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 1-7 of 7 results.

A214319 First differences of A182028.

Original entry on oeis.org

1, 1, 2, 5, 9, 19, 37, 74, 149, 297, 594, 1189, 2377, 4755, 9509, 19018, 38037, 76073, 152147, 304293, 608586, 1217173, 2434345, 4868690, 9737381, 19474761, 38949523, 77899045, 155798090, 311596181, 623192361, 1246384722, 2492769445, 4985538889, 9971077779, 19942155557, 39884311114, 79768622229
Offset: 0

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Author

N. J. A. Sloane, Jul 12 2012

Keywords

Crossrefs

A000225 a(n) = 2^n - 1. (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)

Original entry on oeis.org

0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191, 16383, 32767, 65535, 131071, 262143, 524287, 1048575, 2097151, 4194303, 8388607, 16777215, 33554431, 67108863, 134217727, 268435455, 536870911, 1073741823, 2147483647, 4294967295, 8589934591
Offset: 0

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Author

Keywords

Comments

This is the Gaussian binomial coefficient [n,1] for q=2.
Number of rank-1 matroids over S_n.
Numbers k such that the k-th central binomial coefficient is odd: A001405(k) mod 2 = 1. - Labos Elemer, Mar 12 2003
This gives the (zero-based) positions of odd terms in the following convolution sequences: A000108, A007460, A007461, A007463, A007464, A061922.
Also solutions (with minimum number of moves) for the problem of Benares Temple, i.e., three diamond needles with n discs ordered by decreasing size on the first needle to place in the same order on the third one, without ever moving more than one disc at a time and without ever placing one disc at the top of a smaller one. - Xavier Acloque, Oct 18 2003
a(0) = 0, a(1) = 1; a(n) = smallest number such that a(n)-a(m) == 0 (mod (n-m+1)), for all m. - Amarnath Murthy, Oct 23 2003
Binomial transform of [1, 1/2, 1/3, ...] = [1/1, 3/2, 7/3, ...]; (2^n - 1)/n, n=1,2,3, ... - Gary W. Adamson, Apr 28 2005
Numbers whose binary representation is 111...1. E.g., the 7th term is (2^7) - 1 = 127 = 1111111 (in base 2). - Alexandre Wajnberg, Jun 08 2005
Number of nonempty subsets of a set with n elements. - Michael Somos, Sep 03 2006
For n >= 2, a(n) is the least Fibonacci n-step number that is not a power of 2. - Rick L. Shepherd, Nov 19 2007
Let P(A) be the power set of an n-element set A. Then a(n+1) = the number of pairs of elements {x,y} of P(A) for which x and y are disjoint and for which either x is a subset of y or y is a subset of x. - Ross La Haye, Jan 10 2008
A simpler way to state this is that it is the number of pairs (x,y) where at least one of x and y is the empty set. - Franklin T. Adams-Watters, Oct 28 2011
2^n-1 is the sum of the elements in a Pascal triangle of depth n. - Brian Lewis (bsl04(AT)uark.edu), Feb 26 2008
Sequence generalized: a(n) = (A^n -1)/(A-1), n >= 1, A integer >= 2. This sequence has A=2; A003462 has A=3; A002450 has A=4; A003463 has A=5; A003464 has A=6; A023000 has A=7; A023001 has A=8; A002452 has A=9; A002275 has A=10; A016123 has A=11; A016125 has A=12; A091030 has A=13; A135519 has A=14; A135518 has A=15; A131865 has A=16; A091045 has A=17; A064108 has A=20. - Ctibor O. Zizka, Mar 03 2008
a(n) is also a Mersenne prime A000668 when n is a prime number in A000043. - Omar E. Pol, Aug 31 2008
a(n) is also a Mersenne number A001348 when n is prime. - Omar E. Pol, Sep 05 2008
With offset 1, = row sums of triangle A144081; and INVERT transform of A009545 starting with offset 1; where A009545 = expansion of sin(x)*exp(x). - Gary W. Adamson, Sep 10 2008
Numbers n such that A000120(n)/A070939(n) = 1. - Ctibor O. Zizka, Oct 15 2008
For n > 0, sequence is equal to partial sums of A000079; a(n) = A000203(A000079(n-1)). - Lekraj Beedassy, May 02 2009
Starting with offset 1 = the Jacobsthal sequence, A001045, (1, 1, 3, 5, 11, 21, ...) convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 23 2009
Numbers n such that n=2*phi(n+1)-1. - Farideh Firoozbakht, Jul 23 2009
a(n) = (a(n-1)+1)-th odd numbers = A005408(a(n-1)) for n >= 1. - Jaroslav Krizek, Sep 11 2009
Partial sums of a(n) for n >= 0 are A000295(n+1). Partial sums of a(n) for n >= 1 are A000295(n+1) and A130103(n+1). a(n) = A006127(n) - (n+1). - Jaroslav Krizek, Oct 16 2009
If n is even a(n) mod 3 = 0. This follows from the congruences 2^(2k) - 1 ~ 2*2*...*2 - 1 ~ 4*4*...*4 - 1 ~ 1*1*...*1 - 1 ~ 0 (mod 3). (Note that 2*2*...*2 has an even number of terms.) - Washington Bomfim, Oct 31 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=2,(i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n)=det(A). - Milan Janjic, Jan 26 2010
This is the sequence A(0,1;1,2;2) = A(0,1;3,-2;0) 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) = S(n+1,2), a Stirling number of the second kind. See the example below. - Dennis P. Walsh, Mar 29 2011
Entries of row a(n) in Pascal's triangle are all odd, while entries of row a(n)-1 have alternating parities of the form odd, even, odd, even, ..., odd.
Define the bar operation as an operation on signed permutations that flips the sign of each entry. Then a(n+1) is the number of signed permutations of length 2n that are equal to the bar of their reverse-complements and avoid the set of patterns {(-2,-1), (-1,+2), (+2,+1)}. (See the Hardt and Troyka reference.) - Justin M. Troyka, Aug 13 2011
A159780(a(n)) = n and A159780(m) < n for m < a(n). - Reinhard Zumkeller, Oct 21 2011
This sequence is also the number of proper subsets of a set with n elements. - Mohammad K. Azarian, Oct 27 2011
a(n) is the number k such that the number of iterations of the map k -> (3k +1)/2 == 1 (mod 2) until reaching (3k +1)/2 == 0 (mod 2) equals n. (see the Collatz problem). - Michel Lagneau, Jan 18 2012
For integers a, b, denote by a<+>b the least c >= a such that Hd(a,c) = b (note that, generally speaking, a<+>b differs from b<+>a). Then a(n+1)=a(n)<+>1. Thus this sequence is the Hamming analog of nonnegative integers. - Vladimir Shevelev, Feb 13 2012
Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... apparently A007733. - R. J. Mathar, Aug 10 2012
Start with n. Each n generates a sublist {n-1,n-2,...,1}. Each element of each sublist also generates a sublist. Take the sum of all. E.g., 3->{2,1} and 2->{1}, so a(3)=3+2+1+1=7. - Jon Perry, Sep 02 2012
This is the Lucas U(P=3,Q=2) sequence. - R. J. Mathar, Oct 24 2012
The Mersenne numbers >= 7 are all Brazilian numbers, as repunits in base two. See Proposition 1 & 5.2 in Links: "Les nombres brésiliens". - Bernard Schott, Dec 26 2012
Number of line segments after n-th stage in the H tree. - Omar E. Pol, Feb 16 2013
Row sums of triangle in A162741. - Reinhard Zumkeller, Jul 16 2013
a(n) is the highest power of 2 such that 2^a(n) divides (2^n)!. - Ivan N. Ianakiev, Aug 17 2013
In computer programming, these are the only unsigned numbers such that k&(k+1)=0, where & is the bitwise AND operator and numbers are expressed in binary. - Stanislav Sykora, Nov 29 2013
Minimal number of moves needed to interchange n frogs in the frogs problem (see for example the NRICH 1246 link or the Britton link below). - N. J. A. Sloane, Jan 04 2014
a(n) !== 4 (mod 5); a(n) !== 10 (mod 11); a(n) !== 2, 4, 5, 6 (mod 7). - Carmine Suriano, Apr 06 2014
After 0, antidiagonal sums of the array formed by partial sums of integers (1, 2, 3, 4, ...). - Luciano Ancora, Apr 24 2015
a(n+1) equals the number of ternary words of length n avoiding 01,02. - Milan Janjic, Dec 16 2015
With offset 0 and another initial 0, the n-th term of 0, 0, 1, 3, 7, 15, ... is the number of commas required in the fully-expanded von Neumann definition of the ordinal number n. For example, 4 := {0, 1, 2, 3} := {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}}, which uses seven commas. Also, for n>0, a(n) is the total number of symbols required in the fully-expanded von Neumann definition of ordinal n - 1, where a single symbol (as usual) is always used to represent the empty set and spaces are ignored. E.g., a(5) = 31, the total such symbols for the ordinal 4. - Rick L. Shepherd, May 07 2016
With the quantum integers defined by [n+1]A001045%20are%20given%20by%20q%20=%20i%20*%20sqrt(2)%20for%20i%5E2%20=%20-1.%20Cf.%20A239473.%20-%20_Tom%20Copeland">q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)), the Mersenne numbers are a(n+1) = q^n [n+1]_q with q = sqrt(2), whereas the signed Jacobsthal numbers A001045 are given by q = i * sqrt(2) for i^2 = -1. Cf. A239473. - _Tom Copeland, Sep 05 2016
For n>1: numbers n such that n - 1 divides sigma(n + 1). - Juri-Stepan Gerasimov, Oct 08 2016
This is also the second column of the Stirling2 triangle A008277 (see also A048993). - Wolfdieter Lang, Feb 21 2017
Except for the initial terms, the decimal representation of the x-axis of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 659", "Rule 721" and "Rule 734", based on the 5-celled von Neumann neighborhood initialized with a single on cell. - Robert Price, Mar 14 2017
a(n), n > 1, is the number of maximal subsemigroups of the monoid of order-preserving partial injective mappings on a set with n elements. - James Mitchell and Wilf A. Wilson, Jul 21 2017
Also the number of independent vertex sets and vertex covers in the complete bipartite graph K_{n-1,n-1}. - Eric W. Weisstein, Sep 21 2017
Sum_{k=0..n} p^k is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*p + binomial(i+j-1, i), in this case p=2 (empirical observation). - Tony Foster III, May 11 2019
The rational numbers r(n) = a(n+1)/2^(n+1) = a(n+1)/A000079(n+1) appear also as root of the n-th iteration f^{[n]}(c; x) = 2^(n+1)*x - a(n+1)*c of f(c; x) = f^{[0]}(c; x) = 2*x - c as r(n)*c. This entry is motivated by a riddle of Johann Peter Hebel (1760 - 1826): Erstes Rechnungsexempel(Ein merkwürdiges Rechnungs-Exempel) from 1803, with c = 24 and n = 2, leading to the root r(2)*24 = 21 as solution. See the link and reference. For the second problem, also involving the present sequence, see a comment in A130330. - Wolfdieter Lang, Oct 28 2019
a(n) is the sum of the smallest elements of all subsets of {1,2,..,n} that contain n. For example, a(3)=7; the subsets of {1,2,3} that contain 3 are {3}, {1,3}, {2,3}, {1,2,3}, and the sum of smallest elements is 7. - Enrique Navarrete, Aug 21 2020
a(n-1) is the number of nonempty subsets of {1,2,..,n} which don't have an element that is the size of the set. For example, for n = 4, a(3) = 7 and the subsets are {2}, {3}, {4}, {1,3}, {1,4}, {3,4}, {1,2,4}. - Enrique Navarrete, Nov 21 2020
From Eric W. Weisstein, Sep 04 2021: (Start)
Also the number of dominating sets in the complete graph K_n.
Also the number of minimum dominating sets in the n-helm graph for n >= 3. (End)
Conjecture: except for a(2)=3, numbers m such that 2^(m+1) - 2^j - 2^k - 1 is composite for all 0 <= j < k <= m. - Chai Wah Wu, Sep 08 2021
a(n) is the number of three-in-a-rows passing through a corner cell in n-dimensional tic-tac-toe. - Ben Orlin, Mar 15 2022
From Vladimir Pletser, Jan 27 2023: (Start)
a(n) == 1 (mod 30) for n == 1 (mod 4);
a(n) == 7 (mod 120) for n == 3 (mod 4);
(a(n) - 1)/30 = (a(n+2) - 7)/120 for n odd;
(a(n) - 1)/30 = (a(n+2) - 7)/120 = A131865(m) for n == 1 (mod 4) and m >= 0 with A131865(0) = 0. (End)
a(n) is the number of n-digit numbers whose smallest decimal digit is 8. - Stefano Spezia, Nov 15 2023
Also, number of nodes in a perfect binary tree of height n-1, or: number of squares (or triangles) after the n-th step of the construction of a Pythagorean tree: Start with a segment. At each step, construct squares having the most recent segment(s) as base, and isosceles right triangles having the opposite side of the squares as hypotenuse ("on top" of each square). The legs of these triangles will serve as the segments which are the bases of the squares in the next step. - M. F. Hasler, Mar 11 2024
a(n) is the length of the longest path in the n-dimensional hypercube. - Christian Barrientos, Apr 13 2024
a(n) is the diameter of the n-Hanoi graph. Equivalently, a(n) is the largest minimum number of moves between any two states of the Towers of Hanoi problem (aka problem of Benares Temple described above). - Allan Bickle, Aug 09 2024

Examples

			For n=3, a(3)=S(4,2)=7, a Stirling number of the second kind, since there are 7 ways to partition {a,b,c,d} into 2 nonempty subsets, namely,
  {a}U{b,c,d}, {b}U{a,c,d}, {c}U{a,b,d}, {d}U{a,b,c}, {a,b}U{c,d}, {a,c}U{b,d}, and {a,d}U{b,c}. - _Dennis P. Walsh_, Mar 29 2011
From _Justin M. Troyka_, Aug 13 2011: (Start)
Since a(3) = 7, there are 7 signed permutations of 4 that are equal to the bar of their reverse-complements and avoid {(-2,-1), (-1,+2), (+2,+1)}. These are:
  (+1,+2,-3,-4),
  (+1,+3,-2,-4),
  (+1,-3,+2,-4),
  (+2,+4,-1,-3),
  (+3,+4,-1,-2),
  (-3,+1,-4,+2),
  (-3,-4,+1,+2). (End)
G.f. = x + 3*x^2 + 7*x^3 + 15*x^4 + 31*x^5 + 63*x^6 + 127*x^7 + ...
For the Towers of Hanoi problem with 2 disks, the moves are as follows, so a(2) = 3.
12|_|_ -> 2|1|_ -> _|1|2 -> _|_|12  - _Allan Bickle_, Aug 07 2024
		

References

  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
  • Ralph P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, Addison-Wesley, 2004, p. 134.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, p. 79.
  • Johann Peter Hebel, Gesammelte Werke in sechs Bänden, Herausgeber: Jan Knopf, Franz Littmann und Hansgeorg Schmidt-Bergmann unter Mitarbeit von Ester Stern, Wallstein Verlag, 2019. Band 3, S. 20-21, Loesung, S. 36-37. See also the link below.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 60, 75-83.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 141.
  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, "Tower of Hanoi", Penguin Books, 1987, pp. 112-113.

Crossrefs

Cf. A000043 (Mersenne exponents).
Cf. A000668 (Mersenne primes).
Cf. A001348 (Mersenne numbers with n prime).
Cf. a(n)=A112492(n, 2). Rightmost column of A008969.
a(n) = A118654(n, 1) = A118654(n-1, 3), for n > 0.
Subsequence of A132781.
Smallest number whose base b sum of digits is n: this sequence (b=2), A062318 (b=3), A180516 (b=4), A181287 (b=5), A181288 (b=6), A181303 (b=7), A165804 (b=8), A140576 (b=9), A051885 (b=10).
Cf. A008277, A048993 (columns k=2), A000918, A130330.
Cf. A000225, A029858, A058809, A375256 (Hanoi graphs).

Programs

  • Haskell
    a000225 = (subtract 1) . (2 ^)
    a000225_list = iterate ((+ 1) . (* 2)) 0
    -- Reinhard Zumkeller, Mar 20 2012
    
  • Maple
    A000225 := n->2^n-1; [ seq(2^n-1,n=0..50) ];
    A000225:=1/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation, sequence starting at a(1)
  • Mathematica
    a[n_] := 2^n - 1; Table[a[n], {n, 0, 30}] (* Stefan Steinerberger, Mar 30 2006 *)
    Array[2^# - 1 &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
    NestList[2 # + 1 &, 0, 32] (* Robert G. Wilson v, Feb 28 2011 *)
    2^Range[0, 20] - 1 (* Eric W. Weisstein, Jul 17 2017 *)
    LinearRecurrence[{3, -2}, {1, 3}, 20] (* Eric W. Weisstein, Sep 21 2017 *)
    CoefficientList[Series[1/(1 - 3 x + 2 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
  • PARI
    A000225(n) = 2^n-1  \\ Michael B. Porter, Oct 27 2009
    
  • PARI
    concat(0, Vec(x/((1-2*x)*(1-x)) + O(x^100))) \\ Altug Alkan, Oct 28 2015
    
  • Python
    def A000225(n): return (1<Chai Wah Wu, Jul 06 2022
  • SageMath
    def isMersenne(n): return n == sum([(1 - b) << s for (s, b) in enumerate((n+1).bits())]) # Peter Luschny, Sep 01 2019
    

Formula

G.f.: x/((1-2*x)*(1-x)).
E.g.f.: exp(2*x) - exp(x).
E.g.f. if offset 1: ((exp(x)-1)^2)/2.
a(n) = Sum_{k=0..n-1} 2^k. - Paul Barry, May 26 2003
a(n) = a(n-1) + 2*a(n-2) + 2, a(0)=0, a(1)=1. - Paul Barry, Jun 06 2003
Let b(n) = (-1)^(n-1)*a(n). Then b(n) = Sum_{i=1..n} i!*i*Stirling2(n,i)*(-1)^(i-1). E.g.f. of b(n): (exp(x)-1)/exp(2x). - Mario Catalani (mario.catalani(AT)unito.it), Dec 19 2003
a(n+1) = 2*a(n) + 1, a(0) = 0.
a(n) = Sum_{k=1..n} binomial(n, k).
a(n) = n + Sum_{i=0..n-1} a(i); a(0) = 0. - Rick L. Shepherd, Aug 04 2004
a(n+1) = (n+1)*Sum_{k=0..n} binomial(n, k)/(k+1). - Paul Barry, Aug 06 2004
a(n+1) = Sum_{k=0..n} binomial(n+1, k+1). - Paul Barry, Aug 23 2004
Inverse binomial transform of A001047. Also U sequence of Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005
a(n) = A099393(n-1) - A020522(n-1) for n > 0. - Reinhard Zumkeller, Feb 07 2006
a(n) = A119258(n,n-1) for n > 0. - Reinhard Zumkeller, May 11 2006
a(n) = 3*a(n-1) - 2*a(n-2); a(0)=0, a(1)=1. - Lekraj Beedassy, Jun 07 2006
Sum_{n>0} 1/a(n) = 1.606695152... = A065442, see A038631. - Philippe Deléham, Jun 27 2006
Stirling_2(n-k,2) starting from n=k+1. - Artur Jasinski, Nov 18 2006
a(n) = A125118(n,1) for n > 0. - Reinhard Zumkeller, Nov 21 2006
a(n) = StirlingS2(n+1,2). - Ross La Haye, Jan 10 2008
a(n) = A024036(n)/A000051(n). - Reinhard Zumkeller, Feb 14 2009
a(n) = A024088(n)/A001576(n). -Reinhard Zumkeller, Feb 15 2009
a(2*n) = a(n)*A000051(n); a(n) = A173787(n,0). - Reinhard Zumkeller, Feb 28 2010
For n > 0: A179857(a(n)) = A024036(n) and A179857(m) < A024036(n) for m < a(n). - Reinhard Zumkeller, Jul 31 2010
From Enrique Pérez Herrero, Aug 21 2010: (Start)
a(n) = J_n(2), where J_n is the n-th Jordan Totient function: (A007434, is J_2).
a(n) = Sum_{d|2} d^n*mu(2/d). (End)
A036987(a(n)) = 1. - Reinhard Zumkeller, Mar 06 2012
a(n+1) = A044432(n) + A182028(n). - Reinhard Zumkeller, Apr 07 2012
a(n) = A007283(n)/3 - 1. - Martin Ettl, Nov 11 2012
a(n+1) = A001317(n) + A219843(n); A219843(a(n)) = 0. - Reinhard Zumkeller, Nov 30 2012
a(n) = det(|s(i+2,j+1)|, 1 <= i,j <= n-1), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 06 2013
G.f.: Q(0), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 - 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013
E.g.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - (k+1)/Q(k+1))); (continued fraction).
G.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 23 2013
a(n) = A000203(2^(n-1)), n >= 1. - Ivan N. Ianakiev, Aug 17 2013
a(n) = Sum_{t_1+2*t_2+...+n*t_n=n} n*multinomial(t_1+t_2 +...+t_n,t_1,t_2,...,t_n)/(t_1+t_2 +...+t_n). - Mircea Merca, Dec 06 2013
a(0) = 0; a(n) = a(n-1) + 2^(n-1) for n >= 1. - Fred Daniel Kline, Feb 09 2014
a(n) = A125128(n) - A000325(n) + 1. - Miquel Cerda, Aug 07 2016
From Ilya Gutkovskiy, Aug 07 2016: (Start)
Binomial transform of A057427.
Sum_{n>=0} a(n)/n! = A090142. (End)
a(n) = A000918(n) + 1. - Miquel Cerda, Aug 09 2016
a(n+1) = (A095151(n+1) - A125128(n))/2. - Miquel Cerda, Aug 12 2016
a(n) = (A079583(n) - A000325(n+1))/2. - Miquel Cerda, Aug 15 2016
Convolution of binomial coefficient C(n,a(k)) with itself is C(n,a(k+1)) for all k >= 3. - Anton Zakharov, Sep 05 2016
a(n) = (A083706(n-1) + A000325(n))/2. - Miquel Cerda, Sep 30 2016
a(n) = A005803(n) + A005408(n-1). - Miquel Cerda, Nov 25 2016
a(n) = A279396(n+2,2). - Wolfdieter Lang, Jan 10 2017
a(n) = n + Sum_{j=1..n-1} (n-j)*2^(j-1). See a Jun 14 2017 formula for A000918(n+1) with an interpretation. - Wolfdieter Lang, Jun 14 2017
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k,i). - Wesley Ivan Hurt, Sep 21 2017
a(n+m) = a(n)*a(m) + a(n) + a(m). - Yuchun Ji, Jul 27 2018
a(n+m) = a(n+1)*a(m) - 2*a(n)*a(m-1). - Taras Goy, Dec 23 2018
a(n+1) is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*2 + binomial(i+j-1, i) (empirical observation). - Tony Foster III, May 11 2019
From Peter Bala, Jun 27 2025: (Start)
For n >= 1, a(3*n)/a(n) = A001576(n), a(4*n)/a(n) = A034496(n), a(5*n)/a(n) = A020514(n) a(6*n)/a(n) = A034665(n), a(7*n)/a(n) = A020516(n) and a(8*n)/a(n) = A034674(n).
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n.
Modulo differences in offsets, exp( Sum_{n >= 1} a(k*n)/a(n)*x^n/n ) is the o.g.f. of A006095 (k = 3), A006096 (k = 4), A006097 (k = 5), A006110 (k = 6), A022189 (k = 7), A022190 (k = 8), A022191 (k = 9) and A022192 (k = 10).
The following are all examples of telescoping series:
Sum_{n >= 1} 2^n/(a(n)*a(n+1)) = 1; Sum_{n >= 1} 2^n/(a(n)*a(n+1)*a(n+2)) = 1/9.
In general, for k >= 1, Sum_{n >= 1} 2^n/(a(n)*a(n+1)*...*a(n+k)) = 1/(a(1)*a(2)*...*a(k)*a(k)).
Sum_{n >= 1} 2^n/(a(n)*a(n+2)) = 4/9, since 2^n/(a(n)*a(n+2)) = b(n) - b(n+1), where b(n) = (2/3)*(3*2^(n-1) - 1)/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+2)) = -2/9, since (-2)^n/(a(n)*a(n+2)) = c(n) - c(n+1), where c(n) = (1/3)*(-2)^n/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} 2^n/(a(n)*a(n+4)) = 18/175, since 2^n/(a(n)*a(n+4)) = d(n) - d(n+1), where d(n) = (120*8^n - 140*4^n + 45*2^n - 4)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+4)) = -26/525, since (-2)^n/(a(n)*a(n+4)) = e(n) - e(n+1), where e(n) = (-1)^n*(40*8^n - 24*4^n + 5*2^n)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)). (End)

Extensions

Name partially edited by Eric W. Weisstein, Sep 04 2021

A003849 The infinite Fibonacci word (start with 0, apply 0->01, 1->0, take limit).

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1
Offset: 0

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Author

Keywords

Comments

A Sturmian word.
Define strings S(0)=0, S(1)=01, S(n)=S(n-1)S(n-2); iterate; sequence is S(infinity). If the initial 0 is omitted from S(n) for n>0, we obtain A288582(n+1).
The 0's occur at positions in A022342 (i.e., A000201 - 1), the 1's at positions in A003622.
Replace each run (1;1) with (1;0) in the infinite Fibonacci word A005614 (and add 0 as prefix) A005614 begins: 1,0,1,1,0,1,0,1,1,0,1,1,... changing runs (1,1) with (1,0) produces 1,0,0,1,0,1,0,0,1,0,0,1,... - Benoit Cloitre, Nov 10 2003
Characteristic function of A003622. - Philippe Deléham, May 03 2004
The fraction of 0's in the first n terms approaches 1/phi (see for example Allouche and Shallit). - N. J. A. Sloane, Sep 24 2007
The limiting mean and variance of the first n terms are 2-phi and 2*phi-3, respectively. - Clark Kimberling, Mar 12 2014, Aug 16 2018
Let S(n) be defined as above. Then this sequence is S(1) + Sum_{n=0..} S(n), where the addition of strings represents concatenation. - Isaac Saffold, May 03 2019
The word is a concatenation of three runs: 0, 1, and 00. The limiting proportions of these are respectively 1 - phi/2, 1/2, and (phi - 1)/2. The mean runlength is (phi + 1)/2. - Clark Kimberling, Dec 26 2010
From Amiram Eldar, Mar 10 2021: (Start)
a(n) is the number of the trailing 0's in the dual Zeckendorf representation of (n+1) (A104326).
The asymptotic density of the occurrences of k (0 or 1) is 1/phi^(k+1), where phi is the golden ratio (A001622).
The asymptotic mean of this sequence is 1/phi^2 (A132338). (End)

Examples

			The word is 010010100100101001010010010100...
Over the alphabet {a,b} this is a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, ...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • Jean Berstel, Fibonacci words—a survey, In The book of L, pp. 13-27. Springer Berlin Heidelberg, 1986.
  • J. C. Lagarias, Number Theory and Dynamical Systems, pp. 35-72 of S. A. Burr, ed., The Unreasonable Effectiveness of Number Theory, Proc. Sympos. Appl. Math., 46 (1992). Amer. Math. Soc. - see p. 64.
  • Wolfdieter Lang, The Wythoff and the Zeckendorf representations of numbers are equivalent, in G. E. Bergum et al. (edts.) Application of Fibonacci numbers vol. 6, Kluwer, Dordrecht, 1996, pp. 319-337. [See A317208 for a link.]
  • G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

There are several versions of this sequence in the OEIS. This one and A003842 are probably the most important. See also A008352, A076662, A288581, A288582.
Positions of 1's gives A003622.
Sequences mentioned in the Allouche et al. "Taxonomy" paper, listed by example number: 1: A003849, 2: A010060, 3: A010056, 4: A020985 and A020987, 5: A191818, 6: A316340 and A273129, 18: A316341, 19: A030302, 20: A063438, 21: A316342, 22: A316343, 23: A003849 minus its first term, 24: A316344, 25: A316345 and A316824, 26: A020985 and A020987, 27: A316825, 28: A159689, 29: A049320, 30: A003849, 31: A316826, 32: A316827, 33: A316828, 34: A316344, 35: A043529, 36: A316829, 37: A010060.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a003849 n = a003849_list !! n
    a003849_list = tail $ concat fws where
       fws = [1] : [0] : (zipWith (++) fws $ tail fws)
    -- Reinhard Zumkeller, Nov 01 2013, Apr 07 2012
    
  • Magma
    t1:=[ n le 2 select ["0","0,1"][n] else Self(n-1) cat "," cat Self(n-2) : n in [1..12]]; t1[12];
    
  • Maple
    z := proc(m) option remember; if m=0 then [0] elif m=1 then [0,1] else [op(z(m-1)),op(z(m-2))]; fi; end; z(12);
    M:=19; S[0]:=`0`; S[1]:=`01`; for n from 2 to M do S[n]:=cat(S[n-1], S[n-2]); od:
    t0:=S[M]: l:=length(t0); for i from 1 to l do lprint(i-1,substring(t0,i..i)); od: # N. J. A. Sloane, Nov 01 2006
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* Robert G. Wilson v, Mar 05 2005 *)
    Flatten[Nest[{#, #[[1]]} &, {0, 1}, 9]] (* IWABUCHI Yu(u)ki, Oct 23 2013 *)
    Table[Floor[(n + 2) #] - Floor[(n + 1) #], {n, 0, 120}] &[2 - GoldenRatio] (* Michael De Vlieger, Aug 15 2016 *)
    SubstitutionSystem[{0->{0,1},1->{0}},{0},{10}][[1]] (* Harvey P. Dale, Dec 20 2021 *)
  • PARI
    a(n)=my(k=2);while(fibonacci(k)<=n,k++);while(n>1,while(fibonacci(k--)>n,); n-=fibonacci(k)); n==1 \\ Charles R Greathouse IV, Feb 03 2014
    
  • PARI
    M3849=[2,2,1,0]/*L(k),S(k),L(k-1),S(k-1)*/; A003849(n)={while(n>M3849[1],M3849=vecextract(M3849,[1,2,1,2])+[M3849[3],M3849[4]<M. F. Hasler, Apr 07 2021
    
  • Python
    def fib(n):
        """Return the concatenation of A003849(0..F-1) where F is the smallest
           Fibonacci number > n, so that the result contains a(n) at index n."""
        a, b = '10'
        while len(b)<=n:
            a, b = b, b + a
        return b # Robert FERREOL, Apr 15 2016, edited by M. F. Hasler, Apr 07 2021
    
  • Python
    from math import isqrt
    def A003849(n): return 2-(n+2+isqrt(m:=5*(n+2)**2)>>1)+(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 25 2022

Formula

a(n) = floor((n+2)*r) - floor((n+1)*r) where r=phi/(1+2*phi) and phi is the Golden Ratio. - Benoit Cloitre, Nov 10 2003
a(n) = A003714(n) mod 2 = A014417(n) mod 2. - Philippe Deléham, Jan 04 2004
The first formula by Cloitre is just one of an infinite family of formulas. Using phi^2=1+phi, it follows that r=phi/(1+2*phi)=2-phi. Then from floor(-x)=-floor(x)-1 for non-integer x, it follows that a(n)=2-A014675(n)=2-(floor((n+2)* phi)-floor((n+1)*phi)). - Michel Dekking, Aug 27 2016
a(n) = 1 - A096270(n+1), i.e., A096270 is the complement of this sequence. - A.H.M. Smeets, Mar 31 2024

Extensions

Revised by N. J. A. Sloane, Jul 03 2012

A003842 The infinite Fibonacci word: start with 1, repeatedly apply the morphism 1->12, 2->1, take limit; or, start with S(0)=2, S(1)=1, and for n>1 define S(n)=S(n-1)S(n-2), then the sequence is S(oo).

Original entry on oeis.org

1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1
Offset: 0

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Author

Keywords

Comments

Or, fixed point of the morphism 1->12, 2->1, starting from a(1) = 2.
A Sturmian word, as are all versions of this sequence. This means that if one slides a window of length n along the sequence, one sees exactly n+1 different subwords (see A213975). For a proof, see for example Chap. 2 of Lothaire (2002).
The limiting mean of the first n terms is 3 - phi, where phi is the golden ratio (A001622); the limiting variance is 2 - phi. - Clark Kimberling, Mar 12 2014
The Wikipedia article on L-system Example 1 is "Algae" given by the axiom: A and rules: A -> AB, B -> A. The sequence G(n) = G(n-1)G(n-2) yields this sequence when A -> 1, B -> 2. - Michael Somos, Jan 12 2015
In the limit #1's : #2's = phi : 1. - Frank M Jackson, Mar 12 2018

Examples

			Over the alphabet {a,b} this is the sequence a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, ...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • Jean Berstel, "Fibonacci words—a survey." In The book of L, pp. 13-27. Springer Berlin Heidelberg, 1986.
  • J. Berstel and J. Karhumaki, Combinatorics on words - a tutorial, Bull. EATCS, #79 (2003), pp. 178-228.
  • E. Bombieri and J. Taylor, Which distribution of matter diffracts? An initial investigation, in International Workshop on Aperiodic Crystals (Les Houches, 1986), J. de Physique, Colloq. C3, 47 (1986), C3-19 to C3-28.
  • Aldo de Luca and Stefano Varricchio, Finiteness and regularity in semigroups and formal languages. Monographs in Theoretical Computer Science. An EATCS Series. Springer-Verlag, Berlin, 1999. x+240 pp. ISBN: 3-540-63771-0 MR1696498 (2000g:68001). See p. 25.
  • J. C. Lagarias, Number Theory and Dynamical Systems, pp. 35-72 of S. A. Burr, ed., The Unreasonable Effectiveness of Number Theory, Proc. Sympos. Appl. Math., 46 (1992). Amer. Math. Soc. - see p. 64.
  • G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.

Crossrefs

A003849 is another common version of this sequence.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a003842 n = a003842_list !! n
    a003842_list = tail $ concat fws where
       fws = [2] : [1] : (zipWith (++) fws $ tail fws)
    -- Reinhard Zumkeller, Oct 26 2013
    
  • Mathematica
    Nest[ Flatten[ # /. {1 -> {1, 2}, 2 -> {1}}] &, {1}, 10] (* Robert G. Wilson v, Mar 04 2005 *)
    Table[n + 1 - Floor[((1 + Sqrt[5])/2)*Floor[2*(n + 1)/(1 + Sqrt[5])]], {n, 1, 50}] (* G. C. Greubel, May 18 2017 *)
    SubstitutionSystem[{1->{1,2},2->{1}},{1},{10}][[1]] (* Harvey P. Dale, Nov 19 2022 *)
  • PARI
    for(n=1,50, print1(n+1 - floor(((1+sqrt(5))/2)*floor(2*(n+1)/(1+sqrt(5)))), ", ")) \\ G. C. Greubel, May 18 2017
    
  • Python
    def A003842(length):
        a = [1]
        while len(a)Nicholas Stefan Georgescu, Jun 14 2022
    
  • Python
    def aupto(nn):
        S, Fnm2, Fnm1 = [1, 2], 1, 2
        while len(S) < nn+1:
            S += S[:min(Fnm2, nn+1-len(S))]
            Fnm2, Fnm1 = Fnm1, Fnm1+Fnm2
        return S
    print(aupto(104)) # Michael S. Branicky, Jun 06 2022
    
  • Python
    from math import isqrt
    def A003842(n): return n+2-((m:=(n+2+isqrt(5*(n+2)**2)>>1)-n-2)+isqrt(5*m**2)>>1) # Chai Wah Wu, Aug 26 2022

Formula

Define strings S(0)=2, S(1)=1, S(n)=S(n-1)S(n-2); iterate. Sequence is S(infinity).
a(n) = n + 2 - A120613(n+1). - Benoit Cloitre, Jul 28 2005 [Corrected by N. J. A. Sloane, Jun 30 2018]

Extensions

Entry revised by N. J. A. Sloane, Jul 03 2012

A044432 a(n) is the number whose base-2 representation is d(0)d(1)...d(n), where d=A005614 (the infinite Fibonacci word).

Original entry on oeis.org

1, 2, 5, 11, 22, 45, 90, 181, 363, 726, 1453, 2907, 5814, 11629, 23258, 46517, 93035, 186070, 372141, 744282, 1488565, 2977131, 5954262, 11908525, 23817051, 47634102, 95268205, 190536410, 381072821, 762145643, 1524291286, 3048582573, 6097165147, 12194330294
Offset: 0

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Author

Keywords

Comments

a(n) can also be calculated as floor(2^n * R), where the rabbit constant R=0.709803442861291314641787399444575597012... converges rapidly using the result from Davison described in the comments at A014565. - Federico Provvedi, Oct 24 2018

Crossrefs

Programs

  • Haskell
    a044432 n = a044432_list !! n
    a044432_list = scanl1 (\v b -> 2 * v + b) a005614_list
    -- Reinhard Zumkeller, Apr 07 2012
  • Mathematica
    FromDigits[(Floor[GoldenRatio(#+1)]-Floor[GoldenRatio #]-1)&@Range@#,2]&/@Range@40 (* Federico Provvedi, Oct 19 2018 *)
    Floor[2^#/FromContinuedFraction[2^Fibonacci[Range[0,3*Max[1,Floor[2+Log[(#+1)/11]/ArcSinh[2]]]]]]]&/@Range[200] (* Federico Provvedi, Nov 01 2018 *)

Formula

a(n) = A000225(n+1) - A182028(n). - Reinhard Zumkeller, Apr 07 2012
a(n) = 2*a(n-1) + A005614(n) for n > 0, a(0) = 1. - Reinhard Zumkeller, Apr 07 2012
From Federico Provvedi, Oct 24 2018: (Start)
a(n) = A000079(n) * Sum_{k=0..n} ((floor(phi*(k+1)) - floor(phi*k) - 1)/2^k).
a(n) = floor(2^n*(1-Sum_{n >= 1} (-1)^(n+1)*(1+2^Fibonacci(3*n+1))/((2^(Fibonacci(3*n-1))-1)*(2^(Fibonacci(3*n + 2))-1)))).
a(n) = floor(2^n*R), where R is the rabbit constant.
a(n) = floor(2^n/[1, 2, 2, 4, 8, 32, ..., 2^Fibonacci(3*h)]), with h=1 for n=0, h=floor(2+log((n+1)/11)/arcsinh(2)) for n>0. (End)

Extensions

Offset fixed by Reinhard Zumkeller, Apr 07 2012

A214318 Replace the word A214317(n) with its position in A007931.

Original entry on oeis.org

1, 4, 9, 19, 40, 81, 164, 329, 659, 1320, 2641, 5283, 10568, 21137, 42276, 84553, 169107, 338216, 676433, 1352868, 2705737, 5411475, 10822952, 21645905, 43291811, 86583624, 173167249, 346334500, 692669001, 1385338003, 2770676008, 5541352017, 11082704035
Offset: 1

Views

Author

N. J. A. Sloane, Jul 12 2012

Keywords

Examples

			A214317(5) = 12112 is the 40th term of A007931, so a(5)=40.
		

Crossrefs

Programs

  • Maple
    S:= proc(n) option remember;
          `if`(n<2, [2-n], [S(n-1)[], S(n-2)[]])
        end:
    a:= proc(n) option remember; local k;
          for k while nops(S(k))Alois P. Heinz, Jul 19 2012
  • Mathematica
    nesting = 6; A003849 = Flatten[Nest[{#, #[[1]]}&, {0, 1}, nesting]]; A182028[n_] := FromDigits[Take[A003849, n+1], 2]; a[n_] := A182028[n-1] + 2^n - 1; Table[a[n], {n, 1, Length[A003849]}] (* Jean-François Alcover, Feb 13 2016 *)

Formula

a(n) = A182028(n-1)+2^n-1.
a(n) = 2*a(n-1) + A003842(n-1) for n>1, a(1) = 1. - Alois P. Heinz, Jul 19 2012

Extensions

More terms from Alois P. Heinz, Jul 19 2012

A363070 Take the terms 0..n of the infinite Fibonacci word A003849, regard them as a number in Fibonacci base.

Original entry on oeis.org

0, 1, 2, 3, 6, 10, 17, 28, 45, 74, 120, 194, 315, 510, 826, 1337, 2163, 3501, 5665, 9167, 14833, 24000, 38834, 62835, 101669, 164505, 266175, 430681, 696857, 1127538, 1824396, 2951935, 4776331, 7728267, 12504599, 20232867, 32737467, 52970334, 85707802, 138678137, 224385940, 363064078
Offset: 0

Views

Author

Gandhar Joshi, May 16 2023

Keywords

Examples

			0 ->                            0 -> a(0) = 0,
0,1 ->                         01 -> a(1) = 1,
0,1,0 ->                      010 -> a(2) = 2,
0,1,0,0 ->                   0100 -> a(3) = 3,
0,1,0,0,1 ->                01001 -> a(4) = 6,
0,1,0,0,1,0 ->             010010 -> a(5) = 10,
0,1,0,0,1,0,1 ->          0100101 -> a(6) = 17,
0,1,0,0,1,0,1,0 ->       01001010 -> a(7) = 28,
0,1,0,0,1,0,1,0,0 ->    010010100 -> a(8) = 45,
0,1,0,0,1,0,1,0,0,1 -> 0100101001 -> a(9) = 74.
		

Crossrefs

Cf. A003849 (Fibonacci word), A005713, A189920 (Zeckendorf digits).
Cf. A182028.

Programs

  • Python
    # see linked program
    
  • Python
    def aupto(n): # produces n terms, indices 0..n-1
        F1, F, a = [0], [0, 1], [0, 1]
        while len(F) < n:
            F1, F = F, F+F1
        [a.append(a[-2]+a[-1]+F[i]+F[i-1]) for i in range(2, n)]
        return a
    print(aupto(42)) # Michael S. Branicky, May 17 2023

Formula

a(n) = Sum_{i=0..n} A003849(i)*Fibonacci(n-i+2).
a(n) = a(n-1) + a(n-2) + A003849(n) + A003849(n-1).
a(n) = a(n-1) + a(n-2) + A005713(n-1). - Kevin Ryde, May 20 2023
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