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 1-10 of 28 results. Next

A127962 Binary expansion of A000979(n).

Original entry on oeis.org

11, 1011, 101011, 1010101011, 101010101011, 1010101010101011, 101010101010101011, 1010101010101010101011, 101010101010101010101010101011, 101010101010101010101010101010101010101011
Offset: 1

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Links

Crossrefs

Cf. A000978, A000979, A007088, A124400, A126614, A127936, A127955, A127956, A127957, A127958, A127961.

Programs

  • Mathematica
    b[n_] := FromDigits[IntegerDigits[n, 2]]; b /@ Select[Table[(2^p + 1)/3, {p, Prime[Range[15]]}], PrimeQ] (* Amiram Eldar, Jul 23 2023 *)
  • Python
    from gmpy2 import divexact
from sympy import prime, isprime
A127962 = [int(bin(p)[2:]) for p in (divexact(2**prime(n)+1,3) for n in range(2,10**2)) if isprime(p)] # Chai Wah Wu, Sep 04 2014

Formula

a(n) = A007088(A000979(n)). - Amiram Eldar, Jul 23 2023

Extensions

Edited by N. J. A. Sloane, Jun 11 2007

A194810 Indices k such that A139250(k) = A000979(n).

Original entry on oeis.org

2, 4, 8, 32, 64, 256, 512, 2048, 32768, 2097152, 1073741824, 549755813888, 1125899906842624, 9223372036854775808, 9671406556917033397649408, 39614081257132168796771975168, 633825300114114700748351602688
Offset: 1

Views

Author

Omar E. Pol, Oct 23 2011

Keywords

Comments

Indices k such that the number of toothpicks in the toothpick structure of A139250 after k-th stage equals the n-th Wagstaff prime A000979. All terms of this sequence are powers of 2 (see formulas).
For a picture of the n-th Wagstaff prime as a toothpick structure see the Applegate link "A139250: the movie version", then enter N = a(n) and click "Update", for N = a(n) <= 32768 (due to the resolution of the movie).

Examples

			For n = 5 we have that a(5) = 64, then we can see that the number of toothpicks in the toothpick structure of A139250 after 64 stages is 2731 which coincides with the fifth Wagstaff prime, so we can write A139250(64) = A000979(5) = 2731. See the illustration in the Applegate-Pol-Sloane paper, figure 3: T(64) = 2731 toothpicks.

Links

Crossrefs

Cf. A000978, A000979, A007583, A001045, A127936, A127958, A127962, A127965, A139250, A139253.

Programs

  • Mathematica
    2^Reap[Do[If[PrimeQ[1+Sum[2^(2n-1), {n, m}]], Sow[m]], {m, 100}]][[2, 1]] (* Jean-François Alcover, Oct 06 2018 *)

Formula

a(n) = 2^A127936(n) = 2^(floor(A000978(n)/2)) = 2^(ceiling(log_4(A000979(n)))).
A139250(2^n) = A007583(n), n >= 0.
A139250(a(n)) = A000979(n).

Extensions

More terms from Omar E. Pol, Mar 14 2012

A001045 Jacobsthal sequence (or Jacobsthal numbers): a(n) = a(n-1) + 2*a(n-2), with a(0) = 0, a(1) = 1; also a(n) = nearest integer to 2^n/3.

Original entry on oeis.org

0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, 5726623061, 11453246123
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Don Knuth points out (personal communication) that Jacobsthal may never have seen the actual values of this sequence. However, Horadam uses the name "Jacobsthal sequence", such an important sequence needs a name, and there is a law that says the name for something should never be that of its discoverer. - N. J. A. Sloane, Dec 26 2020
Number of ways to tile a 3 X (n-1) rectangle with 1 X 1 and 2 X 2 square tiles.
Also, number of ways to tile a 2 X (n-1) rectangle with 1 X 2 dominoes and 2 X 2 squares. - Toby Gottfried, Nov 02 2008
Also a(n) counts each of the following four things: n-ary quasigroups of order 3 with automorphism group of order 3, n-ary quasigroups of order 3 with automorphism group of order 6, (n-1)-ary quasigroups of order 3 with automorphism group of order 2 and (n-2)-ary quasigroups of order 3. See the McKay-Wanless (2008) paper. - Ian Wanless, Apr 28 2008
Also the number of ways to tie a necktie using n + 2 turns. So three turns make an "oriental", four make a "four in hand" and for 5 turns there are 3 methods: "Kelvin", "Nicky" and "Pratt". The formula also arises from a special random walk on a triangular grid with side conditions (see Fink and Mao, 1999). - arne.ring(AT)epost.de, Mar 18 2001
Also the number of compositions of n + 1 ending with an odd part (a(2) = 3 because 3, 21, 111 are the only compositions of 3 ending with an odd part). Also the number of compositions of n + 2 ending with an even part (a(2) = 3 because 4, 22, 112 are the only compositions of 4 ending with an even part). - Emeric Deutsch, May 08 2001
Arises in study of sorting by merge insertions and in analysis of a method for computing GCDs - see Knuth reference.
Number of perfect matchings of a 2 X n grid upon replacing unit squares with tetrahedra (C_4 to K_4):
o----o----o----o...
| \/ | \/ | \/ |
| /\ | /\ | /\ |
o----o----o----o... - Roberto E. Martinez II, Jan 07 2002
Also the numerators of the reduced fractions in the alternating sum 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 + ... - Joshua Zucker, Feb 07 2002
Also, if A(n), B(n), C(n) are the angles of the n-orthic triangle of ABC then A(1) = Pi - 2*A, A(n) = s(n)*Pi + (-2)^n*A where s(n) = (-1)^(n-1) * a(n) [1-orthic triangle = the orthic triangle of ABC, n-orthic triangle = the orthic triangle of the (n-1)-orthic triangle]. - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), Jun 05 2002
Also the number of words of length n+1 in the two letters s and t that reduce to the identity 1 by using the relations sss = 1, tt = 1 and stst = 1. The generators s and t and the three stated relations generate the group S3. - John W. Layman, Jun 14 2002
Sums of pairs of consecutive terms give all powers of 2 in increasing order. - Amarnath Murthy, Aug 15 2002
Excess clockwise moves (over counterclockwise) needed to move a tower of size n to the clockwise peg is -(-1)^n*(2^n - (-1)^n)/3; a(n) is its unsigned version. - Wouter Meeussen, Sep 01 2002
Also the absolute value of the number represented in base -2 by the string of n 1's, the negabinary repunit. The Mersenne numbers (A000225 and its subsequences) are the binary repunits. - Rick L. Shepherd, Sep 16 2002
Note that 3*a(n) + (-1)^n = 2^n is significant for Pascal's triangle A007318. It arises from a Jacobsthal decomposition of Pascal's triangle illustrated by 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = (7 + 35 + 1) + (1 + 35 + 7) + (21 + 21) = 43 + 43 + 42 = 3a(7) - 1; 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = (1 + 56 + 28) + (28 + 56 + 1) + (8 + 70 + 8) = 85 + 85 + 86 = 3a(8)+1. - Paul Barry, Feb 20 2003
Number of positive integers requiring exactly n signed bits in the nonadjacent form representation.
Equivalently, number of length-(n-1) words with letters {0, 1, 2} where no two consecutive letters are nonzero, see example and fxtbook link. - Joerg Arndt, Nov 10 2012
Counts walks between adjacent vertices of a triangle. - Paul Barry, Nov 17 2003
Every amphichiral rational knot written in Conway notation is a palindromic sequence of numbers, not beginning or ending with 1. For example, for 4 <= n <= 12, the amphichiral rational knots are: 2 2, 2 1 1 2, 4 4, 3 1 1 3, 2 2 2 2, 4 1 1 4, 3 1 1 1 1 3, 2 3 3 2, 2 1 2 2 1 2, 2 1 1 1 1 1 1 2, 6 6, 5 1 1 5, 4 2 2 4, 3 3 3 3, 2 4 4 2, 3 2 1 1 2 3, 3 1 2 2 1 3, 2 2 2 2 2 2, 2 2 1 1 1 1 2 2, 2 1 2 1 1 2 1 2, 2 1 1 1 1 1 1 1 1 2. For the number of amphichiral rational knots for n=2*k (k=1, 2, 3, ...), we obtain the sequence 0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, ... - Slavik Jablan, Dec 26 2003
a(n+2) counts the binary sequences of total length n made up of codewords from C = {0, 10, 11}. - Paul Barry, Jan 23 2004
Number of permutations with no fixed points avoiding 231 and 132.
The n-th entry (n > 1) of the sequence is equal to the 2,2-entry of the n-th power of the unnormalized 4 X 4 Haar matrix: [1 1 1 0 / 1 1 -1 0 / 1 1 0 1 / 1 1 0 -1]. - Simone Severini, Oct 27 2004
a(n) is the number of Motzkin (n+1)-sequences whose flatsteps all occur at level 1 and whose height is less than or equal to 2. For example, a(4) = 5 counts UDUFD, UFDUD, UFFFD, UFUDD, UUDFD. - David Callan, Dec 09 2004
a(n+1) gives row sums of A059260. - Paul Barry, Jan 26 2005
If (m + n) is odd, then 3*(a(m) + a(n)) is always of the form a^2 + 2*b^2, where a and b both equal powers of 2; consequently every factor of (a(m) + a(n)) is always of the form a^2 + 2*b^2. - Matthew Vandermast, Jul 12 2003
Number of "0,0" in f_{n+1}, where f_0 = "1" and f_{n+1} = a sequence formed by changing all "1"s in f_n to "1,0" and all "0"s in f_n to "0,1". - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Sep 22 2006
All prime Jacobsthal numbers A049883[n] = {3, 5, 11, 43, 683, 2731, 43691, ...} have prime indices except for a(4) = 5. All prime Jacobsthal numbers with prime indices (all but a(4) = 5) are of the form (2^p + 1)/3 - the Wagstaff primes A000979[n]. Indices of prime Jacobsthal numbers are listed in A107036[n] = {3, 4, 5, 7, 11, 13, 17, 19, 23, 31, 43, 61, ...}. For n>1 A107036[n] = A000978[n] Numbers n such that (2^n + 1)/3 is prime. - Alexander Adamchuk, Oct 03 2006
Correspondence: a(n) = b(n)*2^(n-1), where b(n) is the sequence of the arithmetic means of previous two terms defined by b(n) = 1/2*(b(n-1) + b(n-2)) with initial values b(0) = 0, b(1) = 1; the g.f. for b(n) is B(x) := x/(1-(x^1+x^2)/2), so the g.f. A(x) for a(n) satisfies A(x) = B(2*x)/2. Because b(n) converges to the limit lim (1-x)*B(x) = 1/3*(b(0) + 2*b(1)) = 2/3 (for x --> 1), it follows that a(n)/2^(n-1) also converges to 2/3 (see also A103770). - Hieronymus Fischer, Feb 04 2006
Inverse: floor(log_2(a(n))) = n - 2 for n >= 2. Also: log_2(a(n) + a(n-1)) = n - 1 for n >= 1 (see also A130249). Characterization: x is a Jacobsthal number if and only if there is a power of 4 (= c) such that x is a root of p(x) = 9*x*(x-c) + (c-1)*(2*c+1) (see also the indicator sequence A105348). - Hieronymus Fischer, May 17 2007
This sequence counts the odd coefficients in the expansion of (1 + x + x^2)^(2^n - 1), n >= 0. - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Oct 18 2007, Jan 08 2008
2^(n+1) = 2*A005578(n) + 2*a(n) + 2*A000975(n-1). Let A005578(n), a(n), A000975(n-1) = triangle (a, b, c). Then ((S-c), (S-b), (S-a)) = (A005578(n-1), a(n-1), A000975(n-2)). Example: (a, b, c) = (11, 11, 10) = (A005578(5), a(5), A000975(4)). Then ((S-c), (S-b), (S-a)) = (6, 5, 5) = (A005578(4), a(4), A000975(3)). - Gary W. Adamson, Dec 24 2007
Sequence is identical to the absolute values of its inverse binomial transform. A similar result holds for [0,A001045*2^n]. - Paul Curtz, Jan 17 2008
From a(2) on (i.e., 1, 3, 5, 11, 21, ...) also: least odd number such that the subsets of {a(2), ..., a(n)} sum to 2^(n-1) different values, cf. A138000 and A064934. It is interesting to note the pattern of numbers occurring (or not occurring) as such a sum (A003158). - M. F. Hasler, Apr 09 2008
a(n) is the term (5, 1) of n-th power of the 5 X 5 matrix shown in A121231. - Gary W. Adamson, Oct 03 2008
A147612(a(n)) = 1. - Reinhard Zumkeller, Nov 08 2008
a(n+1) = Sum(A153778(i): 2^n <= i < 2^(n+1)). - Reinhard Zumkeller, Jan 01 2009
It appears that a(n) is also the number of integers between 2^n and 2^(n+1) that are divisible by 3 with no remainder. - John Fossaceca (john(AT)fossaceca.net), Jan 31 2009
Number of pairs of consecutive odious (or evil) numbers between 2^(n+1) and 2^(n+2), inclusive. - T. D. Noe, Feb 05 2009
Equals eigensequence of triangle A156319. - Gary W. Adamson, Feb 07 2009
A three-dimensional interpretation of a(n+1) is that it gives the number of ways of filling a 2 X 2 X n hole with 1 X 2 X 2 bricks. - Martin Griffiths, Mar 28 2009
Starting with offset 1 = INVERTi transform of A002605: (1, 2, 6, 16, 44, ...). - Gary W. Adamson, May 12 2009
Convolved with (1, 2, 2, 2, ...) = A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 23 2009
The product of a pair of successive terms is always a triangular number. - Giuseppe Ottonello, Jun 14 2009
Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := -2, A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = (-1)^(n-1)*det(A). - Milan Janjic, Jan 26 2010
Let R denote the irreducible representation of the symmetric group S_3 of dimension 2, and let s and t denote respectively the sign and trivial irreducible representations of dimension 1. The decomposition of R^n into irreducible representations consists of a(n) copies of R and a(n-1) copies of each of s and t. - Andrew Rupinski, Mar 12 2010
As a fraction: 1/88 = 0.0113636363... or 1/9898 = 0.00010103051121... - Mark Dols, May 18 2010
Starting with "1" = the INVERT transform of (1, 0, 2, 0, 4, 0, 8, ...); e.g., a(7) = 43 = (1, 1, 1, 3, 5, 11, 21) dot (8, 0, 4, 0, 2, 0, 1) = (8 + 4 + 10 + 21) = 43. - Gary W. Adamson, Oct 28 2010
Rule 28 elementary cellular automaton (A266508) generates this sequence. - Paul Muljadi, Jan 27 2011
This is a divisibility sequence. - Michael Somos, Feb 06 2011
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(6,2) =
(0 0 1)
(0 2 0)
(2 0 1).
Then a(n+1) = (Trace(U^n))/3, a(n+1) = ((U^n){3, 3})/3, a(n) = ((U^n){1, 3})/3 and a(n) = ((U^(n+1))_{1, 1})/2. (End)
The sequence emerges in using iterated deletion of strictly dominated strategies to establish the best-response solution to the Cournot duopoly problem as a strictly dominant strategy. The best response of firm 1 to firm 2's chosen quantity is given by q*1 = 1/2*(a - c - q_2), where a is reservation price, c is marginal cost, and q_2 is firm two's chosen quantity. Given that q_2 is in [o, a - c], q*_1 must be in [o, 1/2*(a - c)]. Since costs are symmetric, we know q_2 is in [0, 1/2*(a - c)]. Then we know q*_1 is in [1/4*(a - c), 1/2*(a - c)]. Continuing in this way, the sequence of bounds we get (factoring out a - c) is {1/2, 1/4, 3/8, 5/16, ...}; the numerators are the Jacobsthal numbers. - _Michael Chirico, Sep 10 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 3*a(n-1) equals the number of 3-colored compositions of n with all parts greater than or equal to 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
This sequence is connected with the Collatz problem. We consider the array T(i, j) where the i-th row gives the parity trajectory of i, for example for i = 6, the infinite trajectory is 6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1... and T(6, j) = [0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, ..., 1, 0, 0, 1, ...]. Now, we consider the sum of the digits "1" of each column. We obtain the sequence a(n) = Sum_{k = 1..2^n} T(k, n) = Sum {k = 1..2^n} digits "1" of the n-th column. Because a(n) + a(n+1) = 2^n, then a(n+1) = Number of digits "0" among the 2^n elements of the n-th column. - _Michel Lagneau, Jan 11 2012
3!*a(n-1) is apparently the trace of the n-th power of the adjacency matrix of the complete 3-graph, a 3 X 3 matrix with diagonal elements all zero and off-diagonal all ones. The off-diagonal elements for the n-th power are all equal to a(n) while each diagonal element seems to be a(n) + 1 for an even power and a(n) - 1 for an odd. These are related to the lengths of closed paths on the graph (see Delfino and Viti's paper). - Tom Copeland, Nov 06 2012
From Paul Curtz, Dec 11 2012: (Start)
2^n * a(-n) = (-1)^(n-1) * a(n), which extends the sequence to negative indices: ..., -5/16, 3/8, -1/4, 1/2, 0, 1, 1, 3, 5, ...
The "autosequence" property with respect to the binomial transform mentioned in my comment of Jan 17 2008 is still valid if the term a(-1) is added to the array of the sequence and its iterated higher-order differences in subsequent rows:
0 1/2 1/2 3/2 5/2 11/2 ...
1/2 0 1 1 3 5 ...
-1/2 1 0 2 2 6 ...
3/2 -1 2 0 4 4 ...
-5/2 3 -2 4 0 8 ...
11/2 -5 6 -4 8 0 ...
The main diagonal in this array contains 0's. (End)
Assign to a triangle T(n, 0) = 1 and T(n+1, 1) = n; T(r, c) = T(r-1, c-1) + T(r-1, c-2) + T(r-2, c-2). Then T(n+1, n) - T(n, n) = a(n). - J. M. Bergot, May 02 2013
a(n+1) counts clockwise walks on n points on a circle that take steps of length 1 and 2, return to the starting point after two full circuits, and do not duplicate any steps (USAMO 2013, problem 5). - Kiran S. Kedlaya, May 11 2013
Define an infinite square array m by m(n, 0) = m(0, n) = a(n) in top row and left column and m(i, j) = m(i, j-1) + m(i-1, j-1) otherwise, then m(n+1, n+1) = 3^(n-1). - J. M. Bergot, May 10 2013
a(n) is the number of compositions (ordered partitions) of n - 1 into one sort of 1's and two sorts of 2's. Example: the a(4) = 5 compositions of 3 are 1 + 1 + 1, 1 + 2, 1 + 2', 2 + 1 and 2' + 1. - Bob Selcoe, Jun 24 2013
Without 0, a(n)/2^n equals the probability that n will occur as a partial sum in a randomly-generated infinite sequence of 1's and 2's. The limiting ratio is 2/3. - Bob Selcoe, Jul 04 2013
Number of conjugacy classes of Z/2Z X Z/2Z in GL(2,2^(n+1)). - Jared Warner, Aug 18 2013
a(n) is the top left entry of the (n-1)-st power of the 3 X 3 matrix [1, 1, 1, 1, 0, 0, 1, 0, 0]. a(n) is the top left entry of the (n+1)-st power of any of the six 3 X 3 matrices [0, 1, 0; 1, 1, 1; 0, 1, 0], [0, 1, 1; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 1, 1; 1, 1, 0], [0, 1, 1; 1, 0, 1; 0, 1, 1], [0, 0, 1; 0, 0, 1; 1, 1, 1] or [0, 1, 0; 1, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
This is the only integer sequence from the family of homogeneous linear recurrence of order 2 given by a(n) = k*a(n-1) + t*a(n-2) with positive integer coefficients k and t and initial values a(0) = 0 and a(1) = 1 whose ratio a(n+1)/a(n) converges to 2 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is the Lucas sequence U(1, -2). - Felix P. Muga II, Mar 21 2014
sqrt(a(n+1) * a(n-1)) -> a(n) + 3/4 if n is even, and -> a(n) - 3/4 if n is odd, for n >= 2. - Richard R. Forberg, Jun 24 2014
a(n+1) counts closed walks on the end vertices of P_3 containing one loop at the middle vertex. a(n-1) counts closed walks on the middle vertex of P_3 containing one loop at that vertex. - David Neil McGrath, Nov 07 2014
From César Eliud Lozada, Jan 21 2015: (Start)
Let P be a point in the plane of a triangle ABC (with sides a, b, c) and barycentric coordinates P = [x:y:z]. The Complement of P with respect to ABC is defined to be Complement(P) = [b*y + c*z : c*z + a*x : a*x + b*y].
Then, for n >= 1, Complement(Complement(...(Complement(P))..)) = (n times) =
[2*a(n-1)*a*x + (2*a(n-1) - (-1)^n)*(b*y + c*z):
2*a(n-1)*b*y + (2*a(n-1) - (-1)^n)*(c*z + a*x):
2*a(n-1)*c*z + (2*a(n-1) - (-1)^n)*(a*x + b*y)]. (End)
a(n) (n >= 2) is the number of induced hypercubes of the Fibonacci cube Gamma(n-2). See p. 513 of the Klavzar reference. Example: a(5) = 11. Indeed, the Fibonacci cube Gamma(3) is <>- (cycle C(4) with a pendant edge) and the hypercubes are: 5 vertices, 5 edges, and 1 square. - Emeric Deutsch, Apr 07 2016
If the sequence of points {P_i(x_i, y_i)} on the cubic y = a*x^3 + b*x^2 + c*x + d has the property that the segment P_i(x_i, y_i) P_i+1(x_i+1, y_i+1) is always tangent to the cubic at P_i+1(x_i+1, y_i+1) then a(n) = -2^n*a/b*(x_(n+1)-(-1/2)^n*x_1). - Michael Brozinsky, Aug 01 2016
With the quantum integers defined by [n+1]A000225%20are%20given%20by%20q%20=%20sqrt(2).%20Cf.%20A239473.%20-%20_Tom%20Copeland">q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)), the Jacobsthal numbers are a(n+1) = (-1)^n*q^n [n+1]_q with q = i * sqrt(2) for i^2 = -1, whereas the signed Mersenne numbers A000225 are given by q = sqrt(2). Cf. A239473. - _Tom Copeland, Sep 05 2016
Every positive integer has a unique expression as a sum of Jacobsthal numbers in which the index of the smallest summand is odd, with a(1) and a(2) both allowed. See the L. Carlitz, R. Scoville, and V. E. Hoggatt, Jr. reference. - Ira M. Gessel, Dec 31 2016. See A280049 for these expansions. - N. J. A. Sloane, Dec 31 2016
For n > 0, a(n) equals the number of ternary words of length n-1 in which 0 and 1 avoid runs of odd lengths. - Milan Janjic, Jan 08 2017
For n > 0, a(n) equals the number of orbits of the finite group PSL(2,2^n) acting on subsets of size 4 for the 2^n+1 points of the projective line. - Paul M. Bradley, Jan 31 2017
For n > 1, number of words of length n-2 over alphabet {1,2,3} such that no odd letter is followed by an odd letter. - Armend Shabani, Feb 17 2017
Also, the decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017
Also the number of independent vertex sets and vertex covers in the 2 X (n-2) king graph. - Eric W. Weisstein, Sep 21 2017
From César Eliud Lozada, Dec 14 2017: (Start)
Let T(0) be a triangle and let T(1) be the medial triangle of T(0), T(2) the medial triangle of T(1) and, in general, T(n) the medial triangle of T(n-1). The barycentric coordinates of the first vertex of T(n) are [2*a(n-1)/a(n), 1, 1], for n > 0.
Let S(0) be a triangle and let S(1) be the antimedial triangle of S(0), S(2) the antimedial triangle of S(1) and, in general, S(n) the antimedial triangle of S(n-1). The barycentric coordinates of the first vertex of S(n) are [-a(n+1)/a(n), 1, 1], for n > 0. (End)
a(n) is also the number of derangements in S_{n+1} with empty peak set. - Isabella Huang, Apr 01 2018
For n > 0, gcd(a(n), a(n+1)) = 1. - Kengbo Lu, Jul 27 2020
Number of 2-compositions of n+1 with 1 not allowed as a part; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
The number of Hamiltonian paths of the flower snark graph of even order 2n > 2 is 12*a(n-1). - Don Knuth, Dec 25 2020
When set S = {1, 2, ..., 2^n}, n>=0, then the largest subset T of S with the property that if x is in T, then 2*x is not in T, has a(n+1) elements. For example, for n = 4, #S = 16, a(5) = 11 with T = {1, 3, 4, 5, 7, 9, 11, 12, 13, 15, 16} (see Hassan Tarfaoui link, Concours Général 1991). - Bernard Schott, Feb 14 2022
a(n) is the number of words of length n over a binary alphabet whose position in the lexicographic order is one more than a multiple of three. a(3) = 3: aaa, abb, bba. - Alois P. Heinz, Apr 13 2022
Named by Horadam (1988) after the German mathematician Ernst Jacobsthal (1882-1965). - Amiram Eldar, Oct 02 2023
Define the sequence u(n) = (u(n-1) + u(n-2))/u(n-3) with u(0) = 0, u(1) = 1, u(2) = u(3) = -1. Then u(4*n) = -1 + (-1)^n/a(n+1), u(4*n+1) = 2 - (-1)^n/a(n+1), u(4*n+2) = u(4*n+3) = -1. For example, a(3) = 3 and u(8) = -2/3, u(9) = 5/3, u(10) = u(11) = -1. - Michael Somos, Oct 24 2023
From Miquel A. Fiol, May 25 2024: (Start)
Also, a(n) is the number of (3-color) states of a cycle (n+1)-pole C_{n+1} with n+1 terminals (or semiedges).
For instance, for n=3, the a(3)=3 states (3-coloring of the terminals) of C_4 are
a a a a a b
a a b b a b (End)
Also, with offset 1, the cogrowth sequence of the 6-element dihedral group D3. - Sean A. Irvine, Nov 04 2024

Examples

			a(2) = 3 because the tiling of the 3 X 2 rectangle has either only 1 X 1 tiles, or one 2 X 2 tile in one of two positions (together with two 1 X 1 tiles).
From _Joerg Arndt_, Nov 10 2012: (Start)
The a(6)=21 length-5 ternary words with no two consecutive letters nonzero are (dots for 0's)
[ 1]   [ . . . . ]
[ 2]   [ . . . 1 ]
[ 3]   [ . . . 2 ]
[ 4]   [ . . 1 . ]
[ 5]   [ . . 2 . ]
[ 6]   [ . 1 . . ]
[ 7]   [ . 1 . 1 ]
[ 8]   [ . 1 . 2 ]
[ 9]   [ . 2 . . ]
[10]   [ . 2 . 1 ]
[11]   [ . 2 . 2 ]
[12]   [ 1 . . . ]
[13]   [ 1 . . 1 ]
[14]   [ 1 . . 2 ]
[15]   [ 1 . 1 . ]
[16]   [ 1 . 2 . ]
[17]   [ 2 . . . ]
[18]   [ 2 . . 1 ]
[19]   [ 2 . . 2 ]
[20]   [ 2 . 1 . ]
[21]   [ 2 . 2 . ]
(End)
G.f. = x + x^2 + 3*x^3 + 5*x^4 + 11*x^5 + 21*x^6 + 43*x^7 + 85*x^8 + 171*x^9 + ...

References

  • Jathan Austin and Lisa Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:1 (2020), 340-350.
  • Thomas Fink and Yong Mao, The 85 ways to tie a tie, Fourth Estate, London, 1999; Die 85 Methoden eine Krawatte zu binden. Hoffmann und Kampe, Hamburg, 1999.
  • International Mathematical Olympiad 2001, Hong Kong Preliminary Selection Contest Problem #16.
  • Jablan S. and Sazdanovic R., LinKnot: Knot Theory by Computer, World Scientific Press, 2007. See p. 80.
  • Ernst Erich Jacobsthal, Fibonaccische Polynome und Kreisteilungsgleichungen, Sitzungsber. Berliner Math. Gesell. 17 (1919-1920), 43-57.
  • Tanya Khovanova, "Coins and Logic", Chapter 6, The Mathematics of Various Entertaining Subjects: Volume 3 (2019), Jennifer Beineke & Jason Rosenhouse, eds. Princeton University Press, Princeton and Oxford, p. 73. ISBN: 0691182582, 978-0691182582.
  • Donald E. Knuth, Art of Computer Programming, Vol. 3, Sect. 5.3.1, Eq. 13.
  • Thomas Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.
  • Steven Roman, Introduction to Coding and Information Theory, Springer Verlag, 1996, 41-42.
  • P. D. Seymour and D. J. A. Welsh, Combinatorial applications of an inequality form statistical mechanics, Math. Proc. Camb. Phil. Soc. 77 (1975), 485-495. [Although Daykin et al. (1979) claim that the present sequence is studied in this article, it does not seem to be explicitly mentioned. Note that definition of log-convex in (3.1) is wrong. - N. J. A. Sloane, Dec 26 2020]
  • 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).
  • Robert M. Young, Excursions in Calculus, MAA, 1992, p. 239

Links

Crossrefs

Partial sums of this sequence give A000975, where there are additional comments from B. E. Williams and Bill Blewett on the tie problem.
A002487(a(n)) = A000045(n).
Row sums of A059260, A156667 and A134317. Equals A026644(n-2)+1 for n > 1.
a(n) = A073370(n-1, 0), n >= 1 (first column of triangle).
Cf. A266508 (binary), A081857 (base 4), A147612 (characteristic function).
Cf. A000978, A000979, A019322, A066845, A105348, A130249, A130250, A130253, A005578, A002083, A113405, A138000, A064934, A003158, A175286 (Pisano periods), A147613, A156319, A002605, A000225, A052129, A014551 (companion "autosequence"), A015266, A015287, A015305, A015323, A015338, A015356, A015371, A084175, A245489, A283641.
Cf. A049883 = primes in this sequence, A107036 = indices of primes, A129738.
Cf. A091084 (mod 10), A239473, A280049.
Bisections: A002450, A007583.
Cf. A077925 (signed version).

Programs

  • Haskell
    a001045 = (`div` 3) . (+ 1) . a000079
a001045_list = 0 : 1 :
   zipWith (+) (map (2 *) a001045_list) (tail a001045_list)
-- Reinhard Zumkeller, Mar 24 2013, Jan 05 2012, Feb 05 2011
  • Magma
    [n le 2 select n-1 else Self(n-1)+2*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jun 27 2016
  • Maple
    A001045 := proc(n)
  (2^n-(-1)^n)/3 ;
end proc: # R. J. Mathar, Dec 18 2012
  • Mathematica
    Jacob0[n_] := (2^n - (-1)^n)/3; Table[Jacob0[n], {n, 0, 33}] (* Robert G. Wilson v, Dec 05 2005 *)
Array[(2^# - (-1)^#)/3 &, 33, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
LinearRecurrence[{1, 2}, {0, 1}, 40] (* Harvey P. Dale, Nov 30 2011 *)
CoefficientList[Series[x/(1 - x - 2 x^2), {x, 0, 34}], x] (* Robert G. Wilson v, Jul 21 2015 *)
Table[(2^n - (-1)^n)/3, {n, 0, 20}] (* Eric W. Weisstein, Sep 21 2017 *)
Table[Abs[QBinomial[n, 1, -2]], {n, 0, 35}] (* John Keith, Jan 29 2022 *)
  • Maxima
    a[0]:0$
a[n]:=2^(n-1)-a[n-1]$
A001045(n):=a[n]$
makelist(A001045(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    a(n) = (2^n - (-1)^n) / 3
  • PARI
    M=[1,1,0;1,0,1;0,1,1];for(i=0,34,print1((M^i)[2,1],",")) \\ Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
  • PARI
    a=0; for(n=0,34,print1(a,", "); a=2*(a-n%2)+1) \\ K. Spage, Aug 22 2014
  • Python
    # A001045.py
def A001045():
    a, b = 0, 1
    while True:
        yield a
        a, b = b, b+2*a
sequence = A001045()
[next(sequence) for i in range(20)] # David Radcliffe, Jun 26 2016
  • Python
    [(2**n-(-1)**n)//3 for n in range(40)] # Gennady Eremin, Mar 03 2022
  • Sage
    [lucas_number1(n, 1, -2) for n in range(34)]  # Zerinvary Lajos, Apr 22 2009
# Alternatively:
a = BinaryRecurrenceSequence(1,2)
[a(n) for n in (0..34)] # Peter Luschny, Aug 29 2016

Formula

a(n) = 2^(n-1) - a(n-1). a(n) = 2*a(n-1) - (-1)^n = (2^n - (-1)^n)/3.
G.f.: x/(1 - x - 2*x^2) = x/((x+1)*(1-2*x)). Simon Plouffe in his 1992 dissertation
E.g.f.: (exp(2*x) - exp(-x))/3.
a(2*n) = 2*a(2*n-1)-1 for n >= 1, a(2*n+1) = 2*a(2*n)+1 for n >= 0. - Lee Hae-hwang, Oct 11 2002; corrected by Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002
Also a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x, y) = x*F(n-1)(x, y) + y*F(n-2)(x, y), with y=2*x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*3^(k-1). - Paul Barry, Apr 02 2003
The ratios a(n)/2^(n-1) converge to 2/3 and every fraction after 1/2 is the arithmetic mean of the two preceding fractions. - Gary W. Adamson, Jul 05 2003
a(n) = U(n-1, i/(2*sqrt(2)))*(-i*sqrt(2))^(n-1) with i^2=-1. - Paul Barry, Nov 17 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 2^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(2*n) = A002450(n) = (4^n - 1)/3; a(2*n+1) = A007583(n) = (2^(2*n+1) + 1)/3. - Philippe Deléham, Mar 27 2004
a(n) = round(2^n/3) = (2^n + (-1)^(n-1))/3 so lim_{n->infinity} 2^n/a(n) = 3. - Gerald McGarvey, Jul 21 2004
a(n) = Sum_{k=0..n-1} (-1)^k*2^(n-k-1) = Sum_{k=0..n-1}, 2^k*(-1)^(n-k-1). - Paul Barry, Jul 30 2004
a(n+1) = Sum_{k=0..n} binomial(k, n-k)*2^(n-k). - Paul Barry, Oct 07 2004
a(n) = Sum_{k=0..n-1} W(n-k, k)*(-1)^(n-k)*binomial(2*k,k), W(n, k) as in A004070. - Paul Barry, Dec 17 2004
From Paul Barry, Jan 17 2005: (Start)
a(n) = Sum_{k=0..n} k*binomial(n-1, (n-k)/2)*(1+(-1)^(n+k))*floor((2*k+1)/3).
a(n+1) = Sum_{k=0..n} k*binomial(n-1, (n-k)/2)*(1+(-1)^(n+k))*(A042965(k)+0^k). (End)
From Paul Barry, Jan 17 2005: (Start)
a(n+1) = ceiling(2^n/3) + floor(2^n/3) = (ceiling(2^n/3))^2 - (floor(2^n/3))^2.
a(n+1) = A005578(n) + A000975(n-1) = A005578(n)^2 - A000975(n-1)^2. (End)
a(n+1) = Sum_{k=0..n} Sum_{j=0..n} (-1)^(n-j)*binomial(j, k). - Paul Barry, Jan 26 2005
Let M = [1, 1, 0; 1, 0, 1; 0, 1, 1], then a(n) = (M^n)[2, 1], also matrix characteristic polynomial x^3 - 2*x^2 - x + 2 defines the three-step recursion a(0)=0, a(1)=1, a(2)=1, a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n > 2. - Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
a(n) = ceiling(2^(n+1)/3) - ceiling(2^n/3) = A005578(n+1) - A005578(n). - Paul Barry, Oct 08 2005
a(n) = floor(2^(n+1)/3) - floor(2^n/3) = A000975(n) - A000975(n-1). - Paul Barry, Oct 08 2005
From Paul Barry, Feb 20 2003: (Start)
a(n) = Sum_{k=0..floor(n/3)} binomial(n, f(n-1)+3*k);
a(n) = Sum_{k=0..floor(n/3)} binomial(n, f(n-2)+3*k), where f(n)=A080425(n). (End)
From Miklos Kristof, Mar 07 2007: (Start)
a(2*n) = (1/3)*Product_{d|n} cyclotomic(d,4).
a(2*n+1) = (1/3)*Product_{d|2*n+1} cyclotomic(2*d,2). (End)
From Hieronymus Fischer, Apr 23 2007: (Start)
The a(n) are closely related to nested square roots; this is 2*sin(2^(-n)*Pi/2*a(n)) = sqrt(2-sqrt(2-sqrt(2-sqrt(...sqrt(2)))...) {with the '2' n times, n >= 0}.
Also 2*cos(2^(-n)*Pi*a(n)) = sqrt(2-sqrt(2-sqrt(2-sqrt(...sqrt(2)))...) {with the '2' n-1 times, n >= 1} as well as
2*sin(2^(-n)*3/2*Pi*a(n)) = sqrt(2+sqrt(2+sqrt(2+sqrt(...sqrt(2)))...) {with the '2' n times, n >= 0} and
2*cos(2^(-n)*3*Pi*a(n)) = -sqrt(2+sqrt(2+sqrt(2+sqrt(...sqrt(2)))...) {with the '2' n-1 times, n >= 1}.
a(n) = 2^(n+1)/Pi*arcsin(b(n+1)/2) where b(n) is defined recursively by b(0)=2, b(n)=sqrt(2-b(n-1)).
There is a similar formula regarding the arccos function, this is a(n) = 2^n/Pi*arccos(b(n)/2).
With respect to the sequence c(n) defined recursively by c(0)=-2, c(n)=sqrt(2+c(n-1)), the following formulas hold true: a(n) = 2^n/3*(1-(-1)^n*(1-2/Pi*arcsin(c(n+1)/2))); a(n) = 2^n/3*(1-(-1)^n*(1-1/Pi*arccos(-c(n)/2))).
(End)
Sum_{k=0..n} A039599(n,k)*a(k) = A049027(n), for n >= 1. - Philippe Deléham, Jun 10 2007
Sum_{k=0..n} A039599(n,k)*a(k+1) = A067336(n). - Philippe Deléham, Jun 10 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0,] = [A005578(n), a(n), A000975(n-1)]. - Gary W. Adamson, Dec 24 2007
a(n) + a(n+5) = 11*2^n. - Paul Curtz, Jan 17 2008
a(n) = Sum_{k=1..n} K(2, k)*a(n - k), where K(n,k) = k if 0 <= k <= n and K(n,k)=0 otherwise. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, the Fibonacci sequence can be generated in several ways using the K-coefficient.) - Thomas Wieder, Jan 13 2008
a(n) + a(n+2*k+1) = a(2*k+1)*2^n. - Paul Curtz, Feb 12 2008
a(n) = lower left term in the 2 X 2 matrix [0,2; 1,1]^n. - Gary W. Adamson, Mar 02 2008
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-2)^(n-k). -Philippe Deléham, Oct 26 2008
a(n) = sqrt(8*a(n-1)*a(n-2) + 1). E.g., sqrt(3*5*8+1) = 11, sqrt(5*11*8+1) = 21. - Giuseppe Ottonello, Jun 14 2009
Let p[i] = Fibonacci(i-1) and let A be the 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-1) = det(A). - Milan Janjic, May 08 2010
a(p-1) = p*A007663(n)/3 if n > 1, and a(p-1) = p*A096060(n) if n > 2, with p=prime(n). - Jonathan Sondow, Jul 19 2010
Algebraically equivalent to replacing the 5's with 9's in the explicit (Binet) formula for the n-th term in the Fibonacci sequence: The formula for the n-th term in the Fibonacci sequence is F(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)). Replacing the 5's with 9's gives ((1+sqrt(9))^n - (1-sqrt(9))^n)/(2^n*sqrt(9)) = (2^n+(-1)^(n+1))/3 = (2^n-(-1)^(n))/3 = a(n). - Jeffrey R. Goodwin, May 27 2011
For n > 1, a(n) = A000975(n-1) + (1 + (-1)^(n-1))/2. - Vladimir Shevelev, Feb 27 2012
From Sergei N. Gladkovskii, Jun 12 2012: (Start)
G.f.: x/(1-x-2*x^2) = G(0)/3; G(k) = 1 - ((-1)^k)/(2^k - 2*x*4^k/(2*x*2^k - ((-1)^k)/G(k+1))); (continued fraction 3 kind, 3-step).
E.g.f.: G(0)/3; G(k) = 1 - ((-1)^k)/(2^k - 2*x*4^k/(2*x*2^k - ((-1)^k)*(k+1)/G(k+1))); (continued fraction 3rd kind, 3-step). (End)
a(n) = 2^k * a(n-k) + (-1)^(n+k)*a(k). - Paul Curtz, Jean-François Alcover, Dec 11 2012
a(n) = sqrt((A014551(n))^2 + (-1)^(n-1)*2^(n+2))/3. - Vladimir Shevelev, Mar 13 2013
G.f.: Q(0)/3, 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 21 2013
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(2*k+1 + 2*x)/( x*(2*k+2 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 29 2013
G.f.: Q(0) -1, where Q(k) = 1 + 2*x^2 + (k+2)*x - x*(k+1 + 2*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n+2) = Sum_{k=0..n} A108561(n,k)*(-2)^k. - Philippe Deléham, Nov 17 2013
a(n) = (Sum_{k=1..n, k odd} C(n,k)*3^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 2^n for all n in Z. - Michael Somos, Mar 18 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-3)^k = (2^n - (-1)^n)/3 = (-1)^(n-1)*Sum_{k=0..n-1} (-2)^k. Equals (-1)^(n-1)*Phi(n,-2), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
From Peter Bala, Apr 06 2015: (Start)
a(2*n)/a(n) = A014551(n) for n >= 1; a(3*n)/a(n) = 3*A245489(n) for n >= 1.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n.
exp( Sum_{n >= 1} a(3*n)/a(n)*x^n/n ) = Sum_{n >= 0} A084175(n+1)*x^n.
exp( Sum_{n >= 1} a(4*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015266(n+3)*(-x)^n.
exp( Sum_{n >= 1} a(5*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015287(n+4)*x^n.
exp( Sum_{n >= 1} a(6*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015305(n+5)*(-x)^n.
exp( Sum_{n >= 1} a(7*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015323(n+6)*x^n.
exp( Sum_{n >= 1} a(8*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015338(n+7)*(-x)^n.
exp( Sum_{n >= 1} a(9*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015356(n+8)*x^n.
exp( Sum_{n >= 1} a(10*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015371(n+9)*(-x)^n. (End)
a(n) = (1-(-1)^n)/2 + floor((2^n)/3). - Reiner Moewald, Jun 05 2015
a(n+k)^2 - A014551(k)*a(n)*a(n+k) + (-2)^k*a(n)^2 = (-2)^n*a(k)^2, for n >= 0 and k >= 0. - Alexander Samokrutov, Jul 21 2015
Dirichlet g.f.: (PolyLog(s,2) + (1 - 2^(1-s))*zeta(s))/3. - Ilya Gutkovskiy, Jun 27 2016
From Yuchun Ji, Apr 08 2018: (Start)
a(m)*a(n) + a(m-1)*a(n-1) - 2*a(m-2)*a(n-2) = 2^(m+n-3).
a(m+n-1) = a(m)*a(n) + 2*a(m-1)*a(n-1); a(m+n) = a(m+1)*a(n+1) - 4*a(m-1)*a(n-1).
a(2*n-1) = a(n)^2 + 2*a(n-1)^2; a(2*n) = a(n+1)^2 - 4*a(n-1)^2. (End)
a(n+4) = a(n) + 5*2^n, a(0) = 0, a(1..4) = [1,1,3,5]. That is to say, for n > 0, the ones digits of Jacobsthal numbers follow the pattern 1,1,3,5,1,1,3,5,1,1,3,5,.... - Yuchun Ji, Apr 25 2019
a(n) mod 10 = A091084(n). - Alois P. Heinz, Apr 25 2019
The sequence starting with "1" is the second INVERT transform of (1, -1, 3, -5, 11, -21, 43, ...). - Gary W. Adamson, Jul 08 2019
From Kai Wang, Jan 14 2020: (Start)
a(n)^2 - a(n+1)*a(n-1) = (-2)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-2)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-2)^n*a(m-n).
a(m-n) = (-1)^n*(a(m)*A014551(n) - A014551(m)*a(n))/(2^(n+1)).
a(m+n) = (a(m)*A014551(n) + A014551(m)*a(n))/2.
A014551(n)^2 - A014551(n+r)*A014551(n-r) = 9*(-1)^(n-r-1)*2^(n-r)*a(r)^2 .
A014551(m)*A014551(n+1) - A014551(m+1)*A014551(n) = 9*(-1)^(n-1)*2^(n)*a(m-n).
A014551(m-n) = (-1)^(n)*(A014551(m)*A014551(n) - 9*a(m)*a(n))/2^(n+1).
A014551(m+n) = (A014551(m)*A014551(n) + 9*a(m)*a(n))/2.
a(n) = Sum_{i=0..n-1; j=0..n-1; i+2*j=n-1} 2^j*((i+j)!/(i!*j!)). (End)
For n > 0, 1/(2*a(n+1)) = Sum_{m>=n} a(m)/(a(m+1)*a(m+2)). - Kai Wang, Mar 03 2020
For 4 > h >= 0, k >= 0, a(4*k+h) mod 5 = a(h) mod 5. - Kai Wang, May 07 2020
From Kengbo Lu, Jul 27 2020: (Start)
a(n) = 1 + Sum_{k=0..n-1} a(k) if n odd; a(n) = Sum_{k=0..n-1} a(k) if n even.
a(n) = F(n) + Sum_{k=0..n-2} a(k)*F(n-k-1), where F denotes the Fibonacci numbers.
a(n) = b(n) + Sum_{k=0..n-1} a(k)*b(n-k), where b(n) is defined through b(0) = 0, b(1) = 1, b(n) = 2*b(n-2).
a(n) = 1 + 2*Sum_{k=0..n-2} a(k).
a(m+n) = a(m)*a(n+1) + 2*a(m-1)*a(n).
a(2*n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)*2^(i+j). (End)
G.f.: x/(1 - x - 2*x^2) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 2*x)/(1 + k*x) (a telescoping series). - Peter Bala, May 08 2024
a(n) = Sum_{r>=0} F(n-2r, r), where F(n, 0) is the n-th Fibonacci number and F(n,r) = Sum_{j=1..n} F(n+1-j, r-1) F(j, r-1). - Gregory L. Simay, Aug 31 2024
From Peter Bala, Jun 27 2025: (Start)
The following are all examples of telescoping infinite products:
Product_{n >= 1} (1 + 2^n/a(2*n+2)) = 2, since 1 + 2^n/a(2*n+2) = b(n+1)/b(n), where b(n) = 2 - 3/(2^n + 1).
Product_{n >= 1} (1 - 2^n/a(2*n+2)) = 2/5, since 1 - 2^n/a(2*n+2) = c(n+1)/c(n), where c(n) = 2 + 3/(2^n - 1).
Product_{n >= 1} (1 + (-2)^n/a(2*n+2)) = 2/3, since 1 + (-2)^n/a(2*n+2) = d(n+1)/d(n), where d(n) = 2 - 1/(1 + (-2)^n).
Product_{n >= 1} (1 - (-2)^n/a(2*n+2)) = 6/5, since 1 - (-2)^n/a(2*n+2) = e(n+1)/e(n), where e(n) = 2 - 1/(1 - (-2)^n). (End)

Extensions

Thanks to Don Knuth, who pointed out several missing references, including Brocard (1880), which although it was mentioned in the 1973 Handbook of Integer Sequences, was omitted from the 1995 "Encyclopedia". - N. J. A. Sloane, Dec 26 2020

A000043 Mersenne exponents: primes p such that 2^p - 1 is prime. Then 2^p - 1 is called a Mersenne prime.

Original entry on oeis.org

2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161, 74207281
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Equivalently, integers k such that 2^k - 1 is prime.
It is believed (but unproved) that this sequence is infinite. The data suggest that the number of terms up to exponent N is roughly K log N for some constant K.
Length of prime repunits in base 2.
The associated perfect number N=2^(p-1)*M(p) (=A019279*A000668=A000396), has 2p (=A061645) divisors with harmonic mean p (and geometric mean sqrt(N)). - Lekraj Beedassy, Aug 21 2004
In one of his first publications Euler found the numbers up to 31 but erroneously included 41 and 47.
Equals number of bits in binary expansion of n-th Mersenne prime (A117293). - Artur Jasinski, Feb 09 2007
Number of divisors of n-th even perfect number, divided by 2. Number of divisors of n-th even perfect number that are powers of 2. Number of divisors of n-th even perfect number that are multiples of n-th Mersenne prime A000668(n). - Omar E. Pol, Feb 24 2008
Number of divisors of n-th even superperfect number A061652(n). Numbers of divisors of n-th superperfect number A019279(n), assuming there are no odd superperfect numbers. - Omar E. Pol, Mar 01 2008
Differences between exponents when the even perfect numbers are represented as differences of powers of 2, for example: The 5th even perfect number is 33550336 = 2^25 - 2^12 then a(5)=25-12=13 (see A135655, A133033, A090748). - Omar E. Pol, Mar 01 2008
Number of 1's in binary expansion of n-th even perfect number (see A135650). Number of 1's in binary expansion of divisors of n-th even perfect number that are multiples of n-th Mersenne prime A000668(n) (see A135652, A135653, A135654, A135655). - Omar E. Pol, May 04 2008
Indices of the numbers A006516 that are also even perfect numbers. - Omar E. Pol, Aug 30 2008
Indices of Mersenne numbers A000225 that are also Mersenne primes A000668. - Omar E. Pol, Aug 31 2008
The (prime) number p appears in this sequence if and only if there is no prime q<2^p-1 such that the order of 2 modulo q equals p; a special case is that if p=4k+3 is prime and also q=2p+1 is prime then the order of 2 modulo q is p so p is not a term of this sequence. - Joerg Arndt, Jan 16 2011
Primes p such that sigma(2^p) - sigma(2^p-1) = 2^p-1. - Jaroslav Krizek, Aug 02 2013
Integers k such that every degree k irreducible polynomial over GF(2) is also primitive, i.e., has order 2^k-1. Equivalently, the integers k such that A001037(k) = A011260(k). - Geoffrey Critzer, Dec 08 2019
Conjecture: for k > 1, 2^k-1 is (a Mersenne) prime or k = 2^(2^m)+1 (is a Fermat number) if and only if (k-1)^(2^k-2) == 1 (mod (2^k-1)k^2). - Thomas Ordowski, Oct 05 2023
Conjecture: for p prime, 2^p-1 is (a Mersenne) prime or p = 2^(2^m)+1 (is a Fermat number) if and only if (p-1)^(2^p-2) == 1 (mod 2^p-1). - David Barina, Nov 25 2024
Already as of Dec. 2020, all exponents up to 10^8 had been verified, implying that 74207281, 77232917 and 82589933 are indeed the next three terms. As of today, all exponents up to 130439863 have been tested at least once, see the GIMPS Milestones Report. - M. F. Hasler, Apr 11 2025
On June 23. 2025 all exponents up to 74340751 have been verified, confirming that 74207281 is the exponent of the 49th Mersenne Prime. - Rodolfo Ruiz-Huidobro, Jun 23 2025

Examples

			Corresponding to the initial terms 2, 3, 5, 7, 13, 17, 19, 31 ... we get the Mersenne primes 2^2 - 1 = 3, 2^3 - 1 = 7, 2^5 - 1 = 31, 127, 8191, 131071, 524287, 2147483647, ... (see A000668).

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 4.
  • J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, p. 79.
  • R. K. Guy, Unsolved Problems in Number Theory, Section A3.
  • F. Lemmermeyer, Reciprocity Laws From Euler to Eisenstein, Springer-Verlag, 2000, p. 57.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 19.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 47.
  • 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, pages 132-134.
  • B. Tuckerman, The 24th Mersenne prime, Notices Amer. Math. Soc., 18 (Jun, 1971), Abstract 684-A15, p. 608.

Links

Crossrefs

Cf. A000668 (Mersenne primes).
Cf. A028335 (integer lengths of Mersenne primes).
Cf. A000225 (Mersenne numbers).
Cf. A001348 (Mersenne numbers with n prime).
Cf. A016027, A046051, A057429, A057951-A057958, A066408, A117293, A127962, A127963, A127964, A127965, A127961, A000979, A000978, A124400, A124401, A127955, A127956, A127957, A127958, A127936, A134458, A000225, A000396, A090748, A133033, A135655, A006516, A019279, A061652, A133033, A135650, A135652, A135653, A135654, A260073, A050475, A379590.

Programs

  • Mathematica
    MersennePrimeExponent[Range[48]] (* Eric W. Weisstein, Jul 17 2017; updated Oct 21 2024 *)
  • PARI
    isA000043(n) = isprime(2^n-1) \\ Michael B. Porter, Oct 28 2009
  • PARI
    is(n)=my(h=Mod(2,2^n-1)); for(i=1, n-2, h=2*h^2-1); h==0||n==2 \\ Lucas-Lehmer test for exponent e. - Joerg Arndt, Jan 16 2011, and Charles R Greathouse IV, Jun 05 2013
forprime(e=2,5000,if(is(e),print1(e,", "))); /* terms < 5000 */
  • Python
    from sympy import isprime, prime
for n in range(1,100):
    if isprime(2**prime(n)-1):
        print(prime(n), end=', ') # Stefano Spezia, Dec 06 2018

Formula

a(n) = log((1/2)*(1+sqrt(1+8*A000396(n))))/log(2). - Artur Jasinski, Sep 23 2008 (under the assumption there are no odd perfect numbers, Joerg Arndt, Feb 23 2014)
a(n) = A000005(A061652(n)). - Omar E. Pol, Aug 26 2009
a(n) = A000120(A000396(n)), assuming there are no odd perfect numbers. - Omar E. Pol, Oct 30 2013

Extensions

Also in the sequence: p = 74207281. - Charles R Greathouse IV, Jan 19 2016
Also in the sequence: p = 77232917. - Eric W. Weisstein, Jan 03 2018
Also in the sequence: p = 82589933. - Gord Palameta, Dec 21 2018
a(46) = 42643801 and a(47) = 43112609, whose ordinal positions in the sequence are now confirmed, communicated by Eric W. Weisstein, Apr 12 2018
a(48) = 57885161, whose ordinal position in the sequence is now confirmed, communicated by Benjamin Przybocki, Jan 05 2022
Also in the sequence: p = 136279841. - Eric W. Weisstein, Oct 21 2024
As of Jan 31 2025, 48 terms are known, and are shown in the DATA section. Four additional numbers are known to be in the sequence, namely 74207281, 77232917, 82589933, and 136279841, but they may not be the next terms. See the GIMP website for the latest information. - N. J. A. Sloane, Jan 31 2025

A139250 Toothpick sequence (see Comments lines for definition).

Original entry on oeis.org

0, 1, 3, 7, 11, 15, 23, 35, 43, 47, 55, 67, 79, 95, 123, 155, 171, 175, 183, 195, 207, 223, 251, 283, 303, 319, 347, 383, 423, 483, 571, 651, 683, 687, 695, 707, 719, 735, 763, 795, 815, 831, 859, 895, 935, 995, 1083, 1163, 1199, 1215, 1243, 1279, 1319, 1379
Offset: 0

Views

Author

Omar E. Pol, Apr 24 2008

Keywords

Comments

A toothpick is a copy of the closed interval [-1,1]. (In the paper, we take it to be a copy of the unit interval [-1/2, 1/2].)
We start at stage 0 with no toothpicks.
At stage 1 we place a toothpick in the vertical direction, anywhere in the plane.
In general, given a configuration of toothpicks in the plane, at the next stage we add as many toothpicks as possible, subject to certain conditions:
- Each new toothpick must lie in the horizontal or vertical directions.
- Two toothpicks may never cross.
- Each new toothpick must have its midpoint touching the endpoint of exactly one existing toothpick.
The sequence gives the number of toothpicks after n stages. A139251 (the first differences) gives the number added at the n-th stage.
Call the endpoint of a toothpick "exposed" if it does not touch any other toothpick. The growth rule may be expressed as follows: at each stage, new toothpicks are placed so their midpoints touch every exposed endpoint.
This is equivalent to a two-dimensional cellular automaton. The animations show the fractal-like behavior.
After 2^k - 1 steps, there are 2^k exposed endpoints, all located on two lines perpendicular to the initial toothpick. At the next step, 2^k toothpicks are placed on these lines, leaving only 4 exposed endpoints, located at the corners of a square with side length 2^(k-1) times the length of a toothpick. - M. F. Hasler, Apr 14 2009 and others. For proof, see the Applegate-Pol-Sloane paper.
If the third condition in the definition is changed to "- Each new toothpick must have at exactly one of its endpoints touching the midpoint of an existing toothpick" then the same sequence is obtained. The configurations of toothpicks are of course different from those in the present sequence. But if we start with the configurations of the present sequence, rotate each toothpick a quarter-turn, and then rotate the whole configuration a quarter-turn, we obtain the other configuration.
If the third condition in the definition is changed to "- Each new toothpick must have at least one of its endpoints touching the midpoint of an existing toothpick" then the sequence n^2 - n + 1 is obtained, because there are no holes left in the grid.
A "toothpick" of length 2 can be regarded as a polyedge with 2 components, both on the same line. At stage n, the toothpick structure is a polyedge with 2*a(n) components.
Conjecture: Consider the rectangles in the sieve (including the squares). The area of each rectangle (A = b*c) and the edges (b and c) are powers of 2, but at least one of the edges (b or c) is <= 2.
In the toothpick structure, if n >> 1, we can see some patterns that look like "canals" and "diffraction patterns". For example, see the Applegate link "A139250: the movie version", then enter n=1008 and click "Update". See also "T-square (fractal)" in the Links section. - Omar E. Pol, May 19 2009, Oct 01 2011
From Benoit Jubin, May 20 2009: The web page "Gallery" of Chris Moore (see link) has some nice pictures that are somewhat similar to the pictures of the present sequence. What sequences do they correspond to?
For a connection to Sierpiński triangle and Gould's sequence A001316, see the leftist toothpick triangle A151566.
Eric Rowland comments on Mar 15 2010 that this toothpick structure can be represented as a 5-state CA on the square grid. On Mar 18 2010, David Applegate showed that three states are enough. See links.
Equals row sums of triangle A160570 starting with offset 1; equivalent to convolving A160552: (1, 1, 3, 1, 3, 5, 7, ...) with (1, 2, 2, 2, ...). Equals A160762: (1, 0, 2, -2, 2, 2, 2, -6, ...) convolved with 2*n - 1: (1, 3, 5, 7, ...). Starting with offset 1 equals A151548: [1, 3, 5, 7, 5, 11, 17, 15, ...] convolved with A078008 signed (A151575): [1, 0, 2, -2, 6, -10, 22, -42, 86, -170, 342, ...]. - Gary W. Adamson, May 19 2009, May 25 2009
For a three-dimensional version of the toothpick structure, see A160160. - Omar E. Pol, Dec 06 2009
From Omar E. Pol, May 20 2010: (Start)
Observation about the arrangement of rectangles:
It appears there is a nice pattern formed by distinct modular substructures: a central cross surrounded by asymmetrical crosses (or "hidden crosses") of distinct sizes and also by "nuclei" of crosses.
Conjectures: after 2^k stages, for k >= 2, and for m = 1 to k - 1, there are 4^(m-1) substructures of size s = k - m, where every substructure has 4*s rectangles. The total number of substructures is equal to (4^(k-1)-1)/3 = A002450(k-1). For example: If k = 5 (after 32 stages) we can see that:
a) There is a central cross, of size 4, with 16 rectangles.
b) There are four hidden crosses, of size 3, where every cross has 12 rectangles.
c) There are 16 hidden crosses, of size 2, where every cross has 8 rectangles.
d) There are 64 nuclei of crosses, of size 1, where every nucleus has 4 rectangles.
Hence the total number of substructures after 32 stages is equal to 85. Note that in every arm of every substructure, in the potential growth direction, the length of the rectangles are the powers of 2. (See illustrations in the links. See also A160124.) (End)
It appears that the number of grid points that are covered after n-th stage of the toothpick structure, assuming the toothpicks have length 2*k, is equal to (2*k-2)*a(n) + A147614(n), k > 0. See the formulas of A160420 and A160422. - Omar E. Pol, Nov 13 2010
Version "Gullwing": on the semi-infinite square grid, at stage 1, we place a horizontal "gull" with its vertices at [(-1, 2), (0, 1), (1, 2)]. At stage 2, we place two vertical gulls. At stage 3, we place four horizontal gulls. a(n) is also the number of gulls after n-th stage. For more information about the growth of gulls see A187220. - Omar E. Pol, Mar 10 2011
From Omar E. Pol, Mar 12 2011: (Start)
Version "I-toothpick": we define an "I-toothpick" to consist of two connected toothpicks, as a bar of length 2. An I-toothpick with length 2 is formed by two toothpicks with length 1. The midpoint of an I-toothpick is touched by its two toothpicks. a(n) is also the number of I-toothpicks after n-th stage in the I-toothpick structure. The I-toothpick structure is essentially the original toothpick structure in which every toothpick is replaced by an I-toothpick. Note that in the physical model of the original toothpick structure the midpoint of a wooden toothpick of the new generation is superimposed on the endpoint of a wooden toothpick of the old generation. However, in the physical model of the I-toothpick structure the wooden toothpicks are not overlapping because all wooden toothpicks are connected by their endpoints. For the number of toothpicks in the I-toothpick structure see A160164 which also gives the number of gullwing in a gullwing structure because the gullwing structure of A160164 is equivalent to the I-toothpick structure. It also appears that the gullwing sequence A187220 is a supersequence of the original toothpick sequence A139250 (this sequence).
For the connection with the Ulam-Warburton cellular automaton see the Applegate-Pol-Sloane paper and see also A160164 and A187220.
(End)
A version in which the toothpicks are connected by their endpoints: on the semi-infinite square grid, at stage 1, we place a vertical toothpick of length 1 from (0, 0). At stage 2, we place two horizontal toothpicks from (0,1), and so on. The arrangement looks like half of the I-toothpick structure. a(n) is also the number of toothpicks after the n-th. - Omar E. Pol, Mar 13 2011
Version "Quarter-circle" (or Q-toothpick): a(n) is also the number of Q-toothpicks after the n-th stage in a Q-toothpick structure in the first quadrant. We start from (0,1) with the first Q-toothpick centered at (1, 1). The structure is asymmetric. For a similar structure but starting from (0, 0) see A187212. See A187210 and A187220 for more information. - Omar E. Pol, Mar 22 2011
Version "Tree": It appears that a(n) is also the number of toothpicks after the n-th stage in a toothpick structure constructed following a special rule: the toothpicks of the new generation have length 4 when they are placed on the infinite square grid (note that every toothpick has four components of length 1), but after every stage, one (or two) of the four components of every toothpick of the new generation is removed, if such component contains an endpoint of the toothpick and if such endpoint is touching the midpoint or the endpoint of another toothpick. The truncated endpoints of the toothpicks remain exposed forever. Note that there are three sizes of toothpicks in the structure: toothpicks of lengths 4, 3 and 2. A159795 gives the total number of components in the structure after the n-th stage. A153006 (the corner sequence of the original version) gives 1/4 of the total of components in the structure after the n-th stage. - Omar E. Pol, Oct 24 2011
From Omar E. Pol, Sep 16 2012: (Start)
It appears that a(n)/A147614(n) converges to 3/4.
It appears that a(n)/A160124(n) converges to 3/2.
It appears that a(n)/A139252(n) converges to 3.
Also:
It appears that A147614(n)/A160124(n) converges to 2.
It appears that A160124(n)/A139252(n) converges to 2.
It appears that A147614(n)/A139252(n) converges to 4.
(End)
It appears that a(n) is also the total number of ON cells after n-th stage in a quadrant of the structure of the cellular automaton described in A169707 plus the total number of ON cells after n+1 stages in a quadrant of the mentioned structure, without its central cell. See the illustration of the NW-NE-SE-SW version in A169707. See also the connection between A160164 and A169707. - Omar E. Pol, Jul 26 2015
On the infinite Cairo pentagonal tiling consider the symmetric figure formed by two non-adjacent pentagons connected by a line segment joining two trivalent nodes. At stage 1 we start with one of these figures turned ON. The rule for the next stages is that the concave part of the figures of the new generation must be adjacent to the complementary convex part of the figures of the old generation. a(n) gives the number of figures that are ON in the structure after n-th stage. A160164(n) gives the number of ON cells in the structure after n-th stage. - Omar E. Pol, Mar 29 2018
From Omar E. Pol, Mar 06 2019: (Start)
The "word" of this sequence is "ab". For further information about the word of cellular automata see A296612.
Version "triangular grid": a(n) is also the total number of toothpicks of length 2 after n-th stage in the toothpick structure on the infinite triangular grid, if we use only two of the three axes. Otherwise, if we use the three axes, so we have the sequence A296510 which has word "abc".
The normal toothpick structure can be considered a superstructure of the Ulam-Warburton celular automaton since A147562(n) equals here the total number of "hidden crosses" after 4*n stages, including the central cross (beginning to count the crosses when their "nuclei" are totally formed with 4 quadrilaterals). Note that every quadrilateral in the structure belongs to a "hidden cross".
Also, the number of "hidden crosses" after n stages equals the total number of "flowers with six petals" after n-th stage in the structure of A323650, which appears to be a "missing link" between this sequence and A147562.
Note that the location of the "nuclei of the hidden crosses" is very similar (essentially the same) to the location of the "flowers with six petals" in the structure of A323650 and to the location of the "ON" cells in the version "one-step bishop" of the Ulam-Warburton cellular automaton of A147562. (End)
From Omar E. Pol, Nov 27 2020: (Start)
The simplest substructures are the arms of the hidden crosses. Each closed region (square or rectangle) of the structure belongs to one of these arms. The narrow arms have regions of area 1, 2, 4, 8, ... The broad arms have regions of area 2, 4, 8, 16 , ... Note that after 2^k stages, with k >= 3, the narrow arms of the main hidden crosses in each quadrant frame the size of the toothpick structure after 2^(k-1) stages.
Another kind of substructure could be called "bar chart" or "bar graph". This substructure is formed by the rectangles and squares of width 2 that are adjacent to any of the four sides of the toothpick structure after 2^k stages, with k >= 2. The height of these successive regions gives the first 2^(k-1) - 1 terms from A006519. For example: if k = 5 the respective heights after 32 stages are [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1]. The area of these successive regions gives the first 2^(k-1) - 1 terms of A171977. For example: if k = 5 the respective areas are [2, 4, 2, 8, 2, 4, 2, 16, 2, 4, 2, 8, 2, 4, 2].
For a connection to Mersenne primes (A000668) and perfect numbers (A000396) see A153006.
For a representation of the Wagstaff primes (A000979) using the toothpick structure see A194810.
For a connection to stained glass windows and a hidden curve see A336532. (End)
It appears that the graph of a(n) bears a striking resemblance to the cumulative distribution function F(x) for X the random variable taking values in [0,1], where the binary expansion of X is given by a sequence of independent coin tosses with probability 3/4 of being 1 at each bit. It appears that F(n/2^k)*(2^(2k+1)+1)/3 approaches a(n) for k large. - James Coe, Jan 10 2022

Examples

			a(10^10) = 52010594272060810683. - _David A. Corneth_, Mar 26 2015

References

  • D. Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191
  • L. D. Pryor, The Inheritance of Inflorescence Characters in Eucalyptus, Proceedings of the Linnean Society of New South Wales, V. 79, (1954), p. 81, 83.
  • Richard P. Stanley, Enumerative Combinatorics, volume 1, second edition, chapter 1, exercise 95, figure 1.28, Cambridge University Press (2012), p. 120, 166.

Links

Crossrefs

Cf. A000079, A002450, A006519, A139251, A139252, A139253, A147614, A139560, A152968, A152978, A152980, A152998, A153000, A153001, A153003, A153004, A153006, A153007, A000217, A007583, A007683, A000396, A000225, A000668, A006516, A006095, A019988, A160570, A160552, A000969, A001316, A151566, A160406, A160408, A160702, A078008, A151548, A001045, A147562, A160124, A160120, A160160, A160170, A160172, A161206, A161328, A161330, A171977, A194810, A296510, A296612, A299476, A299478, A323650, A336532.

Programs

  • Maple
    G := (x/((1-x)*(1+2*x))) * (1 + 2*x*mul(1+x^(2^k-1)+2*x^(2^k),k=0..20)); # N. J. A. Sloane, May 20 2009, Jun 05 2009
# From N. J. A. Sloane, Dec 25 2009: A139250 is T, A139251 is a.
a:=[0,1,2,4]; T:=[0,1,3,7]; M:=10;
for k from 1 to M do
a:=[op(a),2^(k+1)];
T:=[op(T),T[nops(T)]+a[nops(a)]];
for j from 1 to 2^(k+1)-1 do
a:=[op(a), 2*a[j+1]+a[j+2]];
T:=[op(T),T[nops(T)]+a[nops(a)]];
od: od: a; T;
  • Mathematica
    CoefficientList[ Series[ (x/((1 - x)*(1 + 2x))) (1 + 2x*Product[1 + x^(2^k - 1) + 2*x^(2^k), {k, 0, 20}]), {x, 0, 53}], x] (* Robert G. Wilson v, Dec 06 2010 *)
a[0] = 0; a[n_] := a[n] = Module[{m, k}, m = 2^(Length[IntegerDigits[n, 2]] - 1); k = (2m^2+1)/3; If[n == m, k, k + 2 a[n - m] + a[n - m + 1] - 1]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Oct 06 2018, after David A. Corneth *)
  • PARI
    A139250(n,print_all=0)={my(p=[], /* set of "used" points. Points are written as complex numbers, c=x+iy. Toothpicks are of length 2 */
ee=[[0,1]], /* list of (exposed) endpoints. Exposed endpoints are listed as [c,d] where c=x+iy is the position of the endpoint, and d (unimodular) is the direction */
c,d,ne, cnt=1); print_all && print1("0,1"); n<2 && return(n);
for(i=2,n, p=setunion(p, Set(Mat(ee~)[,1])); /* add endpoints (discard directions) from last move to "used" points */
ne=[]; /* new (exposed) endpoints */
for( k=1, #ee, /* add endpoints of new toothpicks if not among the used points */
setsearch(p, c=ee[k][1]+d=ee[k][2]*I) || ne=setunion(ne,Set([[c,d]]));
setsearch(p, c-2*d) || ne=setunion(ne,Set([[c-2*d,-d]]));
); /* using Set() we have the points sorted, so it's easy to remove those which finally are not exposed because they touch a new toothpick */
forstep( k=#ee=eval(ne), 2, -1, ee[k][1]==ee[k-1][1] && k-- && ee=vecextract(ee,Str("^"k"..",k+1)));
cnt+=#ee; /* each exposed endpoint will give a new toothpick */
print_all && print1(","cnt));cnt} \\ M. F. Hasler, Apr 14 2009
  • PARI
    \\works for n > 0
a(n) = {my(k = (2*msb(n)^2 + 1) / 3); if(n==msb(n),k , k + 2*a(n-msb(n)) + a(n - msb(n) + 1) - 1)}
msb(n)=my(t=0);while(n>>t>0,t++);2^(t-1)\\ David A. Corneth, Mar 26 2015
  • Python
    def msb(n):
    t = 0
    while n>>t > 0:
        t += 1
    return 2**(t - 1)
def a(n):
    k = (2 * msb(n)**2 + 1) / 3
    return 0 if n == 0 else k if n == msb(n) else k + 2*a(n - msb(n)) + a(n - msb(n) + 1) - 1
[a(n) for n in range(101)]  # Indranil Ghosh, Jul 01 2017, after David A. Corneth's PARI script

Formula

a(2^k) = A007583(k), if k >= 0.
a(2^k-1) = A006095(k+1), if k >= 1.
a(A000225(k)) - a((A000225(k)-1)/2) = A006516(k), if k >= 1.
a(A000668(k)) - a((A000668(k)-1)/2) = A000396(k), if k >= 1.
G.f.: (x/((1-x)*(1+2*x))) * (1 + 2*x*Product_{k>=0} (1 + x^(2^k-1) + 2*x^(2^k))). - N. J. A. Sloane, May 20 2009, Jun 05 2009
One can show that lim sup a(n)/n^2 = 2/3, and it appears that lim inf a(n)/n^2 is 0.451... - Benoit Jubin, Apr 15 2009 and Jan 29 2010, N. J. A. Sloane, Jan 29 2010
Observation: a(n) == 3 (mod 4) for n >= 2. - Jaume Oliver Lafont, Feb 05 2009
a(2^k-1) = A000969(2^k-2), if k >= 1. - Omar E. Pol, Feb 13 2010
It appears that a(n) = (A187220(n+1) - 1)/2. - Omar E. Pol, Mar 08 2011
a(n) = 4*A153000(n-2) + 3, if n >= 2. - Omar E. Pol, Oct 01 2011
It appears that a(n) = A160552(n) + (A169707(n) - 1)/2, n >= 1. - Omar E. Pol, Feb 15 2015
It appears that a(n) = A255747(n) + A255747(n-1), n >= 1. - Omar E. Pol, Mar 16 2015
Let n = msb(n) + j where msb(n) = A053644(n) and let a(0) = 0. Then a(n) = (2 * msb(n)^2 + 1)/3 + 2 * a(j) + a(j+1) - 1. - David A. Corneth, Mar 26 2015
It appears that a(n) = (A169707(n) - 1)/4 + (A169707(n+1) - 1)/4, n >= 1. - Omar E. Pol, Jul 24 2015

Extensions

Verified and extended, a(49)-a(53), using the given PARI code by M. F. Hasler, Apr 14 2009
Edited by N. J. A. Sloane, Apr 29 2009, incorporating comments from Omar E. Pol, M. F. Hasler, Rob Pratt, Jaume Oliver Lafont, Franklin T. Adams-Watters, R. J. Mathar, David W. Wilson, David Applegate, Benoit Jubin and others.
Further edited by N. J. A. Sloane, Jan 28 2010

A007583 a(n) = (2^(2*n + 1) + 1)/3.

Original entry on oeis.org

1, 3, 11, 43, 171, 683, 2731, 10923, 43691, 174763, 699051, 2796203, 11184811, 44739243, 178956971, 715827883, 2863311531, 11453246123, 45812984491, 183251937963, 733007751851, 2932031007403, 11728124029611, 46912496118443, 187649984473771, 750599937895083
Offset: 0

Views

Author

Simon Plouffe

Keywords

Comments

Let u(k), v(k), w(k) be the 3 sequences defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)-w(k), v(k+1)=u(k)-v(k)+w(k), w(k+1)=-u(k)+v(k)+w(k); let M(k)=Max(u(k),v(k),w(k)); then a(n)=M(2n)=M(2n-1). - Benoit Cloitre, Mar 25 2002
Also the number of words of length 2n generated by the two letters s and t that reduce to the identity 1 by using the relations ssssss=1, tt=1 and stst=1. The generators s and t along with the three relations generate the dihedral group D6=C2xD3. - Jamaine Paddyfoot (jay_paddyfoot(AT)hotmail.com) and John W. Layman, Jul 08 2002
Binomial transform of A025192. - Paul Barry, Apr 11 2003
Number of walks of length 2n+1 between two adjacent vertices in the cycle graph C_6. Example: a(1)=3 because in the cycle ABCDEF we have three walks of length 3 between A and B: ABAB, ABCB and AFAB. - Emeric Deutsch, Apr 01 2004
Numbers of the form 1 + Sum_{i=1..m} 2^(2*i-1). - Artur Jasinski, Feb 09 2007
Prime numbers of the form 1+Sum[2^(2n-1)] are in A000979. Numbers x such that 1+Sum[2^(2n-1)] is prime for n=1,2,...,x is A127936. - Artur Jasinski, Feb 09 2007
Related to A024493(6n+1), A131708(6n+3), A024495(6n+5). - Paul Curtz, Mar 27 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-6, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). - Milan Janjic, Feb 21 2010
Number of toothpicks in the toothpick structure of A139250 after 2^n stages. - Omar E. Pol, Feb 28 2011
Numbers whose binary representation is "10" repeated (n-1) times with "11" appended on the end, n >= 1. For example 171 = 10101011 (2). - Omar E. Pol, Nov 22 2012
a(n) is the smallest number for which A072219(a(n)) = 2*n+1. - Ramasamy Chandramouli, Dec 22 2012
An Engel expansion of 2 to the base b := 4/3 as defined in A181565, with the associated series expansion 2 = b + b^2/3 + b^3/(3*11) + b^4/(3*11*43) + .... Cf. A007051. - Peter Bala, Oct 29 2013
The positive integer solution (x,y) of 3*x - 2^n*y = 1, n>=0, with smallest x is (a(n/2), 2) if n is even and (a((n-1)/2), 1) if n is odd. - Wolfdieter Lang, Feb 15 2014
The smallest positive number that requires at least n additions and subtractions of powers of 2 to be formed. See Puzzling StackExchange link. - Alexander Cooke Jul 16 2023

References

  • H. W. Gould, Combinatorial Identities, Morgantown, 1972, (1.77), page 10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. also A006054, A006356, A005578, A075680.
Partial sums of A081294.
Cf. A002450, A004171, A007051, A083065, A083066, A083884.
Cf. A000978, A000979, A124400, A124401, A127936, A127955, A127956, A127957, A127958.
Cf. location of records in A007302.

Programs

  • GAP
    List([0..25], n-> (2^(2*n+1) + 1)/3); # G. C. Greubel, Dec 25 2019
  • Haskell
    a007583 = (`div` 3) . (+ 1) . a004171
-- Reinhard Zumkeller, Jan 09 2013
  • Magma
    [(2^(2*n+1) + 1)/3: n in [0..30] ]; // Vincenzo Librandi, Apr 28 2011
  • Maple
    a[0]:=1:for n from 1 to 50 do a[n]:=4*a[n-1]-1 od: seq(a[n], n=0..23); # Zerinvary Lajos, Feb 22 2008, with correction by K. Spage, Aug 20 2014
A007583 := proc(n)
    (2^(2*n+1)+1)/3 ;
end proc: # R. J. Mathar, Feb 19 2015
  • Mathematica
    (* From Michael De Vlieger, Aug 22 2016 *)
Table[(2^(2n+1) + 1)/3, {n, 0, 23}]
Table[1 + 2Sum[4^k, {k, 0, n-1}], {n, 0, 23}]
NestList[4# -1 &, 1, 23]
Table[Sum[Binomial[n+k, 2k]/2^(k-n), {k, 0, n}], {n, 0, 23}]
CoefficientList[Series[(1-2x)/(1-5x+4x^2), {x, 0, 23}], x] (* End *)
  • PARI
    a(n)=sum(k=-n\3,n\3,binomial(2*n+1,n+1+3*k))
  • PARI
    a=1; for(n=1,23, print1(a,", "); a=bitor(a,3*a)) \\ K. Spage, Aug 20 2014
  • PARI
    Vec((1-2*x)/(1-5*x+4*x^2) + O(x^30)) \\ Altug Alkan, Dec 08 2015
  • PARI
    apply( {A007583(n)=2<<(2*n)\/3}, [0..25]) \\ M. F. Hasler, Nov 30 2021
  • Sage
    [(2^(2*n+1) + 1)/3 for n in (0..25)] # G. C. Greubel, Dec 25 2019

Formula

a(n) = 2*A002450(n) + 1.
From Wolfdieter Lang, Apr 24 2001: (Start)
a(n) = Sum_{m = 0..n} A060920(n, m) = A002450(n+1) - 2*A002450(n).
G.f.: (1-2*x)/(1-5*x+4*x^2). (End)
a(n) = Sum_{k = 0..n} binomial(n+k, 2*k)/2^(k - n).
a(n) = 4*a(n-1) - 1, n > 0.
From Paul Barry, Mar 17 2003: (Start)
a(n) = 1 + 2*Sum_{k = 0..n-1} 4^k;
a(n) = A001045(2n+1). (End)
a(n) = A020988(n-1) + 1 = A039301(n+1) - 1 = A083584(n-1) + 2. - Ralf Stephan, Jun 14 2003
a(0) = 1; a(n+1) = a(n) * 4 - 1. - Regis Decamps (decamps(AT)users.sf.net), Feb 04 2004 (correction to lead index by K. Spage, Aug 20 2014)
a(n) = Sum_{i + j + k = n; 0 <= i, j, k <= n} (n+k)!/i!/j!/(2*k)!. - Benoit Cloitre, Mar 25 2004
a(n) = 5*a(n-1) - 4*a(n-2). - Emeric Deutsch, Apr 01 2004
a(n) = 4^n - A001045(2*n). - Paul Barry, Apr 17 2004
a(n) = 2*(A001045(n))^2 + (A001045(n+1))^2. - Paul Barry, Jul 15 2004
a(n) = left and right terms in M^n * [1 1 1] where M = the 3X3 matrix [1 1 1 / 1 3 1 / 1 1 1]. M^n * [1 1 1] = [a(n) A002450(n+1) a(n)] E.g. a(3) = 43 since M^n * [1 1 1] = [43 85 43] = [a(3) A002450(4) a(3)]. - Gary W. Adamson, Dec 18 2004
a(n) = A072197(n) - A020988(n). - Creighton Dement, Dec 31 2004
a(n) = A139250(2^n). - Omar E. Pol, Feb 28 2011
a(n) = A193652(2*n+1). - Reinhard Zumkeller, Aug 08 2011
a(n) = Sum_{k = -floor(n/3)..floor(n/3)} binomial(2*n, n+3*k)/2. - Mircea Merca, Jan 28 2012
a(n) = 2^(2*(n+1)) - A072197(n). - Vladimir Pletser, Apr 12 2014
a(n) == 2*n + 1 (mod 3). Indeed, from Regis Decamps' formula (Feb 04 2004) we have a(i+1) - a(i) == -1 (mod 3), i= 0, 1, ..., n - 1. Summing, we have a(n) - 1 == -n (mod 3), and the formula follows. - Vladimir Shevelev, May 20 2015
For n > 0 a(n) = A133494(0) + 2 * (A133494(n) + Sum_{x = 1..n - 1}Sum_{k = 0..x - 1}(binomial(x - 1, k)*(A133494(k+1) + A133494(n-x+k)))). - J. Conrad, Dec 06 2015
a(n) = Sum_{k = 0..2n} (-2)^k == 1 + Sum_{k = 1..n} 2^(2k-1). - Bob Selcoe, Aug 21 2016
E.g.f.: (1 + 2*exp(3*x))*exp(x)/3. - Ilya Gutkovskiy, Aug 21 2016
A075680(a(n)) = 1, for n > 0. - Ralf Stephan, Jun 17 2025

A000978 Wagstaff numbers: numbers k such that (2^k + 1)/3 is prime.

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 19, 23, 31, 43, 61, 79, 101, 127, 167, 191, 199, 313, 347, 701, 1709, 2617, 3539, 5807, 10501, 10691, 11279, 12391, 14479, 42737, 83339, 95369, 117239, 127031, 138937, 141079, 267017, 269987, 374321, 986191, 4031399
Offset: 1

Views

Author

N. J. A. Sloane, Robert G. Wilson v

Keywords

Comments

It is easy to see that the definition implies that k must be an odd prime. - N. J. A. Sloane, Oct 06 2006
The terms from a(32) on only give probable primes as of 2018. Caldwell lists the largest certified primes. - Jens Kruse Andersen, Jan 10 2018
Prime numbers of the form 1+Sum_{i=1..m} 2^(2i-1). - Artur Jasinski, Feb 09 2007
There is a new conjecture stating that a Wagstaff number is prime under the following condition (based on DiGraph cycles under the LLT): Let p be a prime integer > 3, N(p) = 2^p+1 and W(p) = N(p)/3, S(0) = 3/2 (or 1/4) and S(i+1) = S(i)^2 - 2 (mod N(p)). Then W(p) is prime iff S(p-1) == S(0) (mod W(p)). - Tony Reix, Sep 03 2007
As a member of the DUR team (Diepeveen, Underwood, Reix), and thanks to the LLR tool built by Jean Penne, I've found a new and big Wagstaff PRP: (2^4031399+1)/3 is Vrba-Reix PRP! This Wagstaff number has 1,213,572 digits and today is the 3rd biggest PRP ever found. I've done a second verification on a Nehalem core with the PFGW tool. - Tony Reix, Feb 20 2010
13347311 and 13372531 were found to be terms of this sequence (maybe not the next ones) by Ryan Propper in September 2013. - Max Alekseyev, Oct 07 2013
The next term is larger than 10 million. - Gord Palameta, Mar 22 2019
Ryan Propper found another likely term, 15135397, though it only corresponds to a probable prime. - Charles R Greathouse IV, Jul 01 2021

References

  • J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.
  • 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).
  • S. S. Wagstaff, Jr., personal communication.

Links

Crossrefs

Cf. A107036 (indices of prime Jacobsthal numbers).
Cf. A000979, A124400, A124401, A127955, A127956, A127957, A127958, A127936, A010051, A065091, A001045.

Programs

  • Haskell
    a000978 n = a000978_list !! (n-1)
a000978_list = filter ((== 1) . a010051 . a001045) a065091_list
-- Reinhard Zumkeller, Mar 24 2013
  • Mathematica
    Select[Range[5000], PrimeQ[(2^# + 1)/3] &] (* Michael De Vlieger, Jan 10 2018 *)
Select[Prime[Range[2,500]],PrimeQ[(2^#+1)/3]&] (* Harvey P. Dale, Jun 13 2022 *)
  • PARI
    forprime(p=2,5000,if(ispseudoprime(2^p\/3),print1(p", "))) \\ Charles R Greathouse IV, Jul 15 2011
  • Python
    from gmpy2 import divexact
from sympy import prime, isprime
A000978 = [p for p in (prime(n) for n in range(2,10**2)) if isprime(divexact(2**p+1,3))] # Chai Wah Wu, Sep 04 2014

Formula

a(n) = A107036(n) for n>1. - Alexander Adamchuk, Feb 10 2007

Extensions

a(30) from Kamil Duszenko (kdusz(AT)wp.pl), Feb 03 2003; a(30) was proved prime by Francois Morain with FastECPP. - Tony Reix, Sep 03 2007
a(31)-a(39) from Robert G. Wilson v, Apr 11 2005
a(40) from Vincent Diepeveen (diep(AT)xs4all.nl) added by Alexander Adamchuk, Jun 19 2008
a(41) from Tony Reix, Feb 20 2010

A139253 Toothpick primes: primes in the toothpick sequence A139250.

Original entry on oeis.org

3, 7, 11, 23, 43, 47, 67, 79, 223, 251, 283, 347, 383, 571, 683, 719, 859, 1163, 1279, 1319, 1607, 1667, 1759, 1871, 2011, 2731, 2767, 2843, 2879, 3919, 4523, 4783, 4799, 4903, 5051, 6343, 6607, 7823, 8971, 9643, 11003, 11071, 11483, 11519, 12251, 12907
Offset: 1

Views

Author

Omar E. Pol, Apr 24 2008

Keywords

Comments

All Wagstaff primes A000979 are members of this sequence. - Omar E. Pol, Mar 12 2012

Examples

			a(25) = 2011 because 2011 is the number of toothpicks after 60 stages in the toothpick structure of A139250 and it is also a prime number A000040.

Links

Crossrefs

Intersection of A000040 and A139250.
Cf. A000979, A139251, A139252, A194810.

Programs

  • Mathematica
    toothpicks = With[{terms = 200}, CoefficientList[(x/((1-x)(1+2x)))(1 + 2x Product[1 + x^(2^k-1) + 2 x^(2^k), {k, 0, Log[2, terms] // Ceiling}]) + O[x]^terms, x]];
Select[toothpicks, PrimeQ] (* Jean-François Alcover, Aug 05 2018 *)

Formula

A000040 INTERSECT A139250.

Extensions

More terms from Omar E. Pol, Dec 20 2008
Terms beyond 2731 from R. J. Mathar, Jan 21 2010

A126614 a(n) = (2^prime(n) + 1)/3.

Original entry on oeis.org

3, 11, 43, 683, 2731, 43691, 174763, 2796203, 178956971, 715827883, 45812984491, 733007751851, 2932031007403, 46912496118443, 3002399751580331, 192153584101141163, 768614336404564651, 49191317529892137643, 787061080478274202283, 3148244321913096809131
Offset: 2

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Comments

If p - 1 is squarefree, the multiplicative order of 2 modulo a(n) is 2p. - Vladimir Shevelev, Jul 15 2008
The prime numbers in this sequence are the Wagstaff primes (A000979). - Omar E. Pol, Nov 05 2013

Examples

			a(2) = (2^prime(2) + 1)/3 = (2^3 + 1)/3 = 9/3 = 3.
a(3) = (2^prime(3) + 1)/3 = (2^5 + 1)/3 = 33/3 = 11.
a(4) = (2^prime(4) + 1)/3 = (2^7 + 1)/3 = 129/3 = 43.

Links

Crossrefs

Cf. A000040, A000979, A000978, A001045, A124400.

Programs

Formula

a(n) = A001045(A000040(n)). - Alois P. Heinz, Apr 14 2025

A127936 Numbers k such that 1 + Sum_{i=1..k} 2^(2*i-1) is prime.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 9, 11, 15, 21, 30, 39, 50, 63, 83, 95, 99, 156, 173, 350, 854, 1308, 1769, 2903, 5250, 5345, 5639, 6195, 7239, 21368, 41669, 47684, 58619, 63515, 69468, 70539, 133508, 134993, 187160, 493095
Offset: 1

Views

Author

Artur Jasinski, Feb 08 2007, Feb 09 2007

Keywords

Comments

If this sequence is infinite then so is A124401.
Equals A127965(n)/2.
The sum has the simple closed form 1 + 2/3*(4^n-1). - Stefan Steinerberger, Nov 24 2007
Terms beyond a(30) correspond to probable primes, cf. A000978. - M. F. Hasler, Aug 29 2008

Examples

			a(1)=1 because 1 + 2 = 3 is prime;
a(2)=2 because 1 + 2 + 2^3 = 11 is prime;
a(3)=3 because 1 + 2 + 2^3 + 2^5 = 43 is prime;
a(4)=5 because 1 + 2 + 2^3 + 2^5 + 2^7 + 2^9 = 683 is prime;
...

Crossrefs

Cf. A127962, A127963, A127964, A127965, A127961, A000979, A000978, A124400, A126614, A127955, A127956, A127957, A127958, A127936, A127936, A124401, A010051, A007583.

Programs

Formula

a(n) = floor(A000978(n)/2) = ceiling(log(4)(A000979(n))); A000978(n) = 2 a(n) + 1; A000979(n) = (2*4^a(n)+1)/3. - M. F. Hasler, Aug 29 2008

Extensions

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 11 2007
2 more terms from Stefan Steinerberger, Nov 24 2007
6 more terms from Dmitry Kamenetsky, Jul 12 2008
a(30)-a(40) calculated from A000978 by M. F. Hasler, Aug 29 2008
Showing 1-10 of 28 results. Next

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