearRecurrence[Table[1, {4}], {1, 1, 1, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)

Formula

a(n+4) = a(n) + a(n+1) + a(n+2) + a(n+3).
G.f.: (-1+3*x^3+x^2)/(-1+x+x^2+x^3+x^4) . - R. J. Mathar, Mar 28 2025
Showing 1-6 of 6 results.

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

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