Displaying 1-10 of 13 results found.
0, 1, 9, 73, 585, 4681, 37449, 299593, 2396745, 19173961, 153391689, 1227133513, 9817068105, 78536544841, 628292358729, 5026338869833, 40210710958665, 321685687669321, 2573485501354569, 20587884010836553, 164703072086692425
COMMENTS
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 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)
LINKS
Eric Weisstein's World of Mathematics, Repunit.
FORMULA
a(0) = 0, a(n) = 8*a(n-1) + 1 for n>0.
G.f.: x/((1-8x)*(1-x)). (End)
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
EXAMPLE
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
MAPLE
a:=n->sum(8^(n-j), j=1..n): seq(a(n), n=0..20); # Zerinvary Lajos, Jan 04 2007
MATHEMATICA
Table[(8^n-1)/7, {n, 0, m}]
LinearRecurrence[{9, -8}, {0, 1}, 30] (* Harvey P. Dale, Feb 12 2015 *)
PROG
(Sage) [lucas_number1(n, 9, 8) for n in range(0, 21)] # Zerinvary Lajos, Apr 23 2009 (Sage) [gaussian_binomial(n, 1, 8) for n in range(0, 21)] # Zerinvary Lajos, May 28 2009 (Maxima) A023001(n):=floor((8^n-1)/7)$ (GAP)
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.
+10
47
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
COMMENTS
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. WechslerComment 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 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)
REFERENCES
John Stillwell, Elements of Mathematics: From Euclid to Goedel, Princeton, 2016. See page 100, Post's tag system.
LINKS
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]
EXAMPLE
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.
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
MATHEMATICA
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 *)
CROSSREFS
Cf. A033138, A152111, A284119, A284121, A289670, A289671, A289672, A289673, A289674, A289675, A289676, A289677. 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).
EXTENSIONS
Edited by N. J. A. Sloane, Jul 29 2017 and Oct 23 2017 (adding escape clause in case an infinite trajectory exists)
Number of cubic residues mod 2^n.
+10
11
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
FORMULA
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)).
EXAMPLE
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
MAPLE
A049347 := proc(n) op( (n mod 3)+1, [1, -1, 0]) ; end proc:
Expansion of 1/((1-2*x)*(1-x^4)).
+10
6
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
COMMENTS
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 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
FORMULA
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) = 2*a(n-1) + floor(((n-1) mod 4) /3), with a(0)=1. - Andres Cicuttin, Mar 29 2016 a(n) = 2*a(n-1) + 1 - ceiling((n mod 4)/4), with a(0)=1. - Andres Cicuttin, Mar 29 2016
MATHEMATICA
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;
LinearRecurrence[{2, 0, 0, 1, -2}, {1, 2, 4, 8, 17}, 40] (* Harvey P. Dale, Apr 03 2018 *)
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).
+10
6
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
FORMULA
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(n, k) = k iff 1 <= k <= A300654(n).
EXAMPLE
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
PROG
(PARI) See Links section.
Expansion of (1-x)/(1 - x - x^2 - 2*x^3).
+10
5
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
COMMENTS
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
FORMULA
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 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 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
EXAMPLE
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
MATHEMATICA
CoefficientList[Series[(1-x)/(1-x-x^2-2*x^3), {x, 0, 50}], x] (* Harvey P. Dale, Mar 17 2011 *)
PROG
(PARI) {a(n) = ([0, 1, 1; 1, 1, 0; 0, 2, 0]^n)[1, 1]}; /* Michael Somos, Nov 18 2020 */ (Magma) R<x>:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1-x)/(1-x-x^2-2*x^3) )); // G. C. Greubel, Jun 28 2019 (Sage) ((1-x)/(1-x-x^2-2*x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Jun 28 2019 (GAP) 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
Base-2 digits are, in order, the first n terms of the periodic sequence with initial period 1,0,0,0,0,0.
+10
5
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
COMMENTS
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
FORMULA
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)
MATHEMATICA
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 *)
PROG
(PARI) x='x+O('x^50); Vec(x/(2*x^7 - x^6 - 2*x + 1)) \\ G. C. Greubel, Sep 28 2017
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).
+10
3
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
COMMENTS
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.
FORMULA
a(2^k) = 1 for any k >= 0.
a(2^k - 1) = 2^k + 1 for any k > 1.
EXAMPLE
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...
PROG
(PARI) See Links section.
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).
+10
3
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
COMMENTS
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.
FORMULA
a(2^k) = 1 for any k >= 0.
a(2^k - 1) = 2^k + 1 for any k > 1.
EXAMPLE
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...
PROG
(PARI) See Links section.
a(n) = 2^n-1 + Fibonacci(n).
+10
2
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
FORMULA
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
MAPLE
seq(2^x-1+fibonacci(x), x=1..30);
MATHEMATICA
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 *)
PROG
(Sage) [gaussian_binomial(n, 1, 2)+fibonacci (n) for n in range(1, 31)] # Zerinvary Lajos, May 29 2009 (PARI) Vec(x*(2-4*x+x^2)/((1-x)*(1-2*x)*(1-x-x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016
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