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.

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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

N. J. A. Sloane

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.

Links

Crossrefs

Cf. A000043 (Mersenne exponents).
Cf. A000668 (Mersenne primes).
Cf. A001348 (Mersenne numbers with n prime).
Cf. A000079, A001045, A009545, A016189, A052955, A083329, A085104, A144081.
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. A000203, A239473, A279396.
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

A029858 a(n) = (3^n - 3)/2.

Original entry on oeis.org

0, 3, 12, 39, 120, 363, 1092, 3279, 9840, 29523, 88572, 265719, 797160, 2391483, 7174452, 21523359, 64570080, 193710243, 581130732, 1743392199, 5230176600, 15690529803, 47071589412, 141214768239
Offset: 1

Views

Author

Christian G. Bower

Keywords

Comments

Also the number of 2-block covers of a labeled n-set. a(n) = A055154(n,2). Generally, number of k-block covers of a labeled n-set is T(n,k) = (1/k!)*Sum_{i = 1..k + 1} Stirling1(k + 1,i)*(2^(i - 1) - 1)^n. In particular, T(n,2) = (1/2!)*(3^n - 3), T(n,3) = (1/3!)*(7^n - 6*3^n + 11), T(n,4) = (1/4)!*(15^n - 10*7^n + 35*3^n - 50), ... - Vladeta Jovovic, Jan 19 2001
Conjectured to be the number of integers from 0 to 10^(n-1) - 1 that lack 0, 1, 2, 3, 4, 5 and 6 as a digit. - Alexandre Wajnberg, Apr 25 2005. This is easily verified to be true. - Renzo Benedetti, Sep 25 2008
Number of monic irreducible polynomials of degree 1 in GF(3)[x1,...,xn]. - Max Alekseyev, Jan 23 2006
Also, the greatest number of identical weights among which an odd one can be identified and it can be decided if the odd one is heavier or lighter, using n weighings with a comparing balance. If the odd one only needs to be identified, the sequence starts 4, 13, 40 and is A003462 (3^n - 1)/2, n > 1. - Tanya Khovanova, Dec 11 2006; corrected by Samuel E. Rhoads, Apr 18 2016
Binomial transform yields A134057. Inverse binomial transform yields A062510 with one additional 0 in front. - R. J. Mathar, Jun 18 2008
Numbers k where the recurrence s(0)=0, if s(k-1) >= k then s(k) = s(k-1) - k otherwise s(k) = s(k-1) + k produces s(k) = 0. - Hugo Pfoertner, Jan 05 2012
For n > 1: A008344(a(n)) = a(n). - Reinhard Zumkeller, May 09 2012
Also the number of edges in the (n-1)-Hanoi graph. - Eric W. Weisstein, Jun 18 2017
A level 1 Sierpiński triangle graph is a triangle. Level n+1 is formed from three copies of level n by identifying pairs of corner vertices of each pair of triangles. a(n) is the number of degree 4 vertices in the level n Sierpinski triangle graph. - Allan Bickle, Jul 30 2020
Also the number of minimum vertex cuts in the n-Apollonian network. - Eric W. Weisstein, Dec 20 2020
Also the minimum number of turns in n-dimensional Euclidean space needed to visit all 3^n points of the grid {0, 1, 2}^n, moving in straight lines between turns (repeated visits and direction changes at non-grid points are allowed). - Marco Ripà, Aug 06 2025

Examples

			For the Sierpiński triangle, Level 1 is a triangle, so a(1) = 0.
Level 2 has three corners (degree 2) and three degree 4 vertices, so a(2) = 3.
The level 2 Hanoi graph has 3 triangles joined by 3 edges, so a(2+1) = 12.

Links

Crossrefs

Cf. A055154, A003462, A007051, A034472, A024023.
Cf. A000203, A008776.
Cf. A007283, A029858, A067771, A233774, A233775, A246959 (Sierpiński triangle graphs).
Cf. A000225, A029858, A058809, A375256 (Hanoi graphs).

Programs

Formula

a(n) = 3*a(n-1) + 3. - Alexandre Wajnberg, Apr 25 2005
O.g.f: 3*x^2/((1-x)*(1-3*x)). - R. J. Mathar, Jun 18 2008
a(n) = 3^(n-1) + a(n-1) (with a(1)=0). - Vincenzo Librandi, Nov 18 2010
a(n) = 3*A003462(n-1). - R. J. Mathar, Sep 10 2015
E.g.f.: 3*(-1 + exp(2*x))*exp(x)/2. - Ilya Gutkovskiy, Apr 19 2016
a(n) = A067771(n-1) - 3. - Allan Bickle, Jul 30 2020
a(n) = sigma(A008776(n-2)) for n>=2. - Flávio V. Fernandes, Apr 20 2021

Extensions

Corrected by T. D. Noe, Nov 07 2006

A058809 The sequence lambda(3,n), where lambda is defined in A055203. Number of ways of placing n identifiable positive intervals with a total of exactly three starting and/or finishing points.

Original entry on oeis.org

0, 0, 6, 24, 78, 240, 726, 2184, 6558, 19680, 59046, 177144, 531438, 1594320, 4782966, 14348904, 43046718, 129140160, 387420486, 1162261464, 3486784398, 10460353200, 31381059606, 94143178824, 282429536478, 847288609440
Offset: 0

Views

Author

N. J. A. Sloane, Jan 03 2001

Keywords

Comments

For all n, a(n)=1*3^n-3*1^n+3*0^n-1*0^n [with 0^0=1] where powers are taken of triangular numbers and multiplied by binomial coefficients with alternating signs. - Henry Bottomley, Jan 05 2001
For n>=1, a(n) is the number of facets of the harmonic polytope. See Ardila and Escobar. - Michel Marcus, Jun 08 2020
For n >= 3, this is the number of acyclic orientations of the wheel graph of order n+1. - Peter Kagey, Oct 13 2020
Number of ternary strings of length n with at least 2 different digits. - Enrique Navarrete, Nov 20 2020
A level 1 Hanoi graph is a triangle. Level n+1 is formed from three copies of level n by adding edges between pairs of corner vertices of each pair of triangles. This graph represents the allowable moves in the Towers of Hanoi problem with n disks. a(n) is the number of degree 3 vertices in the level n Hanoi graph. - Allan Bickle, Aug 07 2024

Examples

			a(2)=6 since intervals a-a and b-b can be combined as a-ab-b, a-b-ab, ab-a-b, b-ab-a, b-a-ab, or ab-b-a.
The level 2 Hanoi graph has 9 vertices, 6 with degree 3, so a(2) = 6.

Links

Crossrefs

Cf. A059116, A059117.
Cf. A000225, A029858, A058809, A375256 (Hanoi graphs).

Programs

  • Mathematica
    Join[{0},NestList[3#+6&,0,30]] (* or *) Join[{0},LinearRecurrence[{4,-3},{0,6},30]] (* Harvey P. Dale, Sep 29 2013 *)
  • PARI
    concat([0,0], Vec(6*x^2 / ((1 - x)*(1 - 3*x)) + O(x^30))) \\ Colin Barker, Oct 14 2020

Formula

For n>0, a(n) = 3^n-3 = 3*a(n-1)+6.
a(0)=0, a(1)=0, a(2)=6, a(n) = 4*a(n-1)-3*a(n-2). - Harvey P. Dale, Sep 29 2013
G.f.: 6*x^2 / ((1 - x)*(1 - 3*x)). - Colin Barker, Oct 14 2020

A370933 Number of pairs of antipodal vertices in the level n>1 Sierpiński triangle graph.

Original entry on oeis.org

6, 15, 42, 132, 456, 1680, 6432, 25152, 99456, 395520, 1577472, 6300672, 25184256, 100700160, 402726912, 1610760192, 6442745856, 25770393600, 103080394752, 412319219712, 1649272160256, 6597079203840, 26388297940992, 105553154015232, 422212540563456, 1688850011258880, 6755399743045632
Offset: 2

Views

Author

Allan Bickle, Aug 07 2024

Keywords

Comments

A level 1 Sierpiński triangle graph is a triangle. Level n+1 is formed from three copies of level n by identifying pairs of corner vertices of each pair of triangles.
Antipodal vertices are a pair of vertices at maximum distance in a graph. The diameter of the level n Sierpiński triangle graph is 2^(n-1).

Examples

			3 example graphs:                        o
                                        / \
                                       o---o
                                      / \ / \
                        o            o---o---o
                       / \          / \     / \
             o        o---o        o---o   o---o
            / \      / \ / \      / \ / \ / \ / \
           o---o    o---o---o    o---o---o---o---o
Graph:      S_1        S_2              S_3
For S_2, there are 3 pairs of corners and 3 pairs of a corner and a middle vertex, so a(2) = 6.

Links

Crossrefs

Cf. A007283, A029858, A067771, A193277, A233774, A233775, A246959, A298202 (Sierpiński triangle graphs).
Cf. A375256 (antipodal pairs in Hanoi graphs).

Programs

  • Mathematica
    A370933[n_] := 3*2^(n - 3)*(2^(n - 2) + 3);
Array[A370933, 30, 2] (* or *)
LinearRecurrence[{6, -8}, {6, 15}, 30] (* Paolo Xausa, Sep 23 2024 *)
  • PARI
    a(n) = 3*2^(n-3)*(2^(n-2)+3); \\ Michel Marcus, Aug 08 2024

Formula

a(n) = 3*2^(n-3)*(2^(n-2)+3).
a(n) = 3*A257273(n-2).
a(n) = A375256(n-1) + 3.

Extensions

More terms from Michel Marcus, Aug 08 2024
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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