A033138 -id:A033138 - cp's OEIS frontend

A023001 a(n) = (8^n - 1)/7.

0, 1, 9, 73, 585, 4681, 37449, 299593, 2396745, 19173961, 153391689, 1227133513, 9817068105, 78536544841, 628292358729, 5026338869833, 40210710958665, 321685687669321, 2573485501354569, 20587884010836553, 164703072086692425

Offset: 0

David W. Wilson

Gives the (zero-based) positions of odd terms in A007556 (numbers n such that A007556(a(n)) mod 2 = 1). - Farideh Firoozbakht, Jun 13 2003
{1, 9, 73, 585, 4681, ...} is the binomial transform of A003950. - Philippe Deléham, Jul 22 2005
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=8, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=9, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n-1) = (-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010
This is the sequence A(0,1;7,8;2) = A(0,1;8,0;1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
a(n) is the total number of squares the carpetmaker has removed after the n-th step of a Sierpiński carpet production. - Ivan N. Ianakiev, Oct 22 2013
For n >= 1, a(n) is the total number of holes in a box fractal (start with 8 boxes, 1 hole) after n iterations. See illustration in link. - Kival Ngaokrajang, Jan 27 2015
From Bernard Schott, May 01 2017: (Start)
Except for 0, 1 and 73, all the terms are composite because a(n) = ((2^n - 1) * (4^n + 2^n + 1))/7.
For n >= 3, all terms are Brazilian repunits numbers in base 8, and so belong to A125134.
a(3) = 73 is the only Brazilian prime in base 8, and so it belongs to A085104 and A285017. (End)

			From _Zerinvary Lajos_, Jan 14 2007: (Start)
Octal.............decimal
0....................0
1....................1
11...................9
111.................73
1111...............585
11111.............4681
111111...........37449
1111111.........299593
11111111.......2396745
111111111.....19173961
1111111111...153391689
etc. ...............etc. (End)
a(4) = (8^4 - 1)/7 = 585 = 1111_8 = (2^4 - 1) * (4^4 + 2^4 + 1)/7 = 15 * 273/7 = 15 * 39. - _Bernard Schott_, May 01 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
Roger B. Eggleton, Maximal Midpoint-Free Subsets of Integers, International Journal of Combinatorics Volume 2015, Article ID 216475, 14 pages.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences. - _Wolfdieter Lang_, Oct 18 2010
Kival Ngaokrajang, Illustration of initial terms
Quynh Nguyen, Jean Pedersen, and Hien T. Vu, New Integer Sequences Arising From 3-Period Folding Numbers, Vol. 19 (2016), Article 16.3.1. See Table 1.
Kalika Prasad, Munesh Kumari, Rabiranjan Mohanta, and Hrishikesh Mahato, The sequence of higher order Mersenne numbers and associated binomial transforms, arXiv:2307.08073 [math.NT], 2023.
D. C. Santos, E. A. Costa, and P. M. M. C. Catarino, On Gersenne Sequence: A Study of One Family in the Horadam-Type Sequence, Axioms 14, 203, (2025). See p. 4.
Eric Weisstein's World of Mathematics, Repunit.
Index entries for linear recurrences with constant coefficients, signature (9,-8).

Cf. A007556, A003950, A001158, A033138.

Cf. A125134, A085104, A285017, A220571, A053696.

A023001:=List([0..10^2],n->(8^n-1)/7); # Muniru A Asiru, Oct 03 2017

[(8^n-1)/7: n in [0..20]]; // Vincenzo Librandi, Sep 17 2011

a:=n->sum(8^(n-j),j=1..n): seq(a(n), n=0..20); # Zerinvary Lajos, Jan 04 2007

Table[(8^n-1)/7, {n, 0, m}]
LinearRecurrence[{9,-8},{0,1},30] (* Harvey P. Dale, Feb 12 2015 *)

A023001(n):=floor((8^n-1)/7)$
makelist(A023001(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=(8^n-1)/7 \\ Charles R Greathouse IV, Mar 22 2016

[lucas_number1(n,9,8) for n in range(0, 21)] # Zerinvary Lajos, Apr 23 2009

[gaussian_binomial(n,1,8) for n in range(0,21)] # Zerinvary Lajos, May 28 2009

Also sum of cubes of divisors of 2^(n-1): a(n) = A001158(A000079(n-1)). - Labos Elemer, Apr 10 2003 and Farideh Firoozbakht, Jun 13 2003
a(n) = A033138(3n-2). - Alexandre Wajnberg, May 31 2005
From Philippe Deléham, Oct 12 2006: (Start)
a(0) = 0, a(n) = 8*a(n-1) + 1 for n>0.
G.f.: x/((1-8x)*(1-x)). (End)
From Wolfdieter Lang, Oct 18 2010: (Start)
a(n) = 7*a(n-1) + 8*a(n-2) + 2, a(0)=0, a(1)=1.
a(n) = 8*a(n-1) + a(n-2) - 8*a(n-3) = 9*a(n-1) - 8*a(n-2), a(0)=0, a(1)=1, a(2)=9. Observation by Gary Detlefs. See the W. Lang comment and link. (End)
a(n) = Sum_{k=0..n-1} 8^k. - Doug Bell, May 26 2017
E.g.f.: exp(x)*(exp(7*x) - 1)/7. - Stefano Spezia, Mar 11 2023

A284116 a(n) = largest number of distinct words arising in Post's tag system {00, 1101} applied to a binary word w, over all starting words w of length n, or a(n) = -1 if there is a word w with an unbounded trajectory.

Original entry on oeis.org

4, 7, 6, 7, 22, 23, 24, 25, 30, 31, 34, 421, 422, 423, 422, 423, 424, 2169, 2170, 2171, 2170, 2171, 2172, 2165, 2166, 2167, 24566, 24567, 24568, 24567, 24568, 24569, 24568, 24569, 24570, 253513, 253514, 342079, 342080, 342083, 342084, 342103, 20858070

Offset: 1

Jeffrey Shallit, Mar 20 2017

nonn

Post's tag system {00, 1101} maps a word w over {0,1} to w', where if w begins with 0, w' is obtained by appending 00 to w and deleting the first three letters, or if w begins with 1, w' is obtained by appending 1101 to w and deleting the first three letters.
The empty word is included in the count.
It is an important open question to decide if there is any word whose orbit grows without limit. - N. J. A. Sloane, Jul 30 2017, based on an email from Allan C. Wechsler
Comment from Don Reble, Aug 01 2017: For n <= 57, all words reach the empty word or a cycle. - N. J. A. Sloane, Aug 01 2017
From David A. Corneth, Aug 02 2017: (Start)
A word w can be described by the pair (c, d) where c is the length of w and d is the number represented by the binary word w. Then 0 <= d < 2^c.
Appending a word ww of m letters to w is the same as setting d to 2^m * w + ww. Preserving only the rightmost q digits of w is the same as setting w to w mod 2^q.
Lastly, we're only really interested in the 1st, 4th, 7th, ... leftmost digits. The others could without loss of generality be set to 0. This can be done with bitand(x, y), with y in A033138.
Therefore this problem can be formulated as follows: Let w = (c, d).
Then if d < 2^(c - 1), w' = (c - 1, bitand(4*d, floor(2^(c + 1) / 7)))
else (if (d >= 2^(c - 1)), w' = (c + 1, bitand(16*d + 13, floor(2^(c + 3) / 7))).
To find a(n), it would be enough to check values d in A152111 with n binary digits and c = n.
(End)
a(110) >= 43913328040672, from w = (100)^k, k=110. - N. J. A. Sloane, Oct 23 2017, based on Lars Blomberg's work on A291792.

			Suppose n=1. Then w = 0 ->000 -> w' = empty word, and w = 1 -> 11101 -> w' = 01 -> 0100 -> w'' = 0 -> 000 -> w''' = empty word. So a(1) = 4 by choosing w = 1.
For n = 5 the orbit of the word 10010 begins 10010, 101101, 1011101, ..., 0000110111011101, and the next word in the orbit has already appeared. The orbit consists of 22 distinct words.
From _David A. Corneth_, Aug 02 2017: (Start)
The 5-letter word w = 10100 can be described as (a, b) = (5, 20). This is equivalent to (5, bitand(20, floor(2^7 / 7))) = (5, bitand(20, 18)) = (5, 16).
As 16 >= 2^(5-1), w' = (5 + 1, bitand(16*16 + 13, floor(2^(5 + 3) / 7))) = (6, bitand(279, 36)) = (6, 4). w'' = w = (5, 16) so 10100 ~ 10000 ends in a period. (End)
Words w that achieve a(1) through a(7) are 1, 10, 100, 0001, 10010, 100000, 0001000. - _N. J. A. Sloane_, Aug 17 2017

John Stillwell, Elements of Mathematics: From Euclid to Goedel, Princeton, 2016. See page 100, Post's tag system.

Peter R. J. Asveld, On a Post's System of Tag. Bulletin of the EATCS 36 (1988), 96-102.
Liesbeth De Mol, Tracing unsolvability. A historical, mathematical and philosophical analysis with a special focus on tag systems, Ph.D. Thesis, Universiteit Gent, 2007.
Liesbeth De Mol, Tag systems and Collatz-like functions Theoret. Comput. Sci. 390 (2008), no. 1, 92-101.
Liesbeth De Mol, On the complex behavior of simple tag systems—an experimental approach, Theoret. Comput. Sci. 412 (2011), no. 1-2, 97-112.
Emil L. Post, Formal Reductions of the General Combinatorial Decision Problem, Amer. J. Math. 65, 197-215, 1943. See page 204.
Don Reble, Python program for A289670.
Shigeru Watanabe, Periodicity of Post's normal process of tag, in Jerome Fox, ed., Proceedings of Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press, Polytechnic Institute of Brooklyn, 1963, pp. 83-99. [Annotated scanned copy]

Cf. A033138, A152111, A284119, A284121, A289670, A289671, A289672, A289673, A289674, A289675, A289676, A289677.
Cf. also A290436-A290441, A290741, A290742, A291792, A291793, A291794, A291795, A291796.
For the 3-shift tag systems {00,1101}, {00, 1011}, {00, 1110}, {00, 0111} see A284116, A291067, A291068, A291069 respectively (as well as the cross-referenced entries mentioned there).

Table[nmax = 0;
 For[i = 0, i < 2^n, i++, lst = {};
  w = IntegerString[i, 2, n];
  While[! MemberQ[lst, w],
   AppendTo[lst, w];
   If[w == "", Break[]];
   If[StringTake[w, 1] == "0", w = StringDrop[w <> "00", 3],
    w = StringDrop[w <> "1101", 3]]];
nmax = Max[nmax, Length[lst]]]; nmax, {n, 1, 12}] (* Robert Price, Sep 26 2019 *)
(* Or, using the (c,d) procedure: *)
 Table[nmax = 0;
 For[i = 0, i < 2^n, i++,
  c = n; d = i; lst = {};
  While[! MemberQ[lst, {c, d}],
   AppendTo[lst, {c, d}];
   If[c == 0,  Break[]];
   If[ d < 2^(c - 1),
    d = BitAnd[4*d, 2^(c - 1) - 1]; c--,
    d = BitAnd[16*d + 13, 2^(c + 1) - 1]; c++]];
nmax = Max[nmax, Length[lst]]]; nmax, {n, 1, 12}] (* Robert Price, Sep 26 2019 *)

a(19)-a(43) from Lars Blomberg, Apr 09 2017

Edited by N. J. A. Sloane, Jul 29 2017 and Oct 23 2017 (adding escape clause in case an infinite trajectory exists)

A046630 Number of cubic residues mod 2^n.

Original entry on oeis.org

1, 2, 3, 5, 10, 19, 37, 74, 147, 293, 586, 1171, 2341, 4682, 9363, 18725, 37450, 74899, 149797, 299594, 599187, 1198373, 2396746, 4793491, 9586981, 19173962, 38347923, 76695845, 153391690, 306783379, 613566757, 1227133514, 2454267027

Offset: 0

David W. Wilson

			For n=3, the cubes 0^3, 1^3, 2^3, ..., 7^3 reduced mod 2^3 = 8 are 0,1,0,3,0,5,0,7, five different values, so a(3)=5. - _N. J. A. Sloane_, Sep 30 2018

S. R. Finch and Pascal Sebah, Squares and Cubes Modulo n, arXiv:math/0604465 [math.NT], 2006-2016.
Index entries for linear recurrences with constant coefficients, signature (2,0,1,-2).

Cf. A033138.

A049347 := proc(n) op( (n mod 3)+1,[1,-1,0]) ;end proc:
A046630 := proc(n) 2^(n+2)/7+2/3-(5*A049347(n)+A049347(n-1))/21 ; end proc: # R. J. Mathar, Feb 27 2011

LinearRecurrence[{2, 0, 1, -2}, {1, 2, 3, 5}, 33] (* Jean-François Alcover, Nov 17 2017 *)

a(n)=(4<Charles R Greathouse IV, Jan 03 2013

a(n) = ceiling(2^(n+2)/7) [Finch-Sebah, page 12]. - N. J. A. Sloane, Sep 30 2018
G.f.: (-2*x^3-x^2+1)/((1-2*x)*(1-x^3)).
a(n) = A046530(2^n) = 2^(n+2)/7 + 2/3 - (5*A049347(n)+A049347(n-1))/21. - R. J. Mathar, Feb 27 2011
a(n) = 1 + A033138(n) for n >= 1. - John Keith, Mar 07 2022

A083593 Expansion of 1/((1-2x)(1-x^4)).

Original entry on oeis.org

1, 2, 4, 8, 17, 34, 68, 136, 273, 546, 1092, 2184, 4369, 8738, 17476, 34952, 69905, 139810, 279620, 559240, 1118481, 2236962, 4473924, 8947848, 17895697, 35791394, 71582788, 143165576, 286331153, 572662306, 1145324612, 2290649224

Offset: 0

Paul Barry, May 02 2003

Here we let p = 4 to produce the above sequence, but p can be an arbitrary natural number. By letting p = 2, 3, 6, 7 we produce A000975, A033138, A195904 and A117302. We denote by U[p,n,m] the number of cases in which the first player gets killed in a Russian roulette game when p players use a gun with n chambers and m bullets. They never rotate the cylinder after the game starts. The chambers can be represented by the list {1,2,...,n}.
We are going to calculate the following (0), (1), ..., (t) separately. (0) The first player gets killed when one bullet is in the first chamber and the remaining m-1  bullets are in {2,3,...,n}. We have binomial(n-1,m-1) cases for this. (1) The first gets killed when one bullet is in the (p+1)th chamber and the rest of the bullets are in {p+2,...,n}. We have binomial(n-p-1,m-1) cases for this. We continue to calculate and the last is (t), where t = floor((n-m)/p). (t) The first gets killed when one bullet is in the (pt+1)-st chamber and the remaining bullets are in {pt+2,...,n}. We have binomial(n-pt-1,m-1) cases for this. Therefore U[p,n,m] = Sum_{z=0..floor((n-m)/p)} binomial(n-pz-1,m-1). Let A[p,n] be the number of the cases in which the first player gets killed when p players use a gun with n chambers and the number of the bullets can be from 1 to n. Then A[p,n] = Sum_{m=1..n} U[p,n,m]. - Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006
A001045(n+5) without last digit. - Paul Curtz, Apr 21 2021
a(n) is the number of partitions of n into parts 1 and 4 where there are two colors of part 1 and the order of the colors of parts 1 matters. If the order of colors doesn't matter we get A001972. - Joerg Arndt, Jan 18 2024

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,0,1,-2).

Cf. A000975, A033138, A195904, A117302.

U[p_,n_,m_,v_]:=Block[{t},t=Floor[(1+p-m+n-v)/p];Sum[Binomial[n-v-p*z,m-1],{z,0,t-1}]]; A[p_,n_,v_]:=Sum[U[p,n,k,v],{k,1,n}]; (* Here we let p = 4 to produce the above sequence, but this code can produce A000975, A033138, A195904, A117302 for p=2,3,6,7.*) Table[A[4,n,1], {n,1,20}] (* Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006 *)
CoefficientList[Series[1/((1-2x)(1-x^4)),{x,0,40}],x] (* Vincenzo Librandi, Apr 04 2012 *)
a[n_] := FromDigits[Table[(Mod[j, 4]/4) // Round, {j, 1, n + 3}], 2] (* Andres Cicuttin, Mar 25 2016 *)
a[n_] := a[n] = 2 a[n - 1] + 1 - Ceiling[Mod[n, 4]/4]; a[0] = 1;
Table[a[n], {n, 0, 31}] (* Andres Cicuttin, Mar 27 2016 *)
LinearRecurrence[{2,0,0,1,-2},{1,2,4,8,17},40] (* Harvey P. Dale, Apr 03 2018 *)

Vec(1/((1-2*x)*(1-x^4))+O(x^99)) \\ Charles R Greathouse IV, May 15 2013

a(n)=(16<Charles R Greathouse IV, Mar 27 2016

def A083593(n): return ((32<Chai Wah Wu, Apr 25 2025

a(n) = 2*a(n-1) + a(n-4) - 2*a(n-5).
If n is a multiple of 4, then a(n) = 2*a(n-1) + 1, otherwise a(n) = 2*a(n-1). - Gerald McGarvey, Oct 14 2008
a(n) = floor((2^(n+5) + 1)/30). - Tani Akinari, Jul 09 2013
From Andres Cicuttin, Mar 29 2016: (Start)
a(n) = 2*a(n-1) + floor(((n-1) mod 4)/3), with a(0)=1.
a(n) = 2*a(n-1) + 1 - ceiling((n mod 4)/4), with a(0)=1. (End)
15*a(n) = 2^(n+4) - A133145(n). - R. J. Mathar, Feb 27 2019
E.g.f.: (3*cos(x) - 5*cosh(x) + 32*cosh(2*x) + 6*sin(x) - 10*sinh(x) + 32*sinh(2*x))/30. - Stefano Spezia, Apr 25 2025

A300653 Square array T(n, k) (n >= 1, k >= 1) read by antidiagonals upwards: T(n, k) is the k-th positive number whose binary representation appears as a substring in the binary representation of 1/n (ignoring the radix point and adding trailing zeros if necessary in case of a terminating expansion).

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 4, 8, 1, 2, 5, 8, 16, 1, 2, 4, 10, 16, 32, 1, 2, 3, 8, 21, 32, 64, 1, 2, 5, 4, 16, 42, 64, 128, 1, 2, 4, 10, 6, 32, 85, 128, 256, 1, 2, 4, 9, 21, 9, 64, 170, 256, 512, 1, 2, 3, 8, 18, 42, 12, 128, 341, 512, 1024, 1, 2, 3, 4, 16, 36, 85

Offset: 1

Rémy Sigrist, Mar 10 2018

			Square array begins:
  n\k|    1    2    3    4    5    6    7    8    9   10
  ---+--------------------------------------------------
    1|    1    2    4    8   16   32   64  128  256  512  -->  A000079
    2|    1    2    4    8   16   32   64  128  256  512
    3|    1    2    5   10   21   42   85  170  341  682  -->  A000975
    4|    1    2    4    8   16   32   64  128  256  512
    5|    1    2    3    4    6    9   12   19   25   38
    6|    1    2    5   10   21   42   85  170  341  682
    7|    1    2    4    9   18   36   73  146  292  585  -->  A033138
    8|    1    2    4    8   16   32   64  128  256  512
    9|    1    2    3    4    6    7    8   12   14   17
   10|    1    2    3    4    6    9   12   19   25   38

Rémy Sigrist, PARI program for A300653

Cf. A000079, A000975, A033138, A153055, A300630, A300654.

PARI
```
See Links section.
```

T(n, 1) = 1.
T(n, 2) = 2.
T(n, 3) > 3 iff n belongs to A300630.
T(2*n, k) = T(n, k).
T(1, k) = A000079(k-1).
T(3, k) = A000975(k).
T(7, k) = A033138(k).
T(n, k) = k iff 1 <= k <= A300654(n).
T(n, k) = n for some k iff n belongs to A000079 or to A153055.

A078010 Expansion of (1-x)/(1 - x - x^2 - 2*x^3).

Original entry on oeis.org

1, 0, 1, 3, 4, 9, 19, 36, 73, 147, 292, 585, 1171, 2340, 4681, 9363, 18724, 37449, 74899, 149796, 299593, 599187, 1198372, 2396745, 4793491, 9586980, 19173961, 38347923, 76695844, 153391689, 306783379, 613566756, 1227133513, 2454267027, 4908534052

Offset: 0

N. J. A. Sloane, Nov 17 2002

For n > 0, a(n) is the number of ways to tile a strip of length n with squares, dominoes, and two colors of trominoes, with the restriction that the first tile cannot be a square. - Greg Dresden and Bora Bursalı, Aug 31 2023

			a(6) = 19 = A077947(4) + 2*A077947(3) = 9 + 2*5 = 19.
G.f. = 1 + x^2 + 3*x^3 + 4*x^4 + 9*x^5 + 19*x^6 + 36*x^7 + 73*x^8 + ... - _Michael Somos_, Nov 18 2020

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,2).

Cf. A033138, A077947.

a:=[1,0,1];; for n in [4..50] do a[n]:=a[n-1]+a[n-2]+2*a[n-3]; od; a; # G. C. Greubel, Jun 28 2019

R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1-x)/(1-x-x^2-2*x^3) )); // G. C. Greubel, Jun 28 2019

CoefficientList[Series[(1-x)/(1-x-x^2-2*x^3), {x,0,50}],x]  (* Harvey P. Dale, Mar 17 2011 *)
LinearRecurrence[{1, 1, 2}, {1, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 24 2012 *)

Vec((1-x)/(1-x-x^2-2*x^3)+O(x^50)) \\ Charles R Greathouse IV, Sep 26 2012

{a(n) = ([0, 1, 1; 1, 1, 0; 0, 2, 0]^n)[1, 1]}; /* Michael Somos, Nov 18 2020 */

((1-x)/(1-x-x^2-2*x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Jun 28 2019

a(0)=1, a(1)=0, a(2)=1, a(n) = a(n-1) + a(n-2) + 2*a(n-3) for n > 2. - Philippe Deléham, Sep 19 2006
a(n) + a(n+1) = A122552(n+1). - Philippe Deléham, Sep 25 2006
If p[1]=0, p[2]=1, p[i]=3, (i>2), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n >= 1, a(n)=det A. - Milan Janjic, May 02 2010
For n > 3, a(n) = A077947(n-2) + 2*A077947(n-3), with A077947 beginning (1, 2, 5, 9, 18, 37, ...); "1" has offset 1. - Gary W. Adamson, May 13 2013
a(n) = 2^(n-1) - 3*floor((2^(n-1))/7) - 1, for n >= 1. - Ridouane Oudra, Dec 02 2019
G.f.: (1 - x) / ((1 - 2*x) * (1 + x + x^2)). - Michael Somos, Nov 18 2020

A195904 Base-2 digits are, in order, the first n terms of the periodic sequence with initial period 1,0,0,0,0,0.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 65, 130, 260, 520, 1040, 2080, 4161, 8322, 16644, 33288, 66576, 133152, 266305, 532610, 1065220, 2130440, 4260880, 8521760, 17043521, 34087042, 68174084, 136348168, 272696336, 545392672, 1090785345, 2181570690, 4363141380, 8726282760

Offset: 1

Jeremy Gardiner, Sep 25 2011

Here we let p = 6 to produce the above sequence, but p can be an arbitrary natural number. By letting p = 2, 3, 4, 7 we produce A000975, A033138, A083593 and A117302. We denote by U(p,n,m) the number of cases in which the first player is killed in a Russian roulette game when p players use a gun with n chambers and m bullets. They never rotate the cylinder after the game starts. The chambers can be represented by the list {1,2,...,n}.

We are going to calculate the following (0), (1), ..., (t) separately. (0) The first player is killed when one bullet is in the first chamber and the remaining (m-1) bullets are in {2,3,...,n}. We have binomial(n-1,m-1) cases for this. (1) The first is killed when one bullet is in the (p+1)th chamber and the rest of the bullets are in {p+2,..,n}. We have binomial(n-p-1,m-1) cases for this. We continue to calculate and the last is (t), where t = floor((n-m)/p). (t) The first is killed when one bullet is in the (pt+1)-th chamber and the remaining bullets are in {pt+2,...,n}. We have binomial(n-pt-1,m-1) cases for this. Therefore U(p,n,m) = Sum_{z=0..floor((n-m)/p)} binomial(n-pz-1,m-1). Let A(p,n) be the number of cases in which the first player is killed when p players use a gun with n chambers and the number of the bullets can be from 1 to n. Then A(p,n) = Sum_{m=1..n} U(p,n,m). - Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,0,1,-2).

Cf. A000975, A033138, A083593, A117302.

U[p_, n_, m_, v_]:=Block[{t}, t=Floor[(1+p-m+n-v)/p]; Sum[Binomial[n - v - p*z, m - 1], {z, 0, t - 1}]]; A[p_, n_, v_]:=Sum[U[p, n, k, v], {k, 1, n}]; (* Here we let p = 6 to produce the above sequence, but this code can produce A000975, A033138, A083593, A117302 for p = 2, 3, 4, 7. *) Table[A[6, n, 1], {n, 1, 20}] (* Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006 *)
Rest[CoefficientList[Series[x/(2*x^7 - x^6 - 2*x + 1), {x, 0, 50}], x]] (* G. C. Greubel, Sep 28 2017 *)

x='x+O('x^50); Vec(x/(2*x^7 - x^6 - 2*x + 1)) \\ G. C. Greubel, Sep 28 2017

From Colin Barker, Jun 09 2013: (Start)
a(n) = floor(2^(n+5)/63).
G.f.: x /(2*x^7 - x^6 - 2*x +1).
G.f.: x /((x-1)*(x+1)*(2*x-1)*(x^2-x+1)*( x^2+x+1)). (End)

More terms from Colin Barker, Jun 09 2013

A300428 a(n) is the least positive k such that the binary representation of n appears as a substring in the binary representation of 1/k (ignoring the radix point and adding trailing zeros if necessary in case of a terminating expansion).

Original entry on oeis.org

1, 1, 5, 1, 3, 5, 9, 1, 5, 3, 11, 5, 11, 9, 17, 1, 9, 7, 5, 11, 3, 13, 11, 9, 5, 11, 13, 9, 11, 17, 33, 1, 17, 11, 9, 7, 19, 5, 13, 11, 29, 3, 19, 13, 27, 11, 19, 17, 9, 19, 5, 11, 19, 13, 29, 9, 19, 11, 13, 17, 25, 33, 65, 1, 33, 23, 17, 13, 11, 29, 9, 23, 7

Offset: 1

Rémy Sigrist, Mar 05 2018

In other words, a(n) is the least k > 0 such that floor((2^i) / k) mod A062383(n) = n for some integer i >= 0.
This sequence is similar to A035335 for the base 2.
All terms are odd.
All terms appears infinitely many times (as a(n) equals at least a(2*n) or a(2*n + 1)).
See also A300475 for a similar sequence.

			The first terms, alongside the binary representation of 1/a(n) with the earliest occurrence of the binary representation of n in parentheses, are:
  n  a(n)    bin(1/a(n))
  -- ----    -----------
   1    1    (1).000...
   2    1    (1.0)000...
   3    5    0.00(11)001...
   4    1    (1.00)000...
   5    3    0.0(101)010...
   6    5    0.00(110)011...
   7    9    0.000(111)000...
   8    1    (1.000)000...
   9    5    0.001(1001)100...
  10    3    0.0(1010)101...
  11   11    0.000(1011)101...
  12    5    0.00(1100)110...
  13   11    0.000101(1101)000...
  14    9    0.000(1110)001...
  15   17    0.0000(1111)000...
  16    1    (1.0000)000...
  17    9    0.00011(10001)110...
  18    7    0.00(10010)010...
  19    5    0.001(10011)001...
  20   11    0.0001011(10100)010...

Rémy Sigrist, Table of n, a(n) for n = 1..8191
Rémy Sigrist, PARI program for A300428

Cf. A000975, A033138, A035335, A062383, A300475.

PARI
```
See Links section.
```

a(2^k) = 1 for any k >= 0.
a(2^k - 1) = 2^k + 1 for any k > 1.
a(A000975(k)) = 3 for any k > 2.
a(A033138(k)) = 7 for any k > 4.
a(n) <= A300475(n) for any n > 0.

A300475 a(n) is the least positive k such that the binary representation n appears in front of the binary representation of 1/k (ignoring the radix point and the leading zeros and adding trailing zeros if necessary in case of a terminating expansion).

Original entry on oeis.org

1, 1, 5, 1, 3, 5, 9, 1, 7, 3, 11, 5, 19, 9, 17, 1, 15, 7, 13, 25, 3, 23, 11, 21, 5, 19, 37, 9, 35, 17, 33, 1, 31, 15, 29, 7, 27, 53, 13, 25, 49, 3, 47, 23, 45, 11, 43, 21, 41, 81, 5, 39, 19, 75, 37, 9, 71, 35, 69, 17, 67, 33, 65, 1, 63, 31, 61, 15, 59, 29, 57

Offset: 1

Rémy Sigrist, Mar 06 2018

In other words, a(n) is the least k > 0 such that floor((2^i) / k) = n for some integer i >= 0.
This sequence is similar to A095156 for the base 2.
All terms are odd.
All terms appears infinitely many times (as a(n) equals at least a(2*n) or a(2*n + 1)).
See also A300428 for a similar sequence.

			The first terms, alongside the binary representation of 1/a(n) with the earliest occurrence of the binary representation of n in parentheses, are:
  n  a(n)    bin(1/a(n))
  -- ----    -----------
   1    1    (1).000...
   2    1    (1.0)000...
   3    5    0.00(11)001...
   4    1    (1.00)000...
   5    3    0.0(101)010...
   6    5    0.00(110)011...
   7    9    0.000(111)000...
   8    1    (1.000)000...
   9    7    0.00(1001)001...
  10    3    0.0(1010)101...
  11   11    0.000(1011)101...
  12    5    0.00(1100)110...
  13   19    0.0000(1101)011...
  14    9    0.000(1110)001...
  15   17    0.0000(1111)000...
  16    1    (1.0000)000...
  17   15    0.000(10001)000...
  18    7    0.00(10010)010...
  19   13    0.000(10011)101...
  20   25    0.0000(10100)011...

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program for A300475
Rémy Sigrist, Colored logarithmic scatterplot of the first 1000000 terms (where the color is function of A070939(n * a(n)))

Cf. A000975, A033138, A070939, A095156, A300428.

PARI
```
See Links section.
```

a(2^k) = 1 for any k >= 0.
a(2^k - 1) = 2^k + 1 for any k > 1.
a(A000975(k)) = 3 for any k > 2.
a(A033138(k)) = 7 for any k > 4.
a(n) >= A300428(n).

A101351 a(n) = 2^n-1 + Fibonacci(n).

Original entry on oeis.org

2, 4, 9, 18, 36, 71, 140, 276, 545, 1078, 2136, 4239, 8424, 16760, 33377, 66522, 132668, 264727, 528468, 1055340, 2108097, 4212014, 8417264, 16823583, 33629456, 67230256, 134414145, 268753266, 537385140, 1074573863, 2148829916, 4297145604, 8593459169

Offset: 1

Jorge Coveiro, Dec 25 2004

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-4,-1,2).

Cf. A000045, A033138.

Maple
```
seq(2^x-1+fibonacci(x),x=1..30);
```

Table[2^n-1+Fibonacci[n],{n,30}] (* or *) LinearRecurrence[{4,-4,-1,2},{2,4,9,18},30] (* Harvey P. Dale, Aug 24 2012 *)

Vec(x*(2-4*x+x^2)/((1-x)*(1-2*x)*(1-x-x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

[gaussian_binomial(n,1,2)+fibonacci (n) for n in range(1,31)] # Zerinvary Lajos, May 29 2009

a(n) = 4*a(n-1) -4*a(n-2) -a(n-3) +2*a(n-4). G.f.: x*(2-4*x+x^2)/((x-1) * (2*x-1) * (1-x-x^2)). - R. J. Mathar, Feb 06 2010

a(n) = ((1+sqrt(5))^n-(1-sqrt(5))^n)/(2^n*sqrt(5)) + 2^n - 1. - Colin Barker, Nov 02 2016

Offset changed to 1 by Colin Barker, Nov 02 2016

cp's OEIS Frontend

A023001 a(n) = (8^n - 1)/7.

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A284116 a(n) = largest number of distinct words arising in Post's tag system {00, 1101} applied to a binary word w, over all starting words w of length n, or a(n) = -1 if there is a word w with an unbounded trajectory.

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A046630 Number of cubic residues mod 2^n.

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A083593 Expansion of 1/((1-2*x)*(1-x^4)).

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A078010 Expansion of (1-x)/(1 - x - x^2 - 2*x^3).

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A083593 Expansion of 1/((1-2x)(1-x^4)).