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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A145070 Partial sums of A006127, starting at n=1.

Original entry on oeis.org

3, 9, 20, 40, 77, 147, 282, 546, 1067, 2101, 4160, 8268, 16473, 32871, 65654, 131206, 262295, 524457, 1048764, 2097360, 4194533, 8388859, 16777490, 33554730, 67109187, 134218077, 268435832, 536871316, 1073742257, 2147484111
Offset: 1

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Keywords

Examples

			a(2) = a(1) + 2^2 + 2 = 3 + 4 + 2 = 9; a(3) = a(2) + 2^3 + 3 = 9 + 8 + 3 = 20.
		

Crossrefs

Cf. A006127 (2^n + n), A000325 (2^n - n), A048492 (partial sums of A000325, starting at n=1).

Programs

  • ARIBAS
    a:=0; for n:=1 to 30 do a:=a+2**n+n; write(a, ","); end;
    
  • Maple
    A145070:=n->(-4+2^(2+n)+n+n^2)/2: seq(A145070(n), n=1..50); # Wesley Ivan Hurt, Jan 22 2017
  • Mathematica
    lst={};s=0;Do[s+=2^n+n;AppendTo[lst,s],{n,5!}];lst
    Accumulate[Table[2^n+n,{n,50}]] (* or *) LinearRecurrence[{5,-9,7,-2},{3,9,20,40},50] (* Harvey P. Dale, Aug 22 2020 *)
  • PARI
    Vec(x*(2*x^2-6*x+3) / ((x-1)^3*(2*x-1)) + O(x^100)) \\ Colin Barker, Oct 27 2014

Formula

a(1) = 3; a(n) = a(n-1) + 2^n + n for n > 1.
From Colin Barker, Oct 27 2014: (Start)
a(n) = (-4+2^(2+n)+n+n^2)/2.
a(n) = 5*a(n-1)-9*a(n-2)+7*a(n-3)-2*a(n-4).
G.f.: x*(2*x^2-6*x+3) / ((x-1)^3*(2*x-1)).
(End)

Extensions

Edited by Klaus Brockhaus, Oct 14 2008

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

Original entry on oeis.org

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

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Author

Keywords

Comments

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

Examples

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

References

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

Crossrefs

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

Programs

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

Formula

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

Extensions

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

A000051 a(n) = 2^n + 1.

Original entry on oeis.org

2, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025, 2049, 4097, 8193, 16385, 32769, 65537, 131073, 262145, 524289, 1048577, 2097153, 4194305, 8388609, 16777217, 33554433, 67108865, 134217729, 268435457, 536870913, 1073741825, 2147483649, 4294967297, 8589934593
Offset: 0

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Author

Keywords

Comments

Same as Pisot sequence L(2,3).
Length of the continued fraction for Sum_{k=0..n} 1/3^(2^k). - Benoit Cloitre, Nov 12 2003
See also A004119 for a(n) = 2a(n-1)-1 with first term = 1. - Philippe Deléham, Feb 20 2004
From the second term on (n>=1), in base 2, these numbers present the pattern 1000...0001 (with n-1 zeros), which is the "opposite" of the binary 2^n-2: (0)111...1110 (cf. A000918). - Alexandre Wajnberg, May 31 2005
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (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,3). - Milan Janjic, Jan 27 2010
First differences of A006127. - Reinhard Zumkeller, Apr 14 2011
The odd prime numbers in this sequence form A019434, the Fermat primes. - David W. Wilson, Nov 16 2011
Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... . - R. J. Mathar, Aug 10 2012
Is the mentioned Pisano period lengths (see above) the same as A007733? - Omar E. Pol, Aug 10 2012
Only positive integers that are not 1 mod (2k+1) for any k>1. - Jon Perry, Oct 16 2012
For n >= 1, a(n) is the total length of the segments of the Hilbert curve after n iterations. - Kival Ngaokrajang, Mar 30 2014
Frénicle de Bessy (1657) proved that a(3) = 9 is the only square in this sequence. - Charles R Greathouse IV, May 13 2014
a(n) is the number of distinct possible sums made with at most two elements in {1,...,a(n-1)} for n > 0. - Derek Orr, Dec 13 2014
For n > 0, given any set of a(n) lattice points in R^n, there exist 2 distinct members in this set whose midpoint is also a lattice point. - Melvin Peralta, Jan 28 2017
Also the number of independent vertex sets, irredundant sets, and vertex covers in the (n+1)-star graph. - Eric W. Weisstein, Aug 04 and Sep 21 2017
Also the number of maximum matchings in the 2(n-1)-crossed prism graph. - Eric W. Weisstein, Dec 31 2017
Conjecture: For any integer n >= 0, a(n) is the permanent of the (n+1) X (n+1) matrix with M(j, k) = -floor((j - k - 1)/(n + 1)). This conjecture is inspired by the conjecture of Zhi-Wei Sun in A036968. - Peter Luschny, Sep 07 2021

References

  • Paul Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 60, 244.
  • 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.

Crossrefs

Apart from the initial 1, identical to A094373.
See A008776 for definitions of Pisot sequences.
Column 2 of array A103438.
Cf. A007583 (a((n-1)/2)/3 for odd n).

Programs

  • Haskell
    a000051 = (+ 1) . a000079
    a000051_list = iterate ((subtract 1) . (* 2)) 2
    -- Reinhard Zumkeller, May 03 2012
    
  • Magma
    [2^n+1: n in [0..40]]; // G. C. Greubel, Jan 18 2025
  • Maple
    A000051:=-(-2+3*z)/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation
    a := n -> add(binomial(n,k)*bernoulli(n-k,1)*2^(k+1)/(k+1),k=0..n); # Peter Luschny, Apr 20 2009
  • Mathematica
    Table[2^n + 1, {n,0,40}]
    2^Range[0,40] + 1 (* Eric W. Weisstein, Jul 17 2017 *)
    LinearRecurrence[{3, -2}, {2, 3}, 40] (* Eric W. Weisstein, Sep 21 2017 *)
  • PARI
    a(n)=2^n+1
    
  • PARI
    first(n) = Vec((2 - 3*x)/((1 - x)*(1 - 2*x)) + O(x^n)) \\ Iain Fox, Dec 31 2017
    
  • Python
    def A000051(n): return (1<Chai Wah Wu, Dec 21 2022
    

Formula

a(n) = 2*a(n-1) - 1 = 3*a(n-1) - 2*a(n-2).
G.f.: (2-3*x)/((1-x)*(1-2*x)).
First differences of A052944. - Emeric Deutsch, Mar 04 2004
a(0) = 1, then a(n) = (Sum_{i=0..n-1} a(i)) - (n-2). - Gerald McGarvey, Jul 10 2004
Inverse binomial transform of A007689. Also, V sequence in Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005
a(n) = A127904(n+1) for n>0. - Reinhard Zumkeller, Feb 05 2007
Equals binomial transform of [2, 1, 1, 1, ...]. - Gary W. Adamson, Apr 23 2008
a(n) = A000079(n)+1. - Omar E. Pol, May 18 2008
E.g.f.: exp(x) + exp(2*x). - Mohammad K. Azarian, Jan 02 2009
a(n) = A024036(n)/A000225(n). - Reinhard Zumkeller, Feb 14 2009
From Peter Luschny, Apr 20 2009: (Start)
A weighted binomial sum of the Bernoulli numbers A027641/A027642 with A027641(1)=1 (which amounts to the definition B_{n} = B_{n}(1)).
a(n) = Sum_{k=0..n} C(n,k)*B_{n-k}*2^(k+1)/(k+1). (See also A052584.) (End)
a(n) is the a(n-1)-th odd number for n >= 1. - Jaroslav Krizek, Apr 25 2009
From Reinhard Zumkeller, Feb 28 2010: (Start)
a(n)*A000225(n) = A000225(2*n).
a(n) = A173786(n,0). (End)
If p[i]=Fibonacci(i-4) and if A is 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(n+2) = a(n) + a(n+1) + A000225(n). - Ivan N. Ianakiev, Jun 24 2012
a(A006521(n)) mod A006521(n) = 0. - Reinhard Zumkeller, Jul 17 2014
a(n) = 3*A007583((n-1)/2) for n odd. - Eric W. Weisstein, Jul 17 2017
Sum_{n>=0} 1/a(n) = A323482. - Amiram Eldar, Nov 11 2020

A000302 Powers of 4: a(n) = 4^n.

Original entry on oeis.org

1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736, 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656
Offset: 0

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Author

Keywords

Comments

Same as Pisot sequences E(1, 4), L(1, 4), P(1, 4), T(1, 4). Essentially same as Pisot sequences E(4, 16), L(4, 16), P(4, 16), T(4, 16). See A008776 for definitions of Pisot sequences.
The convolution square root of this sequence is A000984, the central binomial coefficients: C(2n,n). - T. D. Noe, Jun 11 2002
With P(n) being the number of integer partitions of n, p(i) as the number of parts of the i-th partition of n, d(i) as the number of different parts of the i-th partition of n, m(i, j) the multiplicity of the j-th part of the i-th partition of n, one has a(n) = Sum_{i = 1..P(n)} p(i)!/(Product_{j = 1..d(i)} m(i, j)!) * 2^(n-1). - Thomas Wieder, May 18 2005
Sums of rows of the triangle in A122366. - Reinhard Zumkeller, Aug 30 2006
Hankel transform of A076035. - Philippe Deléham, Feb 28 2009
Equals the Catalan sequence: (1, 1, 2, 5, 14, ...), convolved with A032443: (1, 3, 11, 42, ...). - Gary W. Adamson, May 15 2009
Sum of coefficients of expansion of (1 + x + x^2 + x^3)^n.
a(n) is number of compositions of natural numbers into n parts less than 4. For example, a(2) = 16 since there are 16 compositions of natural numbers into 2 parts less than 4.
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 >= 1, a(n) equals the number of 4-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Squares in A002984. - Reinhard Zumkeller, Dec 28 2011
Row sums of Pascal's triangle using the rule that going left increases the value by a factor of k = 3. For example, the first three rows are {1}, {3, 1}, and {9, 6, 1}. Using this rule gives row sums as (k+1)^n. - Jon Perry, Oct 11 2012
First differences of A002450. - Omar E. Pol, Feb 20 2013
Sum of all peak heights in Dyck paths of semilength n+1. - David Scambler, Apr 22 2013
Powers of 4 exceed powers of 2 by A020522 which is the m-th oblong number A002378(m), m being the n-th Mersenne number A000225(n); hence, we may write, a(n) = A000079(n) + A002378(A000225(n)). - Lekraj Beedassy, Jan 17 2014
a(n) is equal to 1 plus the sum for 0 < k < 2^n of the numerators and denominators of the reduced fractions k/2^n. - J. M. Bergot, Jul 13 2015
Binomial transform of A000244. - Tony Foster III, Oct 01 2016
From Ilya Gutkovskiy, Oct 01 2016: (Start)
Number of nodes at level n regular 4-ary tree.
Partial sums of A002001. (End)
Satisfies Benford's law [Berger-Hill, 2011]. - N. J. A. Sloane, Feb 08 2017
Also the number of connected dominating sets in the (n+1)-barbell graph. - Eric W. Weisstein, Jun 29 2017
Side length of the cells at level n in a pyramid scheme where a square grid is decomposed into overlapping 2 X 2 blocks (cf. Kropatsch, 1985). - Felix Fröhlich, Jul 04 2019
a(n-1) is the number of 3-compositions of n; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020

References

  • H. W. Gould, Combinatorial Identities, 1972, eq. (1.93), p. 12.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, eq. (5.39), p. 187.
  • D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.
  • 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. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Crossrefs

Cf. A024036, A052539, A032443, A000351 (Binomial transform).
Cf. A249307.
Cf. A083420.

Programs

Formula

a(n) = 4^n.
a(0) = 1; a(n) = 4*a(n-1).
G.f.: 1/(1-4*x).
E.g.f.: exp(4*x).
a(n) = Sum_{k = 0..n} binomial(2k, k) * binomial(2(n - k), n - k). - Benoit Cloitre, Jan 26 2003 [See Graham et al., eq. (5.39), p. 187. - Wolfdieter Lang, Aug 16 2019]
1 = Sum_{n >= 1} 3/a(n) = 3/4 + 3/16 + 3/64 + 3/256 + 3/1024, ...; with partial sums: 3/4, 15/16, 63/64, 255/256, 1023/1024, ... - Gary W. Adamson, Jun 16 2003
a(n) = A001045(2*n) + A001045(2*n+1). - Paul Barry, Apr 27 2004
A000005(a(n)) = A005408(n+1). - Reinhard Zumkeller, Mar 04 2007
a(n) = Sum_{j = 0..n} 2^(n - j)*binomial(n + j, j). - Peter C. Heinig (algorithms(AT)gmx.de), Apr 06 2007
Hankel transform of A115967. - Philippe Deléham, Jun 22 2007
a(n) = 6*Stirling2(n+1, 4) + 6*Stirling2(n+1, 3) + 3*Stirling2(n+1, 2) + 1 = 2*Stirling2(2^n, 2^n - 1) + Stirling2(n+1, 2) + 1. - Ross La Haye, Jun 26 2008
a(n) = A159991(n)/A001024(n) = A047653(n) + A181765(n). A160700(a(n)) = A010685(n). - Reinhard Zumkeller, May 02 2009
a(n) = A188915(A006127(n)). - Reinhard Zumkeller, Apr 14 2011
a(n) = Sum_{k = 0..n} binomial(2*n+1, k). - Mircea Merca, Jun 25 2011
Sum_{n >= 1} Mobius(n)/a(n) = 0.1710822479183... - R. J. Mathar, Aug 12 2012
a(n) = Sum_{k = 0..n} binomial(2*k + x, k)*binomial(2*(n - k) - x, n - k) for every real number x. - Rui Duarte and António Guedes de Oliveira, Feb 16 2013
a(n) = 5*a(n - 1) - 4*a(n - 2). - Jean-Bernard François, Sep 12 2013
a(n) = (2*n+1) * binomial(2*n,n) * Sum_{j=0..n} (-1)^j/(2*j+1)*binomial(n,j). - Vaclav Kotesovec, Sep 15 2013
a(n) = A000217(2^n - 1) + A000217(2^n). - J. M. Bergot, Dec 28 2014
a(n) = (2^n)^2 = A000079(n)^2. - Doug Bell, Jun 23 2015
a(n) = A002063(n)/3 - A004171(n). - Zhandos Mambetaliyev, Nov 19 2016
a(n) = (1/2) * Product_{k = 0..n} (1 + (2*n + 1)/(2*k + 1)). - Peter Bala, Mar 06 2018
a(n) = A001045(n+1)*A001045(n+2) + A001045(n)^2. - Ezhilarasu Velayutham, Aug 30 2019
a(n) = 1 + 3*Sum_{k=0..n} binomial(2*n, n+k)*(k|9), where (k|9) is the Jacobi symbol. - Greg Dresden, Oct 11 2022
a(n) = Sum_{k = 0..n} binomial(2*n+1, 2*k) = Sum_{k = 0..n} binomial(2*n+1, 2*k+1). - Sela Fried, Mar 23 2023

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A000325 a(n) = 2^n - n.

Original entry on oeis.org

1, 1, 2, 5, 12, 27, 58, 121, 248, 503, 1014, 2037, 4084, 8179, 16370, 32753, 65520, 131055, 262126, 524269, 1048556, 2097131, 4194282, 8388585, 16777192, 33554407, 67108838, 134217701, 268435428, 536870883, 1073741794, 2147483617
Offset: 0

Views

Author

Rosario Salamone (Rosario.Salamone(AT)risc.uni-linz.ac.at)

Keywords

Comments

Number of permutations of degree n with at most one fall; called "Grassmannian permutations" by Lascoux and Schützenberger. - Axel Kohnert (Axel.Kohnert(AT)uni-bayreuth.de)
Number of different permutations of a deck of n cards that can be produced by a single shuffle. [DeSario]
Number of Dyck paths of semilength n having at most one long ascent (i.e., ascent of length at least two). Example: a(4)=12 because among the 14 Dyck paths of semilength 4, the only paths that have more than one long ascent are UUDDUUDD and UUDUUDDD (each with two long ascents). Here U = (1, 1) and D = (1, -1). Also number of ordered trees with n edges having at most one branch node (i.e., vertex of outdegree at least two). - Emeric Deutsch, Feb 22 2004
Number of {12,1*2*,21*}-avoiding signed permutations in the hyperoctahedral group.
Number of 1342-avoiding circular permutations on [n+1].
2^n - n is the number of ways to partition {1, 2, ..., n} into arithmetic progressions, where in each partition all the progressions have the same common difference and have lengths at least 1. - Marty Getz (ffmpg1(AT)uaf.edu) and Dixon Jones (fndjj(AT)uaf.edu), May 21 2005
if b(0) = x and b(n) = b(n-1) + b(n-1)^2*x^(n-2) for n > 0, then b(n) is a polynomial of degree a(n). - Michael Somos, Nov 04 2006
The chromatic invariant of the Mobius ladder graph M_n for n >= 2. - Jonathan Vos Post, Aug 29 2008
Dimension sequence of the dual alternative operad (i.e., associative and satisfying the identity xyz + yxz + zxy + xzy + yzx + zyx = 0) over the field of characteristic 3. - Pasha Zusmanovich, Jun 09 2009
An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 26 and 176, lead to this sequence (without the first leading 1). For the central square these vectors lead to the companion sequence A168604. - Johannes W. Meijer, Aug 15 2010
a(n+1) is also the number of order-preserving and order-decreasing partial isometries (of an n-chain). - Abdullahi Umar, Jan 13 2011
A040001(n) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, May 12 2012
A130103(n+1) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, May 12 2012
The number of labeled graphs with n vertices whose vertex set can be partitioned into a clique and a set of isolated points. - Alex J. Best, Nov 20 2012
For n > 0, a(n) is a B_2 sequence. - Thomas Ordowski, Sep 23 2014
See coefficients of the linear terms of the polynomials of the table on p. 10 of the Getzler link. - Tom Copeland, Mar 24 2016
Consider n points lying on a circle, then for n>=2 a(n-1) is the maximum number of ways to connect two points with non-intersecting chords. - Anton Zakharov, Dec 31 2016
Also the number of cliques in the (n-1)-triangular honeycomb rook graph. - Eric W. Weisstein, Jul 14 2017
From Eric M. Schmidt, Jul 17 2017: (Start)
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) < e(k). [Martinez and Savage, 2.7]
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i), e(j), e(k) pairwise distinct. [Martinez and Savage, 2.7]
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(j) >= e(k) and e(i) != e(k) pairwise distinct. [Martinez and Savage, 2.7]
(End)
Number of F-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are F-equivalent iff the positions of pattern F are identical in these paths. - Sergey Kirgizov, Apr 08 2018
From Gus Wiseman, Feb 10 2019: (Start)
Also the number of connected partitions of an n-cycle. For example, the a(1) = 1 through a(4) = 12 connected partitions are:
{{1}} {{12}} {{123}} {{1234}}
{{1}{2}} {{1}{23}} {{1}{234}}
{{12}{3}} {{12}{34}}
{{13}{2}} {{123}{4}}
{{1}{2}{3}} {{124}{3}}
{{134}{2}}
{{14}{23}}
{{1}{2}{34}}
{{1}{23}{4}}
{{12}{3}{4}}
{{14}{2}{3}}
{{1}{2}{3}{4}}
(End)
Number of subsets of n-set without the single-element subsets. - Yuchun Ji, Jul 16 2019
For every prime p, there are infinitely many terms of this sequence that are divisible by p (see IMO Compendium link and Doob reference). Corresponding indices n are: for p = 2, even numbers A299174; for p = 3, A047257; for p = 5, A349767. - Bernard Schott, Dec 10 2021
Primes are in A081296 and corresponding indices in A048744. - Bernard Schott, Dec 12 2021

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 12*x^4 + 27*x^5 + 58*x^6 + 121*x^7 + ...
		

References

  • Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1983, page 158, 1993.

Crossrefs

Column 1 of triangle A008518.
Row sum of triangles A184049 and A184050.

Programs

  • Haskell
    a000325 n = 2 ^ n - n
    a000325_list = zipWith (-) a000079_list [0..]
    -- Reinhard Zumkeller, Jul 17 2012
    
  • Magma
    [2^n - n: n in [0..35]]; // Vincenzo Librandi, May 13 2011
    
  • Maple
    A000325 := proc(n) option remember; if n <=1 then n+1 else 2*A000325(n-1)+n-1; fi; end;
    g:=1/(1-2*z): gser:=series(g, z=0, 43): seq(coeff(gser, z, n)-n, n=0..31); # Zerinvary Lajos, Jan 09 2009
  • Mathematica
    Table[2^n - n, {n, 0, 39}] (* Alonso del Arte, Sep 15 2014 *)
    LinearRecurrence[{4, -5, 2}, {1, 2, 5}, {0, 20}] (* Eric W. Weisstein, Jul 14 2017 *)
  • PARI
    {a(n) = 2^n - n}; /* Michael Somos, Nov 04 2006 */
    
  • Python
    def A000325(n): return (1<Chai Wah Wu, Jan 11 2023

Formula

a(n+1) = 2*a(n) + n - 1, a(0) = 1. - Reinhard Zumkeller, Apr 12 2003
Binomial transform of 1, 0, 1, 1, 1, .... The sequence starting 1, 2, 5, ... has a(n) = 1 + n + 2*Sum_{k=2..n} binomial(n, k) = 2^(n+1) - n - 1. This is the binomial transform of 1, 1, 2, 2, 2, 2, .... a(n) = 1 + Sum_{k=2..n} C(n, k). - Paul Barry, Jun 06 2003
G.f.: (1-3x+3x^2)/((1-2x)*(1-x)^2). - Emeric Deutsch, Feb 22 2004
A107907(a(n+2)) = A000051(n+2) for n > 0. - Reinhard Zumkeller, May 28 2005
a(n+1) = sum of n-th row of the triangle in A109128. - Reinhard Zumkeller, Jun 20 2005
Row sums of triangle A133116. - Gary W. Adamson, Sep 14 2007
G.f.: 1 / (1 - x / (1 - x / ( 1 - x / (1 + x / (1 - 2*x))))). - Michael Somos, May 12 2012
First difference is A000225. PSUM transform is A084634. - Michael Somos, May 12 2012
a(n) = [x^n](B(x)^n-B(x)^(n-1)), n>0, a(0)=1, where B(x) = (1+2*x+sqrt(1+4*x^2))/2. - Vladimir Kruchinin, Mar 07 2014
E.g.f.: (exp(x) - x)*exp(x). - Ilya Gutkovskiy, Aug 07 2016
a(n) = A125128(n) - A000225(n) + 1. - Miquel Cerda, Aug 12 2016
a(n) = 2*A125128(n) - A095151(n) + 1. - Miquel Cerda, Aug 12 2016
a(n) = A079583(n-1) - A000225(n-1). - Miquel Cerda, Aug 15 2016
a(n)^2 - 4*a(n-1)^2 = (n-2)*(a(n)+2*a(n-1)). - Yuchun Ji, Jul 13 2018
a(n) = 2^(-n) * A186947(n) = 2^n * A002064(-n) for all n in Z. - Michael Somos, Jul 18 2018
a(2^n) = (2^a(n) - 1)*2^n. - Lorenzo Sauras Altuzarra, Feb 01 2022

A003261 Woodall (or Riesel) numbers: n*2^n - 1.

Original entry on oeis.org

1, 7, 23, 63, 159, 383, 895, 2047, 4607, 10239, 22527, 49151, 106495, 229375, 491519, 1048575, 2228223, 4718591, 9961471, 20971519, 44040191, 92274687, 192937983, 402653183, 838860799, 1744830463, 3623878655, 7516192767, 15569256447, 32212254719, 66571993087
Offset: 1

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Author

Keywords

Comments

For n>1, a(n) is base at which zero is reached for the function "write f(j) in base j, read as base j+1 and then subtract 1 to give f(j+1)" starting from f(n) = n^2 - 1. - Henry Bottomley, Aug 06 2000
Sequence corresponds also to the maximum chain length of the classic puzzle whereby, under agreed commercial terms, an asset of unringed golden chain, when judiciously fragmented into as few as n pieces and n-1 opened links (through n-1 cuts), might be used to settle debt sequentially, with a golden link covering for unit cost. Here beside the n-1 opened links, the n fragmented pieces have lengths n, 2*n, 4*n, ..., 2^(n-1)*n. For instance, the chain of original length a(5)=159, if segregated by 4 cuts into 5+1+10+1+20+1+40+1+80, may be used to pay sequentially, i.e., a link-cost at a time, for an equivalent cost up to 159 links, to the same creditor. - Lekraj Beedassy, Feb 06 2003

Examples

			G.f. = x + 7*x^2 + 23*x^3 + 63*x^4 + 159*x^5 + 383*x^6 + 895*x^7 + ... - _Michael Somos_, Nov 04 2018
		

References

  • A. Brousseau, Number Theory Tables. Fibonacci Association, San Jose, CA, 1973, p. 159.
  • K. R. Bhutani and A. B. Levin, "The Problem of Sawing a Chain", Journal of Recreational Mathematics 2002-3 31(1) 32-35.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • M. Gardner, Martin Gardner's Sixth Book of Mathematical Diversions from Scientific American, "Gold Links", Problem 4, pp. 50-51; 57-58, University of Chicago Press, 1983.
  • O. O'Shea, Mathematical Brainteasers with Surprising Solutions, Problem 76, pp. 183-185, Prometheus Books, Guilford, Connecticut, 2020.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 241.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = A036289(n) - 1 = A002064(n) - 2.
Cf. A133653.

Programs

Formula

G.f.: x*(-1-2*x+4*x^2) / ( (x-1)*(-1+2*x)^2 ). - Simon Plouffe in his 1992 dissertation
Binomial transform of A133653 and double binomial transform of [1, 5, -1, 1, -1, 1, ...]. - Gary W. Adamson, Sep 19 2007
a(n) = -(2)^n * A006127(-n) for all n in Z. - Michael Somos, Nov 04 2018
E.g.f.: 1 + exp(x)*(2*exp(x)*x - 1). - Stefano Spezia, Nov 24 2024

A094373 Expansion of (1-x-x^2)/((1-x)*(1-2*x)).

Original entry on oeis.org

1, 2, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025, 2049, 4097, 8193, 16385, 32769, 65537, 131073, 262145, 524289, 1048577, 2097153, 4194305, 8388609, 16777217, 33554433, 67108865, 134217729, 268435457, 536870913, 1073741825, 2147483649, 4294967297, 8589934593
Offset: 0

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Author

Paul Barry, Apr 28 2004

Keywords

Comments

Partial sum of 1,1,1,2,4,8,...
Binomial transform of abs(A073097).
Binomial transform is A094374.
Partial sums are in A006127. - Paul Barry, Aug 05 2004
An elephant sequence, see A175654. For the corner squares four A[5] vectors, with decimal values 2, 8, 32 and 128, lead to this sequence. For the central square these vectors lead to the companion sequence A011782. - Johannes W. Meijer, Aug 15 2010
This sequence has a(0) = 1 and for all n > 0, a(n) = 2^(n-1)+1. Consequently 2*a(n) >= a(n+1) for all n > 0 and the sequence is complete. - Frank M Jackson, Jan 29 2012
Row lengths of the triangle in A198069. - Reinhard Zumkeller, May 26 2013
Take A007843 and count the repeated values. The result is 1,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,.... Build a third sequence, where a(1) = 1 and a(n) equals the length (greater than 1) of the shortest palindromic subsequence of consecutive terms of the second sequence starting with a(n) of the second sequence. The third sequence starts 1,3,5,3,9,3,5,3,17,3,5,3,9,3,5,3,33,.... Conjecturally, in the third sequence: (1) the indices of the first occurrence of each value form the present sequence and (2) for n>1, a(n) is in the a(n-1)-th position. - Ivan N. Ianakiev, Aug 20 2019

Examples

			G.f. = 1 + 2*x + 3*x^2 + 5*x^3 + 9*x^4 + 17*x^5 + 33*x^6 + 65*x^7 + ...
		

Crossrefs

Apart from the initial 1, identical to A000051.
Cf. A135225.
Column k=1 of A152977.
Row n=2 of A238016.

Programs

  • GAP
    a:=[2,3];; for n in [3..40] do a[n]:=3*a[n-1]-2*a[n-2]; od; Concatenation([1], a); # G. C. Greubel, Nov 06 2019
  • Magma
    [(2^n-0^n)/2+1: n in [0..40]]; // Vincenzo Librandi, Jun 10 2011
    
  • Magma
    R:=PowerSeriesRing(Integers(), 35); Coefficients(R!( (1-x-x^2)/((1-x)*(1-2*x)))); // Marius A. Burtea, Oct 25 2019
    
  • Maple
    1, seq((2^n - 0^n)/2 +1, n=1..40); # G. C. Greubel, Nov 06 2019
  • Mathematica
    CoefficientList[Series[(1-x-x^2)/((1-x)*(1-2*x)), {x, 0, 40}], x] (* or *) Join[{1}, LinearRecurrence[{3, -2}, {2, 3}, 40]] (* Vladimir Joseph Stephan Orlovsky, Jan 22 2012 *)
    a[ n_]:= If[n<0, 0, 1 + Quotient[2^n, 2]]; (* Michael Somos, May 26 2014 *)
    a[ n_]:= SeriesCoefficient[(1-x-x^2)/((1-x)(1-2x)), {x, 0, n}]; (* Michael Somos, May 26 2014 *)
    LinearRecurrence[{3,-2},{1,2,3},40] (* Harvey P. Dale, Aug 09 2015 *)
  • PARI
    a(n)=2^n\2+1 \\ Charles R Greathouse IV, Apr 05 2013
    
  • PARI
    Vec((1-x-x^2)/((1-x)*(1-2*x))+O(x^40)) \\ Charles R Greathouse IV, Apr 05 2013
    
  • Sage
    [(2^n - 0^n)/2 + 1 for n in (0..40)] # G. C. Greubel, Nov 06 2019
    

Formula

a(n) = (2^n - 0^n)/2 + 1.
a(n) = 3*a(n-1) - 2*a(n-2).
a(2*n) = 2*a(2*n-1) - 1, n>0.
Row sums of triangle A135225. - Gary W. Adamson, Nov 23 2007
a(n) = A131577(n) + 1. - Paul Curtz, Aug 07 2008
a(n) = 2*a(n-1) - 1 for n>1, a(0)=1, a(1)=2. - Philippe Deléham, Sep 25 2009
E.g.f.: exp(x)*(1 + sinh(x)). - Arkadiusz Wesolowski, Aug 13 2012
G.f.: G(0), where G(k)= 1 + 2^k*x/(1 - x/(x + 2^k*x/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 26 2013
a(n) = 2^(n-1) +1 = A000051(n-1) for n>0. - M. F. Hasler, Sep 22 2013

A052944 a(n) = 2^n + n - 1.

Original entry on oeis.org

0, 2, 5, 10, 19, 36, 69, 134, 263, 520, 1033, 2058, 4107, 8204, 16397, 32782, 65551, 131088, 262161, 524306, 1048595, 2097172, 4194325, 8388630, 16777239, 33554456, 67108889, 134217754, 268435483, 536870940, 1073741853, 2147483678
Offset: 0

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Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Shortest length of bit-string containing all bit-strings of given length n. - Rainer Rosenthal, Apr 30 2003
Such a bitstring can be obtained by taking a length-2^n binary de Bruijn sequence and repeating the n-1 initial symbols at the end. - Joerg Arndt, Mar 16 2015
Bit string definition is equivalent to minimum number of tosses of a coin to achieve all possible outcomes of n tosses. - Maurizio De Leo, Mar 01 2015
Also the indices of Fermat numbers that can be represented as cyclotomic numbers. Specifically, F(a(n)) = cyclotomic(2^2^n,2^2^n). - T. D. Noe, Oct 17 2003
a(n) = A006127(n) - 1. - Reinhard Zumkeller, Apr 13 2011
Randomly select (with uniform distribution) a length n binary word w. a(n) is apparently the expected wait time for the first occurrence of w over all infinite binary sequences. For example: a(4)=19. We consider A005434(4)=4 distinct classes of length 4 binary words that share the same autocorrelation. There are A003000(4)=6 words that have waiting time = 16; 2 words with waiting time =20; 6 words with waiting time = 18; and 2 words with waiting time =30. 1/16*(6*16 + 2*20 + 6*18 + 2*30) = 19. - Geoffrey Critzer, Feb 27 2014

Examples

			a(3) = 10 because "0001110100" has length 10 and contains all possible patterns of 3 bits:
  0001110100
  ----------
  000.......
  .001......
  ......010.
  ..011.....
  .......100
  .....101..
  ....110...
  ...111....
		

References

  • Discussed in newsgroup de.rec.denksport in Apr 2003
  • N. G. de Bruijn: A combinatorial problem. Indagationes Math. 8 (1946), pp. 461-467.

Crossrefs

Programs

Formula

G.f.: (2-3*x)/((1-2*x)*(1-x)^2).
a(n+1) = 2*a(n) - n + 2 with a(0)=0. - Pieter Moree, Mar 06 2004
For n>=1: partial sums of A000051. - Emeric Deutsch, Mar 04 2004
a(0)=0, a(1)=2, a(2)=5, a(n+3) = 4*a(n+2) - 5*a(n+1) + 2*a(n). - Hermann Kremer (Hermann.Kremer(AT)online.de), Mar 16 2004
a(n) = A000225(n) + n. - Zerinvary Lajos, May 29 2009
E.g.f.: U(0), where U(k) = 1 + x/(2^k + 2^k/(x - 1 - x^2*2^(k+1)/(x*2^(k+1) - (k+1)/U(k+1) )));(continued fraction, 3rd kind, 4-step ). - Sergei N. Gladkovskii, Dec 01 2012
G.f.: G(0)*x/(1-x) where G(k) = 1 + 2^k/(1 - x/(x + 2^k/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
G.f.: Q(0)*x/(1-x), 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 24 2013
E.g.f.: exp(2*x) - (1-x)*exp(x). - G. C. Greubel, Oct 18 2019

A052007 Numbers m such that 2^m + m is prime.

Original entry on oeis.org

1, 3, 5, 9, 15, 39, 75, 81, 89, 317, 701, 735, 1311, 1881, 3201, 3225, 11795, 88071, 204129, 678561
Offset: 1

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Comments

Terms >= 701 are currently only strong pseudoprimes.
If m=1 (mod 6) or m=2 (mod 6) then 3 divides 2^m+m. Thus for n > 1, a(n)!=1 (mod 6) and a(n)!=2 (mod 6).
Some of the results were computed using the PrimeFormGW (PFGW) primality-testing program. - Hugo Pfoertner, Nov 14 2019
Keller (see Links) notes that a Mersenne number M(2^m+m) = 2^(2^m+m) - 1 can be written as (2^m)*2^(2^m) - 1, and lists the first twelve terms of this sequence. The last known case where M(2^m+m) is prime is for m=a(4)=9, which gives the prime M(521). - Jeppe Stig Nielsen, Apr 20 2021

Examples

			2^39 + 39 = 549755813927 is prime.
		

Crossrefs

Programs

  • Mathematica
    Do[ If[ PrimeQ[ 2^n + n ], Print[ n ] ], {n, 0, 7000} ]
    v={1}; Do[If[Mod[n, 2]*(Mod[n, 6]-1)!= 0&&PrimeQ[2^n+n], v=Append[v, n]; Print[v]], {n, 2, 20000}]
  • PARI
    is(n)=isprime(2^n+n) \\ Charles R Greathouse IV, Feb 09 2017

Extensions

11795 from Farideh Firoozbakht, Aug 21 2003
88071 from Hugo Pfoertner, Dec 26 2004
More terms from Henri Lifchitz submitted by Ray Chandler, Mar 02 2007

A104745 a(n) = 5^n + n.

Original entry on oeis.org

1, 6, 27, 128, 629, 3130, 15631, 78132, 390633, 1953134, 9765635, 48828136, 244140637, 1220703138, 6103515639, 30517578140, 152587890641, 762939453142, 3814697265643, 19073486328144, 95367431640645, 476837158203146, 2384185791015647, 11920928955078148, 59604644775390649
Offset: 0

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Author

Zak Seidov, Mar 23 2005

Keywords

Comments

Numbers m=5^n+n such that equation x=5^(m-x) has solution x=5^n, see A104744.
No primes of the form 5^n+n for n < 7954. - Thomas Ordowski, Oct 28 2013
a(7954) is prime (5560 digits). - Thomas Ordowski, May 07 2015

Crossrefs

Programs

Formula

From Vincenzo Librandi, Jun 16 2013: (Start)
G.f.: (1-x-4*x^2)/((1-5*x)*(1-x)^2).
a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3). (End)
E.g.f.: exp(x)*(exp(4*x) + x). - Elmo R. Oliveira, Mar 05 2025

Extensions

More terms from Jonathan R. Love (japanada11(AT)yahoo.ca), Mar 09 2007
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