A000078 -id:A000078 - cp's OEIS frontend

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

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

Arie Bos, Dec 06 2014

Index entries for linear recurrences with constant coefficients, signature (1,1,1,1).

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

NB. see A251655 for the program and apply it to 0,1,1,0.

LinearRecurrence[Table[1, {4}], {0, 1, 1, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)

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

Arie Bos, Dec 06 2014

Tamás Lengyel and Diego Marques, The 2-adic Order of Some Generalized Fibonacci Numbers, INTEGERS, 17, 2017, A5.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1).

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

(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

LinearRecurrence[Table[1, {4}], {0, 1, 1, 1}, 36] (* Michael De Vlieger, Dec 09 2014 *)

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

Arie Bos, Dec 07 2014

Index entries for linear recurrences with constant coefficients, signature (1,1,1,1).

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

NB. see A251655 for the program and apply it to 1,1,0,1.

LinearRecurrence[Table[1, {4}], {1, 1, 0, 1}, 36] (* Michael De Vlieger, Dec 09 2014 *)

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

Arie Bos, Dec 07 2014

Index entries for linear recurrences with constant coefficients, signature (1,1,1,1).

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

NB. see A251655 for the program and apply it to 1,1,1,0.

LinearRecurrence[Table[1, {4}], {1, 1, 1, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)

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

A073937 a(n) = a(n-1) - a(n-2) + a(n-3) + a(n-4), a(0)=4, a(1)=1, a(2)=-1, a(3)=1.

Original entry on oeis.org

4, 1, -1, 1, 7, 6, -1, 1, 15, 19, 4, 1, 31, 53, 27, 6, 63, 137, 107, 39, 132, 337, 351, 185, 303, 806, 1039, 721, 791, 1915, 2884, 2481, 2303, 4621, 7683, 7846, 7087, 11545, 19987, 23375, 22020, 30177, 51519, 66737, 67415, 82374, 133215, 184993, 201567, 232163, 348804

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 13 2002

From Kai Wang, Nov 03 2020: (Start)
Let f(x) = x^4 - x^3 - x^2 - x - 1 and {x1,x2,x3,x4} be the roots of f(x). Then a(n) = (x1*x2*x3)^n + (x1*x2*x4)^n + (x1*x3*x4)^n + (x2*x3*x4)^n.
Let g(y) = y^4 - y^3 + y^2 - y - 1 and {y1,y2,y3,y4} be the roots of g(y). Then a(n) = y1^n + y2^n + y3^n + y4^n. (End)

Michael De Vlieger, Table of n, a(n) for n = 0..9024
Kai Wang, Identities, generating functions and Binet formula for generalized k-nacci sequences, 2020.
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for linear recurrences with constant coefficients, signature (1,-1,1,1).

Cf. A000078, A001630, A073817, A074058, A100329.

CoefficientList[Series[(4-3*x+2*x^2-x^3)/(1-x+x^2-x^3-x^4), {x, 0, 50}], x]
LinearRecurrence[{1,-1,1,1},{4,1,-1,1},60] (* Harvey P. Dale, Sep 05 2021 *)

polsym(y^4 - y^3 + y^2 - y - 1, 55) \\ Joerg Arndt, Nov 07 2020

G.f.: (4 - 3*x + 2*x^2 - x^3)/(1 - x + x^2 - x^3 - x^4).
From Kai Wang, Nov 03 2020: (Start)
For n >= 1, a(n) = Sum_{j1,j2,j3,j4>=0; j1+2*j2+3*j3+4*j4=n} (-1)^j2*n*(j1+j2+j3+j4-1)!/(j1!*j2!*j3!*j4!).
For n > 1, a(n) = (-1)^n*(4*A100329(n+1) + 3*A100329(n) + 2*A100329(n-1) + A100329(n-2)). (End)
From Peter Bala, Jan 19 2023: (Start)
a(n) = (-1)^n*A074058(n).
a(n) = trace of M^n, where M is the 4 X 4 matrix [[0, 0, 0, -1], [-1, 0, 0, 1], [0, -1, 0, 1], [0, 0, -1, 1]].
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^k) for positive integers n and r and all primes p. See Zarelua. (End)

A104534 Indices of prime tetranacci numbers.

Original entry on oeis.org

3, 7, 11, 12, 36, 56, 401, 2707, 8417, 14096, 31561, 50696, 53192, 155182

Offset: 1

Eric W. Weisstein, Mar 13 2005

nonn

Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4
Eric Weisstein's World of Mathematics, Tetranacci Number
Eric Weisstein's World of Mathematics, Integer Sequence Primes

Cf. A000078, A104535.

A000078(a(n) + 2) = A104535(n). - Andrew Howroyd, Oct 08 2024

a(14)=155182 found by Eric W. Weisstein, Oct 28 2005

A106277 Number of distinct zeros of x^4-x^3-x^2-x-1 mod prime(n).

Original entry on oeis.org

0, 1, 0, 1, 0, 0, 1, 1, 1, 2, 0, 2, 2, 0, 1, 0, 1, 1, 1, 1, 2, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 4, 0, 1, 0, 2, 2, 2, 2, 4, 1, 1, 1, 0, 0, 1, 1, 2, 0, 0, 0, 1, 0, 0, 2, 1, 0, 1, 1, 0, 2, 2, 2, 0, 0, 2, 1, 0, 1, 2, 0, 0, 2, 0, 1, 0, 2, 1, 1, 2, 1, 2, 0, 1, 0, 1, 2, 0, 2, 1, 0, 0, 1, 2, 1, 1, 0, 2, 1, 2, 1, 3, 0, 0

Offset: 1

T. D. Noe, May 02 2005

nonn

This polynomial is the characteristic polynomial of the Fibonacci and Lucas 4-step sequences, A000078 and A073817. Similar polynomials are treated in Serre's paper. The discriminant of the polynomial is -563 and 563 is the only prime for which the polynomial has 3 distinct zeros. The primes p yielding 4 distinct zeros, A106280, correspond to the periods of the sequences A000078(k) mod p and A073817(k) mod p having length less than p. The Lucas 4-step sequence mod p has one additional prime p for which the period is less than p: the discriminant 563. For this prime, the Fibonacci 4-step sequence mod p has a period of p(p-1).

J.-P. Serre, On a theorem of Jordan, Bull. Amer. Math. Soc., 40 (No. 4, 2003), 429-440, see p. 433.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number

Cf. A106296 (period of the Lucas 4-step sequences mod prime(n)), A106283 (prime moduli for which the polynomial is irreducible).

Table[p=Prime[n]; cnt=0; Do[If[Mod[x^4-x^3-x^2-x-1, p]==0, cnt++ ], {x, 0, p-1}]; cnt, {n, 150}]

from sympy.abc import x
from sympy import Poly, prime
def A106277(n): return len(Poly(x*(x*(x*(x-1)-1)-1)-1, x, modulus=prime(n)).ground_roots()) # Chai Wah Wu, Mar 29 2024

A108564 a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 2, a(4) = 4, for n>3: a(n+1) = SORT[a(n) + a(n-1) + a(n-2) + a(n-3)], where SORT places digits in ascending order and deletes 0's.

Original entry on oeis.org

0, 1, 1, 2, 4, 8, 15, 29, 56, 18, 118, 122, 134, 239, 136, 136, 456, 679, 147, 1148, 234, 228, 1577, 1378, 1347, 345, 4467, 3577, 3679, 1268, 11299, 12389, 23568, 24458, 11477, 12789, 22279, 137, 24668, 35789, 23788, 23488, 13377, 24469, 12258, 23579

Offset: 0

Jonathan Vos Post, Jun 10 2005

Sorted tetranacci numbers, a.k.a. sorted Fibonacci 4-step sequence.

As found by T. D. Noe: Max=4556699. Cycle period=41652. Cycle starts with the 23944th term.

			a(8) = SORT[a(4) + a(5) + a(6) + a(7)] = SORT[108] = 18.
a(10) = SORT[a(6) + a(7) + a(8) + a(9)] = SORT[221] = 122.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Richard I. Hess, Problem 920: sorted Fibonacci sequence, Pi Mu Epsilon Journal, Vol. 10 (Fall 1998) No. 9, pp. 754-755.

Cf. A000078, A069638, A107281, A108565-A108573.

nxt[{a_,b_,c_,d_,e_}]:={b,c,d,e,FromDigits[Sort[DeleteCases[ IntegerDigits[ b+c+d+e],?0]]]}; NestList[nxt,{0,1,1,2,4},50][[All,1]] (* _Harvey P. Dale, Jan 17 2022 *)

A113244 Prime differences of tetranacci numbers.

Original entry on oeis.org

2, 3, 7, 11, 13, 41, 79, 107, 179, 193, 293, 397, 769, 1489, 2099, 2843, 2857, 5507, 5521, 9181, 10463, 10663, 10667, 19079, 39619, 76423, 126743, 146539, 147283, 147311, 281081, 283949, 547229, 771073, 3919171, 3919543, 3919943, 7555879, 7555927, 10644589, 14564477

Offset: 1

Jonathan Vos Post, Oct 19 2005; corrected Oct 20 2005

A113188-A113194 deal with difference sets of Fibonacci numbers and Lucas numbers and primes in those difference sets. A113238-A113239 deal with the difference set of tribonacci numbers and primes in that difference set.

			a(1) = 2 because 4 - 2 = 2 where 4 and 2 are tetranacci numbers.
a(2) = 3 because 4 - 1 = 3 where 4 and 1 are tetranacci numbers.
a(3) = 7 because 8 - 1 = 7 where 8 and 1 are tetranacci numbers.
a(4) = 11 because 15 - 4 = 11 where 15 and 4 are tetranacci numbers.
a(5) = 13 because 15 - 2 = 13 where 15 and 2 are tetranacci numbers.

Cf. A000040, A000078, A113188-A113194, A113238, A113239, A113243.

isA113244 := proc(n)
    isprime(n) and isA113243(n) ;
end proc:
for n from 1 do
    p := ithprime(n) ;
    if isA113244(p) then
        printf("%d\n",p) ;
    end if;
end do: # R. J. Mathar, Oct 04 2014

{a(n)} = intersection of A000040 and A113243. {a(n)} = primes in the difference set of tetranacci sequence A000078, excluding prime tetranacci numbers A104535.

281081 inserted by R. J. Mathar, Oct 04 2014

A175331 Array A092921(n,k) without the first two rows, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 5, 4, 2, 1, 1, 8, 7, 4, 2, 1, 1, 13, 13, 8, 4, 2, 1, 1, 21, 24, 15, 8, 4, 2, 1, 1, 34, 44, 29, 16, 8, 4, 2, 1, 1, 55, 81, 56, 31, 16, 8, 4, 2, 1, 1, 89, 149, 108, 61, 32, 16, 8, 4, 2, 1, 1, 144, 274, 208, 120, 63, 32, 16, 8, 4, 2, 1, 1, 233, 504, 401, 236, 125, 64, 32, 16, 8, 4, 2, 1

Offset: 2

Roger L. Bagula, Dec 03 2010

Antidiagonal sums are A048888. This is a transposed version of A048887, so the bivariate generating function is obtained by swapping the two arguments.

Brlek et al. (2006) call this table "number of psp-polyominoes with flat bottom". - N. J. A. Sloane, Oct 30 2018

			The array starts in row n=2 with columns k >= 1 as:
  1   1   1   1   1   1   1   1   1   1
  1   2   2   2   2   2   2   2   2   2
  1   3   4   4   4   4   4   4   4   4
  1   5   7   8   8   8   8   8   8   8
  1   8  13  15  16  16  16  16  16  16
  1  13  24  29  31  32  32  32  32  32
  1  21  44  56  61  63  64  64  64  64
  1  34  81 108 120 125 127 128 128 128
  1  55 149 208 236 248 253 255 256 256

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 125, 155.

Srecko Brlek, Andrea Frosini, Simone Rinaldi, and Laurent Vuillon, Tilings by translation: enumeration by a rational language approach, The Electronic Journal of Combinatorics, vol.13, (2006). Table 1 is essentially this array. - _N. J. A. Sloane_, Jul 20 2014

Cf. A000045, A048887, A048004, A126198.

A092921 := proc(n,k) if k <= 0 or n <= 0 then 0; elif k = 1 or n = 1 then 1; else add( procname(n-i,k),i=1..k) ; end if; end proc:
A175331 := proc(n,k) A092921(n,k) ; end proc: # R. J. Mathar, Dec 17 2010

f[x_, n_] = (x - x^(m + 1))/(1 - 2*x + x^(m + 1))
a = Table[Table[SeriesCoefficient[
      Series[f[x, m], {x, 0, 10}], n], {n, 0, 10}], {m, 1, 10}];
Table[Table[a[[m, n - m + 1]], {m, 1, n - 1}], {n, 1, 10}];
Flatten[%]

T(n,k) = A092921(n,k), n >= 2.
T(n,2) = A000045(n).
T(n,3) = A000073(n+2).
T(n,4) = A000078(n+2).

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

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

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

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

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A073937 a(n) = a(n-1) - a(n-2) + a(n-3) + a(n-4), a(0)=4, a(1)=1, a(2)=-1, a(3)=1.

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A104534 Indices of prime tetranacci numbers.

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A106277 Number of distinct zeros of x^4-x^3-x^2-x-1 mod prime(n).

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A108564 a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 2, a(4) = 4, for n>3: a(n+1) = SORT[a(n) + a(n-1) + a(n-2) + a(n-3)], where SORT places digits in ascending order and deletes 0's.

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