A001592 -id:A001592 - cp's OEIS frontend

A104144 a(n) = Sum_{k=1..9} a(n-k); a(8) = 1, a(n) = 0 for n < 8.

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1021, 2040, 4076, 8144, 16272, 32512, 64960, 129792, 259328, 518145, 1035269, 2068498, 4132920, 8257696, 16499120, 32965728, 65866496, 131603200, 262947072, 525375999, 1049716729, 2097364960

Offset: 0

Jean Lefort (jlefort.apmep(AT)wanadoo.fr), Mar 07 2005

Sometimes called the Fibonacci 9-step numbers.
For n >= 8, this gives the number of integers written without 0 in base ten, the sum of digits of which is equal to n-7. E.g., a(11) = 8 because we have the 8 numbers: 4, 13, 22, 31, 112, 121, 211, 1111.
The offset for this sequence is fairly arbitrary. - N. J. A. Sloane, Feb 27 2009

T. D. Noe, Table of n, a(n) for n = 0..208
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Taras Goy and Mark Shattuck, Some Toeplitz-Hessenberg Determinant Identities for the Tetranacci Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.6.8.
Kai Wang, Identities for generalized enneanacci numbers, Generalized Fibonacci Sequences (2020).
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1,1,1).

Cf. A000045, A000073, A000078, A001591, A001592, A066178, A079262 (Fibonacci n-step numbers).

Cf. A255529 (Indices of primes in this sequence).

for n from 0 to 50 do k(n):=sum((-1)^i*binomial(n-8-9*i,i)*2^(n-8-10*i),i=0..floor((n-8)/10))-sum((-1)^i*binomial(n-9-9*i,i)*2^(n-9-10*i),i=0..floor((n-9)/10)):od:seq(k(n),n=0..50);a:=taylor((z^8-z^9)/(1-2*z+z^(10)),z=0,51);for p from 0 to 50 do j(p):=coeff(a,z,p):od :seq(j(p),p=0..50); # Richard Choulet, Feb 22 2010

a={1, 0, 0, 0, 0, 0, 0, 0, 0}; Table[s=Plus@@a; a=RotateLeft[a]; a[[ -1]]=s, {n, 50}]
LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
With[{nn=9},LinearRecurrence[Table[1,{nn}],Join[Table[0,{nn-1}],{1}],50]] (* Harvey P. Dale, Aug 17 2013 *)

a(n)=([0,1,0,0,0,0,0,0,0; 0,0,1,0,0,0,0,0,0; 0,0,0,1,0,0,0,0,0; 0,0,0,0,1,0,0,0,0; 0,0,0,0,0,1,0,0,0; 0,0,0,0,0,0,1,0,0; 0,0,0,0,0,0,0,1,0; 0,0,0,0,0,0,0,0,1; 1,1,1,1,1,1,1,1,1]^n*[0;0;0;0;0;0;0;0;1])[1,1] \\ Charles R Greathouse IV, Jun 16 2015

A104144(n,m=9)=(matrix(m,m,i,j,j==i+1||i==m)^n)[1,m] \\ M. F. Hasler, Apr 22 2018

a(n) = Sum_{k=1..9} a(n-k) for n > 8, a(8) = 1, a(n) = 0 for n=0..7.
G.f.: x^8/(1-x-x^2-x^3-x^4-x^5-x^6-x^7-x^8-x^9). - N. J. A. Sloane, Dec 04 2011
Another form of the g.f. f: f(z) = (z^8-z^9)/(1-2*z+z^(10)), then a(n) = Sum_((-1)^i*binomial(n-8-9*i,i)*2^(n-8-10*i), i=0..floor((n-8)/10))-Sum_((-1)^i*binomial(n-9-9*i,i)*2^(n-9-10*i), i=0..floor((n-9)/10)) with Sum_(alpha(i), i=m..n)=0 for m>n. - Richard Choulet, Feb 22 2010
From N. J. A. Sloane, Dec 04 2011: (Start)
Let b be the smallest root (in magnitude) of g(x) := 1-x-x^2-x^3-x^4-x^5-x^6-x^7-x^8-x^9, b = 0.50049311828655225605926845999420216157202861343888...
Let c = -b^8/g'(b) = 0.00099310812055463178382193226558248643030626601288701...
Then a(n) is the nearest integer to c/b^n. (End)

Edited by N. J. A. Sloane, Aug 15 2006 and Nov 11 2006
Incorrect formula deleted by N. J. A. Sloane, Dec 04 2011
Name edited by M. F. Hasler, Apr 22 2018

A159741 a(n) = 8*(2^n - 1).

Original entry on oeis.org

8, 24, 56, 120, 248, 504, 1016, 2040, 4088, 8184, 16376, 32760, 65528, 131064, 262136, 524280, 1048568, 2097144, 4194296, 8388600, 16777208, 33554424, 67108856, 134217720, 268435448, 536870904, 1073741816, 2147483640, 4294967288, 8589934584, 17179869176, 34359738360

Offset: 1

Al Hakanson (hawkuu(AT)gmail.com), Apr 20 2009

Fifth diagonal of the array which contains m-acci numbers in the m-th row.
The base array is constructed from m-acci numbers starting each with 1, 1, and 2 and filling one row of the table (see the examples).
The main and the upper diagonals of the table are the powers of 2, A000079.
The first subdiagonal is essentially A000225, followed by essentially A036563.
The next subdiagonal is this sequence here, followed by A159742, A159743, A159744, A159746, A159747, A159748.
a(n) written in base 2: 1000, 11000, 111000, 1111000, ..., i.e., n times 1 and 3 times 0 (A161770). - Jaroslav Krizek, Jun 18 2009
Also numbers for which n^8/(n+8) is an integer. - Vicente Izquierdo Gomez, Jan 03 2013

			From _R. J. Mathar_, Apr 22 2009: (Start)
The base table is
.1..1....1....1....1....1....1....1....1....1....1....1....1....1
.1..1....1....1....1....1....1....1....1....1....1....1....1....1
.2..2....2....2....2....2....2....2....2....2....2....2....2....2
.0..2....3....4....4....4....4....4....4....4....4....4....4....4
.0..2....5....7....8....8....8....8....8....8....8....8....8....8
.0..2....8...13...15...16...16...16...16...16...16...16...16...16
.0..2...13...24...29...31...32...32...32...32...32...32...32...32
.0..2...21...44...56...61...63...64...64...64...64...64...64...64
.0..2...34...81..108..120..125..127..128..128..128..128..128..128
.0..2...55..149..208..236..248..253..255..256..256..256..256..256
.0..2...89..274..401..464..492..504..509..511..512..512..512..512
.0..2..144..504..773..912..976.1004.1016.1021.1023.1024.1024.1024
.0..2..233..927.1490.1793.1936.2000.2028.2040.2045.2047.2048.2048
.0..2..377.1705.2872.3525.3840.3984.4048.4076.4088.4093.4095.4096
Columns: A000045, A000073, A000078, A001591, A001592 etc. (End)

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A000079, A000225, A036563, A159742, A159743, A159744, A159746, A159747, A159748.

Cf. A000918, A028399, A161770, A173787, A175161, A175166.

[8*(2^n -1): n in [1..50]]; // G. C. Greubel, May 22 2018

T := proc(n,m) option remember ; if n < 0 then 0; elif n <= 1 then 1; elif n = 2 then 2; else add(procname(n-i,m),i=1..m) ; fi: end: A159741 := proc(n) T(n+4,n+1) ; end: seq(A159741(n),n=1..40) ; # R. J. Mathar, Apr 22 2009

Table[8(2^n-1),{n,60}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{3,-2},{8,24},30] (* Harvey P. Dale, Jan 01 2019 *)

a(n)=8*(2^n-1) \\ Charles R Greathouse IV, Sep 24 2015

From R. J. Mathar, Apr 22 2009: (Start)
a(n) = 3*a(n-1) - 2*a(n-2).
a(n) = 8*(2^n-1).
G.f.: 8*x/((2*x-1)*(x-1)). (End)
From Jaroslav Krizek, Jun 18 2009: (Start)
a(n) = Sum_{i=3..(n+2)} 2^i.
a(n) = Sum_{i=1..n} 2^(i+2).
a(n) = a(n-1) + 2^(n+2) for n >= 2. (End)
a(n) = A173787(n+3,3) = A175166(2*n)/A175161(n). - Reinhard Zumkeller, Feb 28 2010
From Elmo R. Oliveira, Jun 15 2025: (Start)
E.g.f.: 8*exp(x)*(exp(x) - 1).
a(n) = 8*A000225(n) = 4*A000918(n+1) = 2*A028399(n+2). (End)

More terms from R. J. Mathar, Apr 22 2009
Edited by Al Hakanson (hawkuu(AT)gmail.com), May 11 2009
Comments claiming negative entries deleted by R. J. Mathar, Aug 24 2009

A122189 Heptanacci numbers: each term is the sum of the preceding 7 terms, with a(0),...,a(6) = 0,0,0,0,0,0,1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 127, 253, 504, 1004, 2000, 3984, 7936, 15808, 31489, 62725, 124946, 248888, 495776, 987568, 1967200, 3918592, 7805695, 15548665, 30972384, 61695880, 122895984, 244804400, 487641600, 971364608, 1934923521

Offset: 0

Roger L. Bagula and Gary W. Adamson, Oct 18 2006

See A066178 (essentially the same sequence) for more about the heptanacci numbers and other generalizations of the Fibonacci numbers (A000045).

Robert Price, Table of n, a(n) for n = 0..1000
Tomás Aguilar-Fraga, Jennifer Elder, Rebecca E. Garcia, Kimberly P. Hadaway, Pamela E. Harris, Kimberly J. Harry, Imhotep B. Hogan, Jakeyl Johnson, Jan Kretschmann, Kobe Lawson-Chavanu, J. Carlos Martínez Mori, Casandra D. Monroe, Daniel Quiñonez, Dirk Tolson III, and Dwight Anderson Williams II, Interval and L-interval Rational Parking Functions, arXiv:2311.14055 [math.CO], 2023. See p. 14.
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Taras Goy and Mark Shattuck, Some Toeplitz-Hessenberg Determinant Identities for the Tetranacci Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.6.8.
Tian-Xiao He, Impulse Response Sequences and Construction of Number Sequence Identities, J. Int. Seq. 16 (2013) #13.8.2.
F. T. Howard and Curtis Cooper, Some identities for r-Fibonacci numbers, Fibonacci Quart. 49 (2011), no. 3, 231-243.
Omar Khadir, László Németh, and László Szalay, Tiling of dominoes with ranked colors, Results in Math. (2024) Vol. 79, Art. No. 253. See p. 2.
Benjamin E. Merkel, Probabilities of Consecutive Events in Coin Flipping, Master's Thesis, Univ. Cincinatti, May 11 2011.
László Németh and László Szalay, Explicit solution of system of two higher-order recurrences, arXiv:2408.12196 [math.NT], 2024. See p. 10.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1).

Cf. A000045 (k=2, Fibonacci numbers), A000073 (k=3, tribonacci) A000078 (k=4, tetranacci) A001591 (k=5, pentanacci) A001592 (k=6, hexanacci), A122189 (k=7, heptanacci).

Cf. A066178, A000322, A248700.

for n from 0 to 50 do k(n):=sum((-1)^i*binomial(n-6-7*i,i)*2^(n-6-8*i),i=0..floor((n-6)/8))-sum((-1)^i*binomial(n-7-7*i,i)*2^(n-7-8*i),i=0..floor((n-7)/8)):od:seq(k(n),n=0..50); a:=taylor((z^6-z^7)/(1-2*z+z^8),z=0,51);for p from 0 to 50 do j(p):=coeff(a,z,p):od :seq(j(p),p=0..50); # Richard Choulet, Feb 22 2010

LinearRecurrence[{1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
a={0,0,0,0,0,0,1} For[n=7, n≤100, n++, sum=Plus@@a; Print[sum]; a=RotateLeft[a]; a[[7]]=sum] (* Robert Price, Dec 04 2014 *)

a(n)=([0,1,0,0,0,0,0; 0,0,1,0,0,0,0; 0,0,0,1,0,0,0; 0,0,0,0,1,0,0; 0,0,0,0,0,1,0; 0,0,0,0,0,0,1; 1,1,1,1,1,1,1]^n*[0;0;0;0;0;0;1])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

G.f.: x^6/(1-x-x^2-x^3-x^4-x^5-x^6-x^7). - R. J. Mathar, Feb 13 2009
G.f.: Sum_{n >= 0} x^(n+5) * [ Product_{k = 1..n} (k + k*x + k*x^2 + k*x^3 + k*x^4 + k*x^5 + x^6)/(1 + k*x + k*x^2 + k*x^3 + k*x^4 + k*x^5 + k*x^6) ]. - Peter Bala, Jan 04 2015
Another form of the g.f.: f(z) = (z^6-z^7)/(1-2*z+z^8), then a(n) = Sum_{i=0..floor((n-6)/8)} (-1)^i*binomial(n-6-7*i,i)*2^(n-6-8*i) - Sum_{i=0..floor((n-7)/8)} (-1)^i*binomial(n-7-7*i,i)*2^(n-7-8*i) with Sum_{i=m..n} alpha(i) = 0 for m>n. - Richard Choulet, Feb 22 2010
Sum_{k=0..6*n} a(k+b)*A063265(n,k) = a(7*n+b), b>=0.
a(n) = 2*a(n-1) - a(n-8). - Joerg Arndt, Sep 24 2020

Edited by N. J. A. Sloane, Nov 20 2007

Wrong Binet-type formula removed by R. J. Mathar, Feb 13 2009

A106273 Discriminant of the polynomial x^n - x^(n-1) - ... - x - 1.

Original entry on oeis.org

1, 5, -44, -563, 9584, 205937, -5390272, -167398247, 6042477824, 249317139869, -11597205023744, -601139006326619, 34383289858207744, 2151954708695291177, -146323302326154543104, -10742330662077208945103, 846940331265064719417344, 71373256668946058057974997

Offset: 1

T. D. Noe, May 02 2005

sign

This polynomial is the characteristic polynomial of the Fibonacci and Lucas n-step sequences. These discriminants are prime for n=2, 4, 6, 26, 158 (A106274). It appears that the term a(2n+1) always has a factor of 2^(2n). With that factor removed, the discriminants are prime for odd n=3, 5, 7, 21, 99, 405. See A106275 for the combined list.
a(n) is the determinant of an r X r Hankel matrix whose entries are w(i+j) where w(n) = x1^n + x2^n + ... + xr^n where x1,x2,...xr are the roots of the titular characteristic polynomial. E.g., A000032 for n=2, A001644 for n=3, A073817 for n=4, A074048 for n=5, A074584 for n=6, A104621 for n=7, ... - Kai Wang, Jan 17 2021
Luca proves that a(n) is a term of the corresponding k-nacci sequence only for n=2 and 3. - Michel Marcus, Apr 12 2025

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Mohammad K. Azarian, On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials, International Journal of Pure and Applied Mathematics 36(2), 2007, pp. 251-257. MR2312537. Zbl 1133.11012.
Michael Baake and Uwe Grimm, Fourier transform of Rauzy fractals and point spectrum of 1D Pisot inflation tilings, arXiv:1907.11012 [math.MG], 2019.
Herbert Batte and Florian Luca, The Discriminant of the Characteristic Polynomial of the kth Fibonacci sequence is not a member of the kth Lucas sequence, arXiv:2504.02514 [math.NT], 2025.
Florian Luca, On the discriminant of the k-generalized Fibonacci polynomial, II, Fibonacci Quart. 62 (2024), no. 3, 193-200.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number
Eric Weisstein's World of Mathematics, Polynomial Discriminant

Cf. A086797 (discriminant of the polynomial x^n-x-1), A000045, A000073, A000078, A001591, A001592 (Fibonacci n-step sequences), A000032, A001644, A073817, A074048, A074584, A104621, A105754, A105755 (Lucas n-step sequences), A086937, A106276, A106277, A106278 (number of distinct zeros of these polynomials for n=2, 3, 4, 5).

Discriminant[p_?PolynomialQ, x_] := With[{n=Exponent[p, x]}, Cancel[((-1)^(n(n-1)/2) Resultant[p, D[p, x], x])/Coefficient[p, x, n]^(2n-1)]]; Table[Discriminant[x^n-Sum[x^i, {i, 0, n-1}], x], {n, 20}]

{a(n)=(-1)^(n*(n+1)/2)*((n+1)^(n+1)-2*(2*n)^n)/(n-1)^2}  \\ Max Alekseyev, May 05 2005

a(n)=poldisc('x^n-sum(k=0,n-1,'x^k)); \\ Joerg Arndt, May 04 2013

a(n) = (-1)^(n*(n+1)/2) * ((n+1)^(n+1)-2*(2*n)^n)/(n-1)^2. - Max Alekseyev, May 05 2005

A122265 10th-order Fibonacci numbers: a(n+1) = a(n)+...+a(n-9) with a(0) = ... = a(8) = 0, a(9) = 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1023, 2045, 4088, 8172, 16336, 32656, 65280, 130496, 260864, 521472, 1042432, 2083841, 4165637, 8327186, 16646200, 33276064, 66519472, 132973664, 265816832, 531372800, 1062224128

Offset: 0

Roger L. Bagula and Gary W. Adamson, Oct 18 2006

nonn

The (1,10)-entry of the matrix M^n, where M is the 10 X 10 matrix {{0,1,0,0,0, 0,0,0,0,0},{0,0,1,0,0,0,0,0,0,0},{0,0,0,1,0,0,0,0,0,0},{0,0,0,0,1,0,0,0,0,0}, {0,0,0,0,0,1,0,0,0,0},{0,0,0,0,0,0,1,0,0,0},{0,0,0,0,0,0,0,1,0,0},{0,0,0,0,0, 0,0,0,1,0},{0,0,0,0,0,0,0,0,0,1},{1,1,1,1,1,1,1,1,1,1}}.

Robert Price, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Kai Wang, Identities for generalized enneanacci numbers, Generalized Fibonacci Sequences (2020).
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1,1,1,1).

Cf. A000322, A001591, A001592, A079262.
Cf. A001591, A001592, A122189, A079262, A104144, A168082, A168083, A168084. - Richard Choulet, Feb 22 2010
Cf. A257227, A257228 for primes in this sequence.

with(linalg): p:=-1-x-x^2-x^3-x^4-x^5-x^6-x^7-x^8-x^9+x^10: M[1]:=transpose(companion(p,x)): for n from 2 to 40 do M[n]:=multiply(M[n-1],M[1]) od: seq(M[n][1,10],n=1..40);
k:=10:for n from 0 to 50 do l(n):=sum((-1)^i*binomial(n-k+1-k*i,i)*2^(n-k+1-(k+1)*i),i=0..floor((n-k+1)/(k+1)))-sum((-1)^i*binomial(n-k-k*i,i)*2^(n-k-(k+1)*i),i=0..floor((n-k)/(k+1))):od:seq(l(n),n=0..50);k:=10:a:=taylor((z^(k-1)-z^(k))/(1-2*z+z^(k+1)),z=0,51);for p from 0 to 50 do j(p):=coeff(a,z,p):od :seq(j(p),p=0..50); # Richard Choulet, Feb 22 2010

M = {{0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}}; v[1] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}; v[n_] := v[n] = M.v[n - 1]; a = Table[Floor[v[n][[1]]], {n, 1, 50}]
a={1,0,0,0,0,0,0,0,0,0};Flatten[Prepend[Table[s=Plus@@a;a=RotateLeft[a];a[[ -1]]=s,{n,60}],Table[0,{m,Length[a]-1}]]] (* Vladimir Joseph Stephan Orlovsky, Nov 18 2009 *)
LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, 50]  (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
With[{nn=10},LinearRecurrence[Table[1,{nn}],Join[Table[0,{nn-1}],{1}],50]] (* Harvey P. Dale, Aug 17 2013 *)

a(n) = Sum_{j=1..10} a(n-j) for n>=10; a(n) = 0 for 0<=n<=8, a(9) = 1 (follows from the minimal polynomial of M; a Maple program based on this recurrence relation is much slower than the given Maple program, based on the definition).
G.f.: -x^9/(-1+x^10+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
Another form of the g.f. f: f(z)=(z^(k-1)-z^(k))/(1-2*z+z^(k+1)) with k=10. Then a(n)=sum((-1)^i*binomial(n-k+1-k*i,i)*2^(n-k+1-(k+1)*i),i=0..floor((n-k+1)/(k+1)))-sum((-1)^i*binomial(n-k-k*i,i)*2^(n-k-(k+1)*i),i=0..floor((n-k)/(k+1))) with k=10 and sum(alpha(i),i=m..n)=0 for m>n. - Richard Choulet, Feb 22 2010

Edited by N. J. A. Sloane, Oct 29 2006 and Mar 05 2011

A118427 Decimal expansion of hexanacci constant.

Original entry on oeis.org

1, 9, 8, 3, 5, 8, 2, 8, 4, 3, 4, 2, 4, 3, 2, 6, 3, 3, 0, 3, 8, 5, 6, 2, 9, 2, 9, 3, 3, 9, 1, 4, 2, 5, 7, 5, 2, 7, 3, 0, 0, 8, 0, 8, 6, 5, 5, 6, 8, 8, 2, 1, 7, 5, 3, 2, 1, 6, 3, 5, 9, 0, 6, 5, 6, 5, 6, 7, 0, 2, 2, 7, 8, 0, 1, 4, 1, 7, 2, 4, 0, 2, 9, 8, 6, 5, 7, 5, 0, 7, 0, 2, 2, 6, 8, 9, 9, 7, 9, 7, 3, 2, 7, 7, 5

Offset: 1

Eric W. Weisstein, Apr 27 2006

The continued fraction expansion starts 1, 1, 59, 1, 10, 2, 1, 6, 2, 1, 6, 1, 1, 7, 1, 71, 7, 1, 6, 8, ... - R. J. Mathar, Mar 09 2012
For n>=7, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. - Vladimir Shevelev, Mar 21 2014
Note that we have: c + c^(-6) = 2, and the k-nacci constant approaches 2 when k approaches infinity (Martin Gardner). - Bernard Schott, May 06 2022

			1.9835828434243263303...

Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, Simon & Schuster, NY, 1961.

S. Litsyn and Vladimir Shevelev, Irrational Factors Satisfying the Little Fermat Theorem, International Journal of Number Theory, vol.1, no.4 (2005), 499-512.
Vladimir Shevelev, A property of n-bonacci constant, Seqfan (Mar 23 2014)
Eric Weisstein's World of Mathematics, Hexanacci Number
Eric Weisstein's World of Mathematics, Hexanacci Constant
Eric Weisstein's World of Mathematics, Hexanacci Number
Index entries for algebraic numbers, degree 6.

Cf. A001592.

k-nacci constants: A001622 (Fibonacci), A058265 (tribonacci), A086088 (tetranacci), A103814 (pentanacci), this sequence (hexanacci), A118428 (heptanacci).

RealDigits[ Root[ x^6 - x^5 - x^4 - x^3 - x^2 - x - 1, 2] , 10, 105] // First (* Jean-François Alcover, Feb 07 2013 *)

A172316 7th column of the array A172119.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 127, 252, 500, 992, 1968, 3904, 7744, 15361, 30470, 60440, 119888, 237808, 471712, 935680, 1855999, 3681528, 7302616, 14485344, 28732880, 56994048, 113052416, 224248833, 444816138, 882329660

Offset: 0

Richard Choulet, Jan 31 2010

			a(3) = binomial(3,3)*2^3 = 8.
a(7) = binomial(7,7)*2^7 - binomial(1,0)*2^0 = 127.

O. Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see p. 356 with r = 6.
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,0,0,-1)

Cf. A000071, A001949, A008937, A107066, A172119.

Partial sums of A001592.

for k from 0 to 20 do for n from 0 to 30 do b(n):=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))):od:k: seq(b(n),n=0..30):od; k:=6:taylor(1/(1-2*z+z^(k+1)),z=0,30);

G.f.: 1/(1 - 2*z + z^7).
Recurrence formula: a(n+7) = 2*a(n+6) - a(n).
a(n) = Sum_{j=0..floor(n/(k+1))} ((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j)) with k=6.

A061676 Triangle T(n,k) of number of ways of throwing k standard dice to produce a total of n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 0, 6, 15, 20, 15, 6, 1, 0, 5, 21, 35, 35, 21, 7, 1, 0, 4, 25, 56, 70, 56, 28, 8, 1, 0, 3, 27, 80, 126, 126, 84, 36, 9, 1, 0, 2, 27, 104, 205, 252, 210, 120, 45, 10, 1, 0, 1, 25, 125, 305, 456, 462, 330, 165, 55, 11, 1

Offset: 1

Henry Bottomley, Apr 01 2002

			Rows start:
1;
1,1;
1,2,1;
1,3,3,1;
1,4,6,4,1;
1,5,10,10,5,1;
0,6,15,20,15,6,1;
0,5,21,35,35,21,7,1;
etc.
T(8,2)=5 since 8 =2+6 =3+5 =4+4 =5+3 =6+2.

First 21 terms as A007318 (see formula). Cf. A001592, A069713.

Cf. A030528 (2-sided dice), A078803 (3-sided), A213887 (4-sided), A213888 (5-sided).

pts := 6; # A213889 and A061676
g := 1/(1-t*z*add(z^i,i=0..pts-1)) ;
for n from 1 to 13 do
    for k from 1 to n do
        coeftayl(g,z=0,n) ;
        coeftayl(%,t=0,k) ;
        printf("%d ",%) ;
    end do:
    printf("\n") ;
end do: # R. J. Mathar, May 28 2025

T(n, k)=T(n-1, k-1)+T(n-2, k-1)+T(n-3, k-1)+T(n-4, k-1)+T(n-5, k-1)+T(n-6, k-1) starting with T(0, 0)=1. T(n, k)=T(7k-n, k); if n>6k or n6k-6, T(n, k)=C(7k-n-1, k-1); T([7k/2], k)=A018901(k).

A124168 Union of all n-Fibonacci sequences, that is, all sequences s(0) = s(1) = ... = s(n-2) = 0, s(n-1) = 1 and for k >= n, s(k) = s(k-1) + ... + s(k-n).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 13, 15, 16, 21, 24, 29, 31, 32, 34, 44, 55, 56, 61, 63, 64, 81, 89, 108, 120, 125, 127, 128, 144, 149, 208, 233, 236, 248, 253, 255, 256, 274, 377, 401, 464, 492, 504, 509, 511, 512, 610, 773, 912, 927, 976, 987, 1004, 1016, 1021, 1023, 1024

Offset: 1

Carlos Alves, Dec 03 2006

nonn

Note that an n-Fibonacci sequence contains the numbers 2^k numbers for kA001792 (for n large)...

Noe and Post conjectured that the only positive terms that are common to any two distinct n-step Fibonacci sequences are the powers of 2 that begin each sequence and 13 (in 2- and 3-step) and 504 (in 3- and 7-step). Perhaps we should also include 8 (in 2- and 4-step). - T. D. Noe, Dec 05 2006

T. D. Noe, Table of n, a(n) for n=1..1000
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.

Cf. A000045, A000073, A000078, A001591, A001592, A124257.

Cf. A227880 (primes here).

NFib25[nfb_] := Transpose[NestList[Join[Drop[ #, {1}], {Plus @@ #}] &, Map[If[ # == nfb, 1, 0] &, Range[nfb]], 25]][[ -1]]; Union[Flatten[Map[NFib25, Range[2, 20]]]][[Range[100]]]
NFib[nfb_, lim_] := Module[{f = 2^Range[0, nfb - 1]}, While[f[[-1]] <= lim, AppendTo[f, Total[Take[f, -nfb]]]]; Most[f]]; lim = 12; Union[Flatten[Table[NFib[i, 2^lim], {i, 2, lim + 1}]]] (* T. D. Noe, Oct 25 2013 *)

Union(A000045, A000073, A000078, A001591, A001592, ...)

Edited by N. J. A. Sloane, Dec 15 2006

A168084 Fibonacci 13-step numbers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8191, 16381, 32760, 65516, 131024, 262032, 524032, 1048000, 2095872, 4191488, 8382464, 16763904, 33525760, 67047424, 134086657, 268156933, 536281106, 1072496696

Offset: 1

Vladimir Joseph Stephan Orlovsky, Nov 18 2009

G. C. Greubel, Table of n, a(n) for n = 1..1000
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1).

Cf. A000078, A001591, A001592, A122189, A079262, A168083.

k:=13:a:=taylor((z^(k-1)-z^(k))/(1-2*z+z^(k+1)),z=0,51);for p from 0 to 50 do j(p):=coeff(a,z,p):od :seq(j(p),p=0..50); k:=13:for n from 0 to 50 do l(n):=sum((-1)^i*binomial(n-k+1-k*i,i)*2^(n-k+1-(k+1)*i),i=0..floor((n-k+1)/(k+1)))-sum((-1)^i*binomial(n-k-k*i,i)*2^(n-k-(k+1)*i),i=0..floor((n-k)/(k+1))):od:seq(l(n),n=0..50); # Richard Choulet, Feb 22 2010

a={1,0,0,0,0,0,0,0,0,0,0,0,0};Flatten[Prepend[Table[s=Plus@@a;a=RotateLeft[a];a[[ -1]]=s,{n,60}],Table[0,{m,Length[a]-1}]]]
LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, 50]
With[{nn=13},LinearRecurrence[Table[1,{nn}],Join[Table[0,{nn-1}],{1}],50]] (* Harvey P. Dale, Aug 17 2013 *)

Another form of the g.f. f: f(z)=(z^(k-1)-z^(k))/(1-2*z+z^(k+1)) with k=13. then a(n)=sum((-1)^i*binomial(n-k+1-k*i,i)*2^(n-k+1-(k+1)*i),i=0..floor((n-k+1)/(k+1)))-sum((-1)^i*binomial(n-k-k*i,i)*2^(n-k-(k+1)*i),i=0..floor((n-k)/(k+1))) with k=13 and convention sum(alpha(i),i=m..n)=0 for m>n. - Richard Choulet, Feb 22 2010

cp's OEIS Frontend

A104144 a(n) = Sum_{k=1..9} a(n-k); a(8) = 1, a(n) = 0 for n < 8.

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A159741 a(n) = 8*(2^n - 1).

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A122189 Heptanacci numbers: each term is the sum of the preceding 7 terms, with a(0),...,a(6) = 0,0,0,0,0,0,1.

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A106273 Discriminant of the polynomial x^n - x^(n-1) - ... - x - 1.

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A122265 10th-order Fibonacci numbers: a(n+1) = a(n)+...+a(n-9) with a(0) = ... = a(8) = 0, a(9) = 1.

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A118427 Decimal expansion of hexanacci constant.

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A172316 7th column of the array A172119.

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