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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A356447 Integers k such that (k+1)*(2*k-1) does not divide the central binomial coefficient B(k) = binomial(2*k,k) = A000984(k).

Original entry on oeis.org

2, 5, 8, 11, 14, 26, 29, 32, 35, 38, 41, 80, 83, 86, 89, 92, 95, 107, 110, 113, 116, 119, 122, 242, 245, 248, 251, 254, 257, 269, 272, 275, 278, 281, 284, 323, 326, 329, 332, 335, 338, 350, 353, 356, 359, 362, 365, 728, 731, 734, 737, 740, 743, 755, 758, 761
Offset: 1

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Author

Valerio De Angelis, Aug 07 2022

Keywords

Comments

It is well known that B(k) divided by (k+1) is an integer (the Catalan numbers A000108). It is also easy to see that (2k-1) divides B(k). So we ask when the product (k+1)*(2k-1) divides B(k). The terms of this sequence are the positive integers k such that (k+1)*(2k-1) does not divide B(k).
A necessary and sufficient condition for an integer k to be a term of this sequence is: k is congruent to 2 (mod 3), and at least one of (k+1) or (k-1) has no 2's in its base-3 expansion. In particular, this sequence has density 0. This is proved in the Stack Exchange post cited below.
Other equivalent conditions are:
1) k is congruent to 2 (mod 3), and its base-3 expansion either has no 2's, or is of form u12, or u02^i for some i>=1, where u has no 2's and 2^i means a string of i consecutive 2's.
2) the base 3 expansion of k+1 is either u0 or u20, where u has no 2's.

Examples

			k = 95 is a term, since it is k == 2 (mod 3) and k-1 = 94 has base-3 expansion 10111 which has no digit 2's. It can be checked that B(k) = binomial(190,95) is not divisible by (k+1)*(2*k-1) = 18144.
As another example, a(18)=107, also congruent to 2 (mod 3), and 107+1=108 has base-3 expansion 11000. It can be checked that binomial(2*107,107) is not divisible by (107+1)*(2*107-1).
125 is not a term of the sequence, because even though it's congruent to 2 (mod 3), the base-3 expansions of 125+1=126 and 125-1=124 are 11200 and 11121. It can be checked that binomial(2*125,125) is divisible by (125+1)*(2*125-1).
		

Crossrefs

Programs

  • Mathematica
    kmax=762; a={}; For[k=1, k<=kmax, k++, If[Not[Divisible[Binomial[2k, k], (k+1)(2k-1)]], AppendTo[a, k]]]; a (* Stefano Spezia, Aug 12 2022 *)
  • PARI
    isok(k) = Mod(binomial(2*k,k), (k+1)*(2*k-1)) != 0; \\ Michel Marcus, Aug 10 2022
    
  • Python
    def A356447(n):
        a, b = divmod(n,3)
        return (int(bin(a)[2:],3)*3+b)*3-1 # Chai Wah Wu, Jul 29 2025

Formula

a(n) = 3*A096304(n) - 1.

A374891 Obverse convolution (1)**A000984; see Comments.

Original entry on oeis.org

2, 6, 42, 882, 62622, 15843366, 14655113550, 50311004817150, 647552943001537650, 31484671641677762080650, 5817013478501458288734652050, 4103513269179719224996951799587650, 11096544131445222000310082187517540861050
Offset: 0

Views

Author

Clark Kimberling, Jul 31 2024

Keywords

Comments

See A374848 for the definition of obverse convolution and a guide to related sequences. This is a divisibility sequence (see Formula).

Crossrefs

Programs

  • Mathematica
    s[n_] := 1; t[n_] := Binomial[2 n, n];
    u[n_] := Product[s[k] + t[n - k], {k, 0, n}]
    Table[u[n], {n, 0, 20}]

Formula

a(n+1) = a(n)*A244174(n+1) for n>=0 (conjectured) = a(n)*A323230(n+2) for n>=0 (conjectured).
a(n) ~ c * A007685(n), where c = Product_{k=0..oo} (1 + 1/binomial(2*k,k)) = 3.74782908533723753117687910314018231428739915473496578523053032212205053... - Vaclav Kotesovec, Jul 31 2024

A374893 Obverse convolution (n)**A000984; see Comments.

Original entry on oeis.org

1, 4, 54, 2240, 294000, 127534176, 189691951680, 994556567347200, 18760374808177860000, 1292781708803753625600000, 329281303607781567523264143360, 312866979311142515971112580848148480, 1117139877664193310898467728367016525532160
Offset: 0

Views

Author

Clark Kimberling, Sep 14 2024

Keywords

Comments

See A374848 for the definition of obverse convolution and a guide to related sequences.

Crossrefs

Programs

  • Mathematica
    s[n_] := n; t[n_] := Binomial[2 n, n];
    u[n_] := Product[s[k] + t[n - k], {k, 0, n}]
    Table[u[n], {n, 0, 14}]

A066397 Reversion of 1 / sqrt(1 - 4*x) (the o.g.f. for the central binomial coefficients A000984) as if it were an e.g.f.

Original entry on oeis.org

1, -2, 6, -20, 50, 168, -4732, 54024, -356670, -1558040, 106069172, -2197188864, 26605890220, 22266781600, -12120090377400, 402165029201744, -7732409047854318, 38209542402620232, 4126723132306766900, -211906442415502467360, 5955735723962955009156
Offset: 0

Views

Author

N. J. A. Sloane, Dec 24 2001

Keywords

Crossrefs

Cf. A000984.

Programs

  • Maple
    read transforms; t1 := 1 / sqrt(1 - 4*x); t2 := series(t1,x,50); t2 := seriestolist(%); REVEGF(%); # corrected by Sean A. Irvine, Oct 12 2023

Extensions

Name corrected by Sean A. Irvine, Oct 12 2023

A086359 Fixed point if [decimal-digit-sum]-function at initial-value=A000984(n)=C[2n,n] is iterated.

Original entry on oeis.org

2, 6, 2, 7, 9, 6, 3, 9, 2, 4, 3, 7, 2, 9, 9, 9, 9, 6, 3, 9, 3, 6, 9, 9, 9, 9, 2, 4, 3, 4, 5, 9, 3, 6, 9, 7, 5, 6, 2, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 6, 3, 9, 3, 6, 9, 9, 9, 9, 3, 6, 9, 6, 3, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 2, 4, 3, 4, 5, 9, 3, 6, 9, 4, 8, 6, 5, 4, 9, 9, 9, 9, 3, 6, 9, 6, 3, 9, 9
Offset: 1

Views

Author

Labos Elemer, Jul 21 2003

Keywords

Examples

			n=10, C[20,10]=184756, iteration list={184756,31,4},a(10)=4.
		

Crossrefs

Programs

  • Mathematica
    sud[x_] := Apply[Plus, DeleteCases[IntegerDigits[x], 0]] Table[FixedPoint[sud, Binomial[2*w, w]], {w, 1, 128}]

Formula

a(n)=A010888[C[2n, n]]=fixed-point of A007953[C[2n, n]]; It equals C[2n, n] modulo(9); at r=0 use 9.

A096971 G.f. satisfies: A(x) = A( x^2*A000108(x^2) )*x*A000984(x^2), where A000108(x) is the g.f. for the Catalan sequence and A000984(x) = d/dx x*A000108(x).

Original entry on oeis.org

1, 1, 4, 13, 49, 181, 685, 2605, 9988, 38479, 148879, 577930, 2249698, 8777614, 34315012, 134377393, 526994773, 2069403898, 8135377102, 32014655626, 126099239329, 497083313908, 1960943833567, 7740893831005, 30576064032568
Offset: 0

Views

Author

Paul D. Hanna, Jul 16 2004

Keywords

Examples

			A(x) = x^-1 + x + 4*x^3 + 13*x^5 + 49*x^7 + 181*x^9 + 685*x^11 +...
		

Programs

  • PARI
    a(n)=if(n==0 || n==1,1,sum(j=0,(n+1)\2,binomial(2*n-2*j-1,n-1)*a(j)))

Formula

a(n) = Sum_{j=0..[(n+1)/2]} C(2*n-2*j-1, n-1)*a(j), a(0)=a(1)=1. G.f. satisfies: A(x) = A( (1-sqrt(1-4*x^2))/2 )*x/sqrt(1-4*x^2), where A(x) = Sum_{n>=0} a(n)*x^(2*n-1).

A173017 A006987(a(n))=A000984(n).

Original entry on oeis.org

1, 4, 7, 19, 25, 56, 73, 144, 180, 346, 430, 826
Offset: 2

Views

Author

Dylan Hamilton, Nov 06 2010

Keywords

A208890 a(n) = A000984(n)*A004981(n), the term-wise product of the coefficients in (1-4*x)^(-1/2) and (1-8*x)^(-1/4).

Original entry on oeis.org

1, 4, 60, 1200, 27300, 668304, 17153136, 455083200, 12372574500, 342766138000, 9638583800560, 274341178587840, 7887308884400400, 228685287180840000, 6678543795015960000, 196260140322869011200, 5798873833602270315300, 172160337343624495866000
Offset: 0

Views

Author

Paul D. Hanna, Mar 04 2012

Keywords

Comments

The sequences A000984 and A004981 are related by the aesthetic identity:
Sum_{n>=0} A000984(n)^3 *x^n = ( Sum_{n>=0} A004981(n)^2 *x^n )^2.

Examples

			G.f.: A(x) = 1 + 4*x + 60*x^2 + 1200*x^3 + 27300*x^4 + 668304*x^5 +...
The terms are the term-wise products of the sequences:
A000984 = [1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, ...],
A004981 = [1, 2, 10, 60, 390, 2652, 18564, 132600, 961350, ...].
Related sequences:
A^2: [1, 8, 136, 2880, 67800, 1699008, 44368704, 1193107968, ...],
A^4: [1, 16, 336, 7936, 200176, 5266176, 142657536, 3948773376, ...],
A^8: [1, 32, 928, 26624, 767200, 22270976, 651331072, 19178651648, ...].
		

Crossrefs

Programs

  • PARI
    {A000984(n)=polcoeff((1-4*x +x*O(x^n))^(-1/2),n)}
    {A004981(n)=polcoeff((1-8*x +x*O(x^n))^(-1/4),n)}
    {a(n)=A000984(n)*A004981(n)}
    for(n=0,20,print1(a(n),", "))

A000108 Catalan numbers: C(n) = binomial(2n,n)/(n+1) = (2n)!/(n!(n+1)!).

Original entry on oeis.org

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152, 69533550916004, 263747951750360, 1002242216651368, 3814986502092304
Offset: 0

Views

Author

Keywords

Comments

These were formerly sometimes called Segner numbers.
A very large number of combinatorial interpretations are known - see references, esp. R. P. Stanley, "Catalan Numbers", Cambridge University Press, 2015. This is probably the longest entry in the OEIS, and rightly so.
The solution to Schröder's first problem: number of ways to insert n pairs of parentheses in a word of n+1 letters. E.g., for n=2 there are 2 ways: ((ab)c) or (a(bc)); for n=3 there are 5 ways: ((ab)(cd)), (((ab)c)d), ((a(bc))d), (a((bc)d)), (a(b(cd))).
Consider all the binomial(2n,n) paths on squared paper that (i) start at (0, 0), (ii) end at (2n, 0) and (iii) at each step, either make a (+1,+1) step or a (+1,-1) step. Then the number of such paths that never go below the x-axis (Dyck paths) is C(n). [Chung-Feller]
Number of noncrossing partitions of the n-set. For example, of the 15 set partitions of the 4-set, only [{13},{24}] is crossing, so there are a(4)=14 noncrossing partitions of 4 elements. - Joerg Arndt, Jul 11 2011
Noncrossing partitions are partitions of genus 0. - Robert Coquereaux, Feb 13 2024
a(n-1) is the number of ways of expressing an n-cycle (123...n) in the symmetric group S_n as a product of n-1 transpositions (u_1,v_1)*(u_2,v_2)*...*(u_{n-1},v_{n-1}) where u_iA000272. - Joerg Arndt and Greg Stevenson, Jul 11 2011
a(n) is the number of ordered rooted trees with n nodes, not including the root. See the Conway-Guy reference where these rooted ordered trees are called plane bushes. See also the Bergeron et al. reference, Example 4, p. 167. - Wolfdieter Lang, Aug 07 2007
As shown in the paper from Beineke and Pippert (1971), a(n-2)=D(n) is the number of labeled dissections of a disk, related to the number R(n)=A001761(n-2) of labeled planar 2-trees having n vertices and rooted at a given exterior edge, by the formula D(n)=R(n)/(n-2)!. - M. F. Hasler, Feb 22 2012
Shifts one place left when convolved with itself.
For n >= 1, a(n) is also the number of rooted bicolored unicellular maps of genus 0 on n edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 15 2001
Number of ways of joining 2n points on a circle to form n nonintersecting chords. (If no such restriction imposed, then the number of ways of forming n chords is given by (2n-1)!! = (2n)!/(n!*2^n) = A001147(n).)
Arises in Schubert calculus - see Sottile reference.
Inverse Euler transform of sequence is A022553.
With interpolated zeros, the inverse binomial transform of the Motzkin numbers A001006. - Paul Barry, Jul 18 2003
The Hankel transforms of this sequence or of this sequence with the first term omitted give A000012 = 1, 1, 1, 1, 1, 1, ...; example: Det([1, 1, 2, 5; 1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132]) = 1 and Det([1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132; 14, 42, 132, 429]) = 1. - Philippe Deléham, Mar 04 2004
a(n) equals the sum of squares of terms in row n of triangle A053121, which is formed from successive self-convolutions of the Catalan sequence. - Paul D. Hanna, Apr 23 2005
Also coefficients of the Mandelbrot polynomial M iterated an infinite number of times. Examples: M(0) = 0 = 0*c^0 = [0], M(1) = c = c^1 + 0*c^0 = [1 0], M(2) = c^2 + c = c^2 + c^1 + 0*c^0 = [1 1 0], M(3) = (c^2 + c)^2 + c = [0 1 1 2 1], ... ... M(5) = [0 1 1 2 5 14 26 44 69 94 114 116 94 60 28 8 1], ... - Donald D. Cross (cosinekitty(AT)hotmail.com), Feb 04 2005
The multiplicity with which a prime p divides C_n can be determined by first expressing n+1 in base p. For p=2, the multiplicity is the number of 1 digits minus 1. For p an odd prime, count all digits greater than (p+1)/2; also count digits equal to (p+1)/2 unless final; and count digits equal to (p-1)/2 if not final and the next digit is counted. For example, n=62, n+1 = 223_5, so C_62 is not divisible by 5. n=63, n+1 = 224_5, so 5^3 | C_63. - Franklin T. Adams-Watters, Feb 08 2006
Koshy and Salmassi give an elementary proof that the only prime Catalan numbers are a(2) = 2 and a(3) = 5. Is the only semiprime Catalan number a(4) = 14? - Jonathan Vos Post, Mar 06 2006
The answer is yes. Using the formula C_n = binomial(2n,n)/(n+1), it is immediately clear that C_n can have no prime factor greater than 2n. For n >= 7, C_n > (2n)^2, so it cannot be a semiprime. Given that the Catalan numbers grow exponentially, the above consideration implies that the number of prime divisors of C_n, counted with multiplicity, must grow without limit. The number of distinct prime divisors must also grow without limit, but this is more difficult. Any prime between n+1 and 2n (exclusive) must divide C_n. That the number of such primes grows without limit follows from the prime number theorem. - Franklin T. Adams-Watters, Apr 14 2006
The number of ways to place n indistinguishable balls in n numbered boxes B1,...,Bn such that at most a total of k balls are placed in boxes B1,...,Bk for k=1,...,n. For example, a(3)=5 since there are 5 ways to distribute 3 balls among 3 boxes such that (i) box 1 gets at most 1 ball and (ii) box 1 and box 2 together get at most 2 balls:(O)(O)(O), (O)()(OO), ()(OO)(O), ()(O)(OO), ()()(OOO). - Dennis P. Walsh, Dec 04 2006
a(n) is also the order of the semigroup of order-decreasing and order-preserving full transformations (of an n-element chain) - now known as the Catalan monoid. - Abdullahi Umar, Aug 25 2008
a(n) is the number of trivial representations in the direct product of 2n spinor (the smallest) representations of the group SU(2) (A(1)). - Rutger Boels (boels(AT)nbi.dk), Aug 26 2008
The invert transform appears to converge to the Catalan numbers when applied infinitely many times to any starting sequence. - Mats Granvik, Gary W. Adamson and Roger L. Bagula, Sep 09 2008, Sep 12 2008
Limit_{n->oo} a(n)/a(n-1) = 4. - Francesco Antoni (francesco_antoni(AT)yahoo.com), Nov 24 2008
Starting with offset 1 = row sums of triangle A154559. - Gary W. Adamson, Jan 11 2009
C(n) is the degree of the Grassmannian G(1,n+1): the set of lines in (n+1)-dimensional projective space, or the set of planes through the origin in (n+2)-dimensional affine space. The Grassmannian is considered a subset of N-dimensional projective space, N = binomial(n+2,2) - 1. If we choose 2n general (n-1)-planes in projective (n+1)-space, then there are C(n) lines that meet all of them. - Benji Fisher (benji(AT)FisherFam.org), Mar 05 2009
Starting with offset 1 = A068875: (1, 2, 4, 10, 18, 84, ...) convolved with Fine numbers, A000957: (1, 0, 1, 2, 6, 18, ...). a(6) = 132 = (1, 2, 4, 10, 28, 84) dot (18, 6, 2, 1, 0, 1) = (18 + 12 + 8 + 10 + 0 + 84) = 132. - Gary W. Adamson, May 01 2009
Convolved with A032443: (1, 3, 11, 42, 163, ...) = powers of 4, A000302: (1, 4, 16, ...). - Gary W. Adamson, May 15 2009
Sum_{k>=1} C(k-1)/2^(2k-1) = 1. The k-th term in the summation is the probability that a random walk on the integers (beginning at the origin) will arrive at positive one (for the first time) in exactly (2k-1) steps. - Geoffrey Critzer, Sep 12 2009
C(p+q)-C(p)*C(q) = Sum_{i=0..p-1, j=0..q-1} C(i)*C(j)*C(p+q-i-j-1). - Groux Roland, Nov 13 2009
Leonhard Euler used the formula C(n) = Product_{i=3..n} (4*i-10)/(i-1) in his 'Betrachtungen, auf wie vielerley Arten ein gegebenes polygonum durch Diagonallinien in triangula zerschnitten werden könne' and computes by recursion C(n+2) for n = 1..8. (Berlin, 4th September 1751, in a letter to Goldbach.) - Peter Luschny, Mar 13 2010
Let A179277 = A(x). Then C(x) is satisfied by A(x)/A(x^2). - Gary W. Adamson, Jul 07 2010
a(n) is also the number of quivers in the mutation class of type B_n or of type C_n. - Christian Stump, Nov 02 2010
From Matthew Vandermast, Nov 22 2010: (Start)
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1) equals the number of ways to choose 0 or more balls of each color while satisfying the following conditions: 1. No two colors are chosen the same positive number of times. 2. For any two colors (c, d) that are chosen at least once, color c is chosen more times than color d iff color c appears more times in the original set than color d.
If the second requirement is lifted, the number of acceptable ways equals A000110(n+1). See related comments for A016098, A085082. (End)
Deutsch and Sagan prove the Catalan number C_n is odd if and only if n = 2^a - 1 for some nonnegative integer a. Lin proves for every odd Catalan number C_n, we have C_n == 1 (mod 4). - Jonathan Vos Post, Dec 09 2010
a(n) is the number of functions f:{1,2,...,n}->{1,2,...,n} such that f(1)=1 and for all n >= 1 f(n+1) <= f(n)+1. For a nice bijection between this set of functions and the set of length 2n Dyck words, see page 333 of the Fxtbook (see link below). - Geoffrey Critzer, Dec 16 2010
Postnikov (2005) defines "generalized Catalan numbers" associated with buildings (e.g., Catalan numbers of Type B, see A000984). - N. J. A. Sloane, Dec 10 2011
Number of permutations in S(n) for which length equals depth. - Bridget Tenner, Feb 22 2012
a(n) is also the number of standard Young tableau of shape (n,n). - Thotsaporn Thanatipanonda, Feb 25 2012
a(n) is the number of binary sequences of length 2n+1 in which the number of ones first exceed the number of zeros at entry 2n+1. See the example below in the example section. - Dennis P. Walsh, Apr 11 2012
Number of binary necklaces of length 2*n+1 containing n 1's (or, by symmetry, 0's). All these are Lyndon words and their representatives (as cyclic maxima) are the binary Dyck words. - Joerg Arndt, Nov 12 2012
Number of sequences consisting of n 'x' letters and n 'y' letters such that (counting from the left) the 'x' count >= 'y' count. For example, for n=3 we have xxxyyy, xxyxyy, xxyyxy, xyxxyy and xyxyxy. - Jon Perry, Nov 16 2012
a(n) is the number of Motzkin paths of length n-1 in which the (1,0)-steps come in 2 colors. Example: a(4)=14 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have 8 paths of shape HHH, 2 paths of shape UHD, 2 paths of shape UDH, and 2 paths of shape HUD. - José Luis Ramírez Ramírez, Jan 16 2013
If p is an odd prime, then (-1)^((p-1)/2)*a((p-1)/2) mod p = 2. - Gary Detlefs, Feb 20 2013
Conjecture: For any positive integer n, the polynomial Sum_{k=0..n} a(k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
a(n) is the size of the Jones monoid on 2n points (cf. A225798). - James Mitchell, Jul 28 2013
For 0 < p < 1, define f(p) = Sum_{n>=0} a(n)*(p*(1-p))^n, then f(p) = min{1/p, 1/(1-p)}, so f(p) reaches its maximum value 2 at p = 0.5, and p*f(p) is constant 1 for 0.5 <= p < 1. - Bob Selcoe, Nov 16 2013 [Corrected by Jianing Song, May 21 2021]
No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
From Alexander Adamchuk, Dec 27 2013: (Start)
Prime p divides a((p+1)/2) for p > 3. See A120303(n) = Largest prime factor of Catalan number.
Reciprocal Catalan Constant C = 1 + 4*sqrt(3)*Pi/27 = 1.80613.. = A121839.
Log(Phi) = (125*C - 55) / (24*sqrt(5)), where C = Sum_{k>=1} (-1)^(k+1)*1/a(k). See A002390 = Decimal expansion of natural logarithm of golden ratio.
3-d analog of the Catalan numbers: (3n)!/(n!(n+1)!(n+2)!) = A161581(n) = A006480(n) / ((n+1)^2*(n+2)), where A006480(n) = (3n)!/(n!)^3 De Bruijn's S(3,n). (End)
For a relation to the inviscid Burgers's, or Hopf, equation, see A001764. - Tom Copeland, Feb 15 2014
From Fung Lam, May 01 2014: (Start)
One class of generalized Catalan numbers can be defined by g.f. A(x) = (1-sqrt(1-q*4*x*(1-(q-1)*x)))/(2*q*x) with nonzero parameter q. Recurrence: (n+3)*a(n+2) -2*q*(2*n+3)*a(n+1) +4*q*(q-1)*n*a(n) = 0 with a(0)=1, a(1)=1.
Asymptotic approximation for q >= 1: a(n) ~ (2*q+2*sqrt(q))^n*sqrt(2*q*(1+sqrt(q))) /sqrt(4*q^2*Pi*n^3).
For q <= -1, the g.f. defines signed sequences with asymptotic approximation: a(n) ~ Re(sqrt(2*q*(1+sqrt(q)))*(2*q+2*sqrt(q))^n) / sqrt(q^2*Pi*n^3), where Re denotes the real part. Due to Stokes' phenomena, accuracy of the asymptotic approximation deteriorates at/near certain values of n.
Special cases are A000108 (q=1), A068764 to A068772 (q=2 to 10), A240880 (q=-3).
(End)
Number of sequences [s(0), s(1), ..., s(n)] with s(n)=0, Sum_{j=0..n} s(j) = n, and Sum_{j=0..k} s(j)-1 >= 0 for k < n-1 (and necessarily Sum_{j=0..n-1} s(j)-1 = 0). These are the branching sequences of the (ordered) trees with n non-root nodes, see example. - Joerg Arndt, Jun 30 2014
Number of stack-sortable permutations of [n], these are the 231-avoiding permutations; see the Bousquet-Mélou reference. - Joerg Arndt, Jul 01 2014
a(n) is the number of increasing strict binary trees with 2n-1 nodes that avoid 132. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
In a one-dimensional medium with elastic scattering (zig-zag walk), first recurrence after 2n+1 scattering events has the probability C(n)/2^(2n+1). - Joachim Wuttke, Sep 11 2014
The o.g.f. C(x) = (1 - sqrt(1-4x))/2, for the Catalan numbers, with comp. inverse Cinv(x) = x*(1-x) and the functions P(x) = x / (1 + t*x) and its inverse Pinv(x,t) = -P(-x,t) = x / (1 - t*x) form a group under composition that generates or interpolates among many classic arrays, such as the Motzkin (Riordan, A005043), Fibonacci (A000045), and Fine (A000957) numbers and polynomials (A030528), and enumerating arrays for Motzkin, Dyck, and Łukasiewicz lattice paths and different types of trees and non-crossing partitions (A091867, connected to sums of the refined Narayana numbers A134264). - Tom Copeland, Nov 04 2014
Conjecture: All the rational numbers Sum_{i=j..k} 1/a(i) with 0 < min{2,k} <= j <= k have pairwise distinct fractional parts. - Zhi-Wei Sun, Sep 24 2015
The Catalan number series A000108(n+3), offset n=0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset n=0 (empirical observation). - Tony Foster III, Sep 05 2016
Hankel transforms of the Catalan numbers with the first 2, 4, and 5 terms omitted give A001477, A006858, and A091962, respectively, without the first 2 terms in all cases. More generally, the Hankel transform of the Catalan numbers with the first k terms omitted is H_k(n) = Product_{j=1..k-1} Product_{i=1..j} (2*n+j+i)/(j+i) [see Cigler (2011), Eq. (1.14) and references therein]; together they form the array A078920/A123352/A368025. - Andrey Zabolotskiy, Oct 13 2016
Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear. See S. J. Miller, ed., 2015, p. 5. - N. J. A. Sloane, Feb 09 2017
Coefficients of the generating series associated to the Magmatic and Dendriform operadic algebras. Cf. p. 422 and 435 of the Loday et al. paper. - Tom Copeland, Jul 08 2018
Let M_n be the n X n matrix with M_n(i,j) = binomial(i+j-1,2j-2); then det(M_n) = a(n). - Tony Foster III, Aug 30 2018
Also the number of Catalan trees, or planted plane trees (Bona, 2015, p. 299, Theorem 4.6.3). - N. J. A. Sloane, Dec 25 2018
Number of coalescent histories for a caterpillar species tree and a matching caterpillar gene tree with n+1 leaves (Rosenberg 2007, Corollary 3.5). - Noah A Rosenberg, Jan 28 2019
Finding solutions of eps*x^2+x-1 = 0 for eps small, that is, writing x = Sum_{n>=0} x_{n}*eps^n and expanding, one finds x = 1 - eps + 2*eps^2 - 5*eps^3 + 14*eps^3 - 42*eps^4 + ... with x_{n} = (-1)^n*C(n). Further, letting x = 1/y and expanding y about 0 to find large roots, that is, y = Sum_{n>=1} y_{n}*eps^n, one finds y = 0 - eps + eps^2 - 2*eps^3 + 5*eps^3 - ... with y_{n} = (-1)^n*C(n-1). - Derek Orr, Mar 15 2019
Permutations of length n that produce a bipartite permutation graph of order n [see Knuth (1973), Busch (2006), Golumbic and Trenk (2004)]. - Elise Anderson, R. M. Argus, Caitlin Owens, Tessa Stevens, Jun 27 2019
For n > 0, a random selection of n + 1 objects (the minimum number ensuring one pair by the pigeonhole principle) from n distinct pairs of indistinguishable objects contains only one pair with probability 2^(n-1)/a(n) = b(n-1)/A098597(n), where b is the 0-offset sequence with the terms of A120777 repeated (1,1,4,4,8,8,64,64,128,128,...). E.g., randomly selecting 6 socks from 5 pairs that are black, blue, brown, green, and white, results in only one pair of the same color with probability 2^(5-1)/a(5) = 16/42 = 8/21 = b(4)/A098597(5). - Rick L. Shepherd, Sep 02 2019
See Haran & Tabachnikov link for a video discussing Conway-Coxeter friezes. The Conway-Coxeter friezes with n nontrivial rows are generated by the counts of triangles at each vertex in the triangulations of regular n-gons, of which there are a(n). - Charles R Greathouse IV, Sep 28 2019
For connections to knot theory and scattering amplitudes from Feynman diagrams, see Broadhurst and Kreimer, and Todorov. Eqn. 6.12 on p. 130 of Bessis et al. becomes, after scaling, -12g * r_0(-y/(12g)) = (1-sqrt(1-4y))/2, the o.g.f. (expressed as a Taylor series in Eqn. 7.22 in 12gx) given for the Catalan numbers in Copeland's (Sep 30 2011) formula below. (See also Mizera p. 34, Balduf pp. 79-80, Keitel and Bartosch.) - Tom Copeland, Nov 17 2019
Number of permutations in S_n whose principal order ideals in the weak order are modular lattices. - Bridget Tenner, Jan 16 2020
Number of permutations in S_n whose principal order ideals in the weak order are distributive lattices. - Bridget Tenner, Jan 16 2020
Legendre gives the following formula for computing the square root modulo 2^m:
sqrt(1 + 8*a) mod 2^m = (1 + 4*a*Sum_{i=0..m-4} C(i)*(-2*a)^i) mod 2^m
as cited by L. D. Dickson, History of the Theory of Numbers, Vol. 1, 207-208. - Peter Schorn, Feb 11 2020
a(n) is the number of length n permutations sorted to the identity by a consecutive-132-avoiding stack followed by a classical-21-avoiding stack. - Kai Zheng, Aug 28 2020
Number of non-crossing partitions of a 2*n-set with n blocks of size 2. Also number of non-crossing partitions of a 2*n-set with n+1 blocks of size at most 3, and without cyclical adjacencies. The two partitions can be mapped by rotated Kreweras bijection. - Yuchun Ji, Jan 18 2021
Named by Riordan (1968, and earlier in Mathematical Reviews, 1948 and 1964) after the French and Belgian mathematician Eugène Charles Catalan (1814-1894) (see Pak, 2014). - Amiram Eldar, Apr 15 2021
For n >= 1, a(n-1) is the number of interpretations of x^n is an algebra where power-associativity is not assumed. For example, for n = 4 there are a(3) = 5 interpretations: x(x(xx)), x((xx)x), (xx)(xx), (x(xx))x, ((xx)x)x. See the link "Non-associate powers and a functional equation" from I. M. H. Etherington and the page "Nonassociative Product" from Eric Weisstein's World of Mathematics for detailed information. See also A001190 for the case where multiplication is commutative. - Jianing Song, Apr 29 2022
Number of states in the transition diagram associated with the Laplacian system over the complete graph K_N, corresponding to ordered initial conditions x_1 < x_2 < ... < x_N. - Andrea Arlette España, Nov 06 2022
a(n) is the number of 132-avoiding stabilized-interval-free permutations of size n+1. - Juan B. Gil, Jun 22 2023
Number of rooted polyominoes composed of n triangular cells of the hyperbolic regular tiling with Schläfli symbol {3,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {3,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
a(n) is the number of extremely lucky Stirling permutations of order n; i.e., the number of Stirling permutations of order n that have exactly n lucky cars. (see Colmenarejo et al. reference) - Bridget Tenner, Apr 16 2024

Examples

			From _Joerg Arndt_ and Greg Stevenson, Jul 11 2011: (Start)
The following products of 3 transpositions lead to a 4-cycle in S_4:
(1,2)*(1,3)*(1,4);
(1,2)*(1,4)*(3,4);
(1,3)*(1,4)*(2,3);
(1,4)*(2,3)*(2,4);
(1,4)*(2,4)*(3,4). (End)
G.f. = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + ...
For n=3, a(3)=5 since there are exactly 5 binary sequences of length 7 in which the number of ones first exceed the number of zeros at entry 7, namely, 0001111, 0010111, 0011011, 0100111, and 0101011. - _Dennis P. Walsh_, Apr 11 2012
From _Joerg Arndt_, Jun 30 2014: (Start)
The a(4) = 14 branching sequences of the (ordered) trees with 4 non-root nodes are (dots denote zeros):
01:  [ 1 1 1 1 . ]
02:  [ 1 1 2 . . ]
03:  [ 1 2 . 1 . ]
04:  [ 1 2 1 . . ]
05:  [ 1 3 . . . ]
06:  [ 2 . 1 1 . ]
07:  [ 2 . 2 . . ]
08:  [ 2 1 . 1 . ]
09:  [ 2 1 1 . . ]
10:  [ 2 2 . . . ]
11:  [ 3 . . 1 . ]
12:  [ 3 . 1 . . ]
13:  [ 3 1 . . . ]
14:  [ 4 . . . . ]
(End)
		

References

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Crossrefs

A row of A060854.
See A001003, A001190, A001699, A000081 for other ways to count parentheses.
Enumerates objects encoded by A014486.
A diagonal of any of the essentially equivalent arrays A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Cf. A051168 (diagonal of the square array described).
Cf. A033552, A176137 (partitions into Catalan numbers).
Cf. A000753, A000736 (Boustrophedon transforms).
Cf. A120303 (largest prime factor of Catalan number).
Cf. A121839 (reciprocal Catalan constant), A268813.
Cf. A038003, A119861, A119908, A120274, A120275 (odd Catalan number).
Cf. A002390 (decimal expansion of natural logarithm of golden ratio).
Coefficients of square root of the g.f. are A001795/A046161.
For a(n) mod 6 see A259667.
For a(n) in base 2 see A264663.
Hankel transforms with first terms omitted: A001477, A006858, A091962, A078920, A123352, A368025.
Cf. A332602 (conjectured production matrix).
Polyominoes: A001683(n+2) (oriented), A000207 (unoriented), A369314 (chiral), A208355(n-1) (achiral), A001764 {4,oo}.

Programs

  • GAP
    A000108:=List([0..30],n->Binomial(2*n,n)/(n+1)); # Muniru A Asiru, Feb 17 2018
  • Haskell
    import Data.List (genericIndex)
    a000108 n = genericIndex a000108_list n
    a000108_list = 1 : catalan [1] where
       catalan cs = c : catalan (c:cs) where
          c = sum $ zipWith (*) cs $ reverse cs
    -- Reinhard Zumkeller, Nov 12 2011
    a000108 = map last $ iterate (scanl1 (+) . (++ [0])) [1]
    -- David Spies, Aug 23 2015
    
  • Magma
    C:= func< n | Binomial(2*n,n)/(n+1) >; [ C(n) : n in [0..60]];
    
  • Magma
    [Catalan(n): n in [0..40]]; // Vincenzo Librandi, Apr 02 2011
    
  • Maple
    A000108 := n->binomial(2*n,n)/(n+1);
    G000108 := (1 - sqrt(1 - 4*x)) / (2*x);
    spec := [ A, {A=Prod(Z,Sequence(A))}, unlabeled ]: [ seq(combstruct[count](spec, size=n+1), n=0..42) ];
    with(combstruct): bin := {B=Union(Z,Prod(B,B))}: seq(count([B,bin,unlabeled],size=n+1), n=0..25); # Zerinvary Lajos, Dec 05 2007
    gser := series(G000108, x=0, 42): seq(coeff(gser, x, n), n=0..41); # Zerinvary Lajos, May 21 2008
    seq((2*n)!*coeff(series(hypergeom([],[2],x^2),x,2*n+2),x,2*n),n=0..30); # Peter Luschny, Jan 31 2015
    A000108List := proc(m) local A, P, n; A := [1, 1]; P := [1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), A[-1]]);
    A := [op(A), P[-1]] od; A end: A000108List(31); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[(2 n)!/n!/(n + 1)!, {n, 0, 20}]
    Table[4^n Gamma[n + 1/2]/(Sqrt[Pi] Gamma[n + 2]), {n, 0, 20}] (* Eric W. Weisstein, Oct 31 2024 *)
    Table[Hypergeometric2F1[1 - n, -n, 2, 1], {n, 0, 20}] (* Richard L. Ollerton, Sep 13 2006 *)
    Table[CatalanNumber @ n, {n, 0, 20}] (* Robert G. Wilson v, Feb 15 2011 *)
    CatalanNumber[Range[0, 20]] (* Eric W. Weisstein, Oct 31 2024 *)
    CoefficientList[InverseSeries[Series[x/Sum[x^n, {n, 0, 31}], {x, 0, 31}]]/x, x] (* Mats Granvik, Nov 24 2013 *)
    CoefficientList[Series[(1 - Sqrt[1 - 4 x])/(2 x), {x, 0, 20}], x] (* Stefano Spezia, Aug 31 2018 *)
  • Maxima
    A000108(n):=binomial(2*n,n)/(n+1)$ makelist(A000108(n),n,0,30); /* Martin Ettl, Oct 24 2012 */
    
  • MuPAD
    combinat::dyckWords::count(n) $ n = 0..38 // Zerinvary Lajos, Apr 14 2007
    
  • PARI
    a(n)=binomial(2*n,n)/(n+1) \\ M. F. Hasler, Aug 25 2012
    
  • PARI
    a(n) = (2*n)! / n! / (n+1)!
    
  • PARI
    a(n) = my(A, m); if( n<0, 0, m=1; A = 1 + x + O(x^2); while(m<=n, m*=2; A = sqrt(subst(A, x, 4*x^2)); A += (A - 1) / (2*x*A)); polcoeff(A, n));
    
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( serreverse( x / (1 + x)^2 + x * O(x^n)), n))}; /* Michael Somos */
    
  • PARI
    (recur(a,b)=if(b<=2,(a==2)+(a==b)+(a!=b)*(1+a/2), (1+a/b)*recur(a,b-1))); a(n)=recur(n,n); \\ R. J. Cano, Nov 22 2012
    
  • PARI
    x='x+O('x^40); Vec((1-sqrt(1-4*x))/(2*x)) \\ Altug Alkan, Oct 13 2015
    
  • Python
    from gmpy2 import divexact
    A000108 = [1, 1]
    for n in range(1, 10**3):
        A000108.append(divexact(A000108[-1]*(4*n+2),(n+2))) # Chai Wah Wu, Aug 31 2014
    
  • Python
    # Works in Sage also.
    A000108 = [1]
    for n in range(1000):
        A000108.append(A000108[-1]*(4*n+2)//(n+2)) # Günter Rote, Nov 08 2023
    
  • Sage
    [catalan_number(i) for i in range(27)] # Zerinvary Lajos, Jun 26 2008
    
  • Sage
    # Generalized algorithm of L. Seidel
    def A000108_list(n) :
        D = [0]*(n+1); D[1] = 1
        b = True; h = 1; R = []
        for i in range(2*n-1) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1; R.append(D[1])
            else :
                for k in range(1,h, 1) : D[k] += D[k+1]
            b = not b
        return R
    A000108_list(31) # Peter Luschny, Jun 02 2012
    

Formula

a(n) = binomial(2*n, n)/(n+1) = (2*n)!/(n!*(n+1)!) = A000984(n)/(n+1).
Recurrence: a(n) = 2*(2*n-1)*a(n-1)/(n+1) with a(0) = 1.
Recurrence: a(n) = Sum_{k=0..n-1} a(k)a(n-1-k).
G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x), and satisfies A(x) = 1 + x*A(x)^2.
a(n) = Product_{k=2..n} (1 + n/k).
a(n+1) = Sum_{i} binomial(n, 2*i)*2^(n-2*i)*a(i). - Touchard
It is known that a(n) is odd if and only if n=2^k-1, k=0, 1, 2, 3, ... - Emeric Deutsch, Aug 04 2002, corrected by M. F. Hasler, Nov 08 2015
Using the Stirling approximation in A000142 we get the asymptotic expansion a(n) ~ 4^n / (sqrt(Pi * n) * (n + 1)). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 13 2001
Integral representation: a(n) = (1/(2*Pi))*Integral_{x=0..4} x^n*sqrt((4-x)/x). - Karol A. Penson, Apr 12 2001
E.g.f.: exp(2*x)*(I_0(2*x)-I_1(2*x)), where I_n is Bessel function. - Karol A. Penson, Oct 07 2001
a(n) = polygorial(n, 6)/polygorial(n, 3). - Daniel Dockery (peritus(AT)gmail.com), Jun 24 2003
G.f. A(x) satisfies ((A(x) + A(-x)) / 2)^2 = A(4*x^2). - Michael Somos, Jun 27 2003
G.f. A(x) satisfies Sum_{k>=1} k(A(x)-1)^k = Sum_{n>=1} 4^{n-1}*x^n. - Shapiro, Woan, Getu
a(n+m) = Sum_{k} A039599(n, k)*A039599(m, k). - Philippe Deléham, Dec 22 2003
a(n+1) = (1/(n+1))*Sum_{k=0..n} a(n-k)*binomial(2k+1, k+1). - Philippe Deléham, Jan 24 2004
a(n) = Sum_{k>=0} A008313(n, k)^2. - Philippe Deléham, Feb 14 2004
a(m+n+1) = Sum_{k>=0} A039598(m, k)*A039598(n, k). - Philippe Deléham, Feb 15 2004
a(n) = Sum_{k=0..n} (-1)^k*2^(n-k)*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
Sum_{n>=0} 1/a(n) = 2 + 4*Pi/3^(5/2) = F(1,2;1/2;1/4) = A268813 = 2.806133050770763... (see L'Univers de Pi link). - Gerald McGarvey and Benoit Cloitre, Feb 13 2005
a(n) = Sum_{k=0..floor(n/2)} ((n-2*k+1)*binomial(n, n-k)/(n-k+1))^2, which is equivalent to: a(n) = Sum_{k=0..n} A053121(n, k)^2, for n >= 0. - Paul D. Hanna, Apr 23 2005
a((m+n)/2) = Sum_{k>=0} A053121(m, k)*A053121(n, k) if m+n is even. - Philippe Deléham, May 26 2005
E.g.f. Sum_{n>=0} a(n) * x^(2*n) / (2*n)! = BesselI(1, 2*x) / x. - Michael Somos, Jun 22 2005
Given g.f. A(x), then B(x) = x * A(x^3) satisfies 0 = f(x, B(X)) where f(u, v) = u - v + (u*v)^2 or B(x) = x + (x * B(x))^2 which implies B(-B(x)) = -x and also (1 + B^3) / B^2 = (1 - x^3) / x^2. - Michael Somos, Jun 27 2005
a(n) = a(n-1)*(4-6/(n+1)). a(n) = 2a(n-1)*(8a(n-2)+a(n-1))/(10a(n-2)-a(n-1)). - Franklin T. Adams-Watters, Feb 08 2006
Sum_{k>=1} a(k)/4^k = 1. - Franklin T. Adams-Watters, Jun 28 2006
a(n) = A047996(2*n+1, n). - Philippe Deléham, Jul 25 2006
Binomial transform of A005043. - Philippe Deléham, Oct 20 2006
a(n) = Sum_{k=0..n} (-1)^k*A116395(n,k). - Philippe Deléham, Nov 07 2006
a(n) = (1/(s-n))*Sum_{k=0..n} (-1)^k (k+s-n)*binomial(s-n,k) * binomial(s+n-k,s) with s a nonnegative free integer [H. W. Gould].
a(k) = Sum_{i=1..k} |A008276(i,k)| * (k-1)^(k-i) / k!. - André F. Labossière, May 29 2007
a(n) = Sum_{k=0..n} A129818(n,k) * A007852(k+1). - Philippe Deléham, Jun 20 2007
a(n) = Sum_{k=0..n} A109466(n,k) * A127632(k). - Philippe Deléham, Jun 20 2007
Row sums of triangle A124926. - Gary W. Adamson, Oct 22 2007
Limit_{n->oo} (1 + Sum_{k=0..n} a(k)/A004171(k)) = 4/Pi. - Reinhard Zumkeller, Aug 26 2008
a(n) = Sum_{k=0..n} A120730(n,k)^2 and a(k+1) = Sum_{n>=k} A120730(n,k). - Philippe Deléham, Oct 18 2008
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, the present sequence is Phi([1]) (also Phi([1,1])). - Gary W. Adamson, Oct 27 2008
a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}...Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n) where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any l_i < l_(i+1) and l_(i+1) <> 0 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
a(n) = A000680(n)/A006472(n+1). - Mark Dols, Jul 14 2010; corrected by M. F. Hasler, Nov 08 2015
Let A(x) be the g.f., then B(x)=x*A(x) satisfies the differential equation B'(x)-2*B'(x)*B(x)-1=0. - Vladimir Kruchinin, Jan 18 2011
Complement of A092459; A010058(a(n)) = 1. - Reinhard Zumkeller, Mar 29 2011
G.f.: 1/(1-x/(1-x/(1-x/(...)))) (continued fraction). - Joerg Arndt, Mar 18 2011
With F(x) = (1-2*x-sqrt(1-4*x))/(2*x) an o.g.f. in x for the Catalan series, G(x) = x/(1+x)^2 is the compositional inverse of F (nulling the n=0 term). - Tom Copeland, Sep 04 2011
With H(x) = 1/(dG(x)/dx) = (1+x)^3 / (1-x), the n-th Catalan number is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)), and H(x) is the o.g.f. for A115291. - Tom Copeland, Sep 04 2011
From Tom Copeland, Sep 30 2011: (Start)
With F(x) = (1-sqrt(1-4*x))/2 an o.g.f. in x for the Catalan series, G(x)= x*(1-x) is the compositional inverse and this relates the Catalan numbers to the row sums of A125181.
With H(x) = 1/(dG(x)/dx) = 1/(1-2x), the n-th Catalan number (offset 1) is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)). (End)
G.f.: (1-sqrt(1-4*x))/(2*x) = G(0) where G(k) = 1 + (4*k+1)*x/(k+1-2*x*(k+1)*(4*k+3)/(2*x*(4*k+3)+(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: exp(2*x)*(BesselI(0,2*x) - BesselI(1,2*x)) = G(0) where G(k) = 1 + (4*k+1)*x/((k+1)*(2*k+1)-x*(k+1)*(2*k+1)*(4*k+3)/(x*(4*k+3)+(k+1)*(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: Hypergeometric([1/2],[2],4*x) which coincides with the e.g.f. given just above, and also by Karol A. Penson further above. - Wolfdieter Lang, Jan 13 2012
A076050(a(n)) = n + 1 for n > 0. - Reinhard Zumkeller, Feb 17 2012
a(n) = A208355(2*n-1) = A208355(2*n) for n > 0. - Reinhard Zumkeller, Mar 04 2012
a(n+1) = A214292(2*n+1,n) = A214292(2*n+2,n). - Reinhard Zumkeller, Jul 12 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = k*(4*x+1) + 2*x + 2 - x*(2*k+3)*(2*k+4)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Sep 20 2012
G.f.: hypergeom([1/2,1],[2],4*x). - Joerg Arndt, Apr 06 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,1,-1/2-n,-1)/(n+1). - Karol A. Penson, Jul 28 2013
For n > 0: a(n) = sum of row n in triangle A001263. - Reinhard Zumkeller, Oct 10 2013
a(n) = binomial(2n,n-1)/n and a(n) mod n = binomial(2n,n) mod n = A059288(n). - Jonathan Sondow, Dec 14 2013
a(n-1) = Sum_{t1+2*t2+...+n*tn=n} (-1)^(1+t1+t2+...+tn)*multinomial(t1+t2 +...+tn,t1,t2,...,tn)*a(1)^t1*a(2)^t2*...*a(n)^tn. - Mircea Merca, Feb 27 2014
a(n) = Sum_{k=1..n} binomial(n+k-1,n)/n if n > 0. Alexander Adamchuk, Mar 25 2014
a(n) = -2^(2*n+1) * binomial(n-1/2, -3/2). - Peter Luschny, May 06 2014
a(n) = (4*A000984(n) - A000984(n+1))/2. - Stanislav Sykora, Aug 09 2014
a(n) = A246458(n) * A246466(n). - Tom Edgar, Sep 02 2014
a(n) = (2*n)!*[x^(2*n)]hypergeom([],[2],x^2). - Peter Luschny, Jan 31 2015
a(n) = 4^(n-1)*hypergeom([3/2, 1-n], [3], 1). - Peter Luschny, Feb 03 2015
a(2n) = 2*A000150(2n); a(2n+1) = 2*A000150(2n+1) + a(n). - John Bodeen, Jun 24 2015
a(n) = Sum_{t=1..n+1} n^(t-1)*abs(Stirling1(n+1, t)) / Sum_{t=1..n+1} abs(Stirling1(n+1, t)), for n > 0, see (10) in Cereceda link. - Michel Marcus, Oct 06 2015
a(n) ~ 4^(n-2)*(128 + 160/N^2 + 84/N^4 + 715/N^6 - 10180/N^8)/(N^(3/2)*Pi^(1/2)) where N = 4*n+3. - Peter Luschny, Oct 14 2015
a(n) = Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)*binomial(n+1-k,k)*a(n-k) if n > 0; and a(0) = 1. - David Pasino, Jun 29 2016
Sum_{n>=0} (-1)^n/a(n) = 14/25 - 24*arccsch(2)/(25*sqrt(5)) = 14/25 - 24*A002390/(25*sqrt(5)) = 0.353403708337278061333... - Ilya Gutkovskiy, Jun 30 2016
C(n) = (1/n) * Sum_{i+j+k=n-1} C(i)*C(j)*C(k)*(k+1), n >= 1. - Yuchun Ji, Feb 21 2016
C(n) = 1 + Sum_{i+j+kYuchun Ji, Sep 01 2016
a(n) = A001700(n) - A162551(n) = binomial(2*n+1,n+1). - 2*binomial(2*n,n-1). - Taras Goy, Aug 09 2018
G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x) = 2F1(1/2,1;2;4*x). G.f. A(x) satisfies A = 1 + x*A^2. - R. J. Mathar, Nov 17 2018
C(n) = 1 + Sum_{i=0..n-1} A000245(i). - Yuchun Ji, Jan 10 2019
From A.H.M. Smeets, Apr 11 2020: (Start)
(1+sqrt(1+4*x))/2 = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4; and sqrt(x+sqrt(x+sqrt(x+...))) = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4 and x <> 0. (End)
a(3n+1)*a(5n+4)*a(15n+10) = a(3n+2)*a(5n+2)*a(15n+11). The first case of Catalan product equation of a triple partition of 23n+15. - Yuchun Ji, Sep 27 2020
a(n) = 4^n * (-1)^(n+1) * 3F2[{n + 1,n + 1/2,n}, {3/2,1}, -1], n >= 1. - Sergii Voloshyn, Oct 22 2020
a(n) = 2^(1 + 2 n) * (-1)^(n)/(1 + n) * 3F2[{n, 1/2 + n, 1 + n}, {1/2, 1}, -1], n >= 1. - Sergii Voloshyn, Nov 08 2020
a(n) = (1/Pi)*4^(n+1)*Integral_{x=0..Pi/2} cos(x)^(2*n)*sin(x)^2 dx. - Greg Dresden, May 30 2021
From Peter Bala, Aug 17 2021: (Start)
G.f. A(x) satisfies A(x) = 1/sqrt(1 - 4*x) * A( -x/(1 - 4*x) ) and (A(x) + A(-x))/2 = 1/sqrt(1 - 4*x) * A( -2*x/(1 - 4*x) ); these are the cases k = 0 and k = -1 of the general formula 1/sqrt(1 - 4*x) * A( (k-1)*x/(1 - 4*x) ) = Sum_{n >= 0} ((k^(n+1) - 1)/(k - 1))*Catalan(n)*x^n.
2 - sqrt(1 - 4*x)/A( k*x/(1 - 4*x) ) = 1 + Sum_{n >= 1} (1 + (k + 1)^n) * Catalan(n-1)*x^n. (End)
Sum_{n>=0} a(n)*(-1/4)^n = 2*(sqrt(2)-1) (A163960). - Amiram Eldar, Mar 22 2022
0 = a(n)*(16*a(n+1) - 10*a(n+2)) + a(n+1)*(2*a(n+1) + a(n+2)) for all n>=0. - Michael Somos, Dec 12 2022
G.f.: (offset 1) 1/G(x), with G(x) = 1 - 2*x - x^2/G(x) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 01 2023
a(n) = K^(2n+1, n, 1) for all n >= 0, where K^(n, s, x) is the Krawtchouk polynomial defined to be Sum_{k=0..s} (-1)^k * binomial(n-x, s-k) * binomial(x, k). - Vladislav Shubin, Aug 17 2023
From Peter Bala, Feb 03 2024: (Start)
The g.f. A(x) satisfies the following functional equations:
A(x) = 1 + x/(1 - 4*x) * A(-x/(1 - 4*x))^2,
A(x^2) = 1/(1 - 2*x) * A(- x/(1 - 2*x))^2 and, for arbitrary k,
1/(1 - k*x) * A(x/(1 - k*x))^2 = 1/(1 - (k+4)*x) * A(-x/(1 - (k+4)*x))^2. (End)
a(n) = A363448(n) + A363449(n). - Julien Rouyer, Jun 28 2024

A000079 Powers of 2: a(n) = 2^n.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592
Offset: 0

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Comments

2^0 = 1 is the only odd power of 2.
Number of subsets of an n-set.
There are 2^(n-1) compositions (ordered partitions) of n (see for example Riordan). This is the unlabeled analog of the preferential labelings sequence A000670.
This is also the number of weakly unimodal permutations of 1..n + 1, that is, permutations with exactly one local maximum. E.g., a(4) = 16: 12345, 12354, 12453, 12543, 13452, 13542, 14532 and 15432 and their reversals. - Jon Perry, Jul 27 2003 [Proof: see next line! See also A087783.]
Proof: n must appear somewhere and there are 2^(n-1) possible choices for the subset that precedes it. These must appear in increasing order and the rest must follow n in decreasing order. QED. - N. J. A. Sloane, Oct 26 2003
a(n+1) is the smallest number that is not the sum of any number of (distinct) earlier terms.
Same as Pisot sequences E(1, 2), L(1, 2), P(1, 2), T(1, 2). See A008776 for definitions of Pisot sequences.
With initial 1 omitted, same as Pisot sequences E(2, 4), L(2, 4), P(2, 4), T(2, 4). - David W. Wilson
Not the sum of two or more consecutive numbers. - Lekraj Beedassy, May 14 2004
Least deficient or near-perfect numbers (i.e., n such that sigma(n) = A000203(n) = 2n - 1). - Lekraj Beedassy, Jun 03 2004. [Comment from Max Alekseyev, Jan 26 2005: All the powers of 2 are least deficient numbers but it is not known if there exists a least deficient number that is not a power of 2.]
Almost-perfect numbers referred to as least deficient or slightly defective (Singh 1997) numbers. Does "near-perfect numbers" refer to both almost-perfect numbers (sigma(n) = 2n - 1) and quasi-perfect numbers (sigma(n) = 2n + 1)? There are no known quasi-perfect or least abundant or slightly excessive (Singh 1997) numbers.
The sum of the numbers in the n-th row of Pascal's triangle; the sum of the coefficients of x in the expansion of (x+1)^n.
The Collatz conjecture (the hailstone sequence will eventually reach the number 1, regardless of which positive integer is chosen initially) may be restated as (the hailstone sequence will eventually reach a power of 2, regardless of which positive integer is chosen initially).
The only hailstone sequence which doesn't rebound (except "on the ground"). - Alexandre Wajnberg, Jan 29 2005
With p(n) as the number of integer partitions of n, p(i) is the number of parts of the i-th partition of n, d(i) is the number of different parts of the i-th partition of n, m(i,j) is 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)!)). - Thomas Wieder, May 18 2005
The number of binary relations on an n-element set that are both symmetric and antisymmetric. Also the number of binary relations on an n-element set that are symmetric, antisymmetric and transitive.
The first differences are the sequence itself. - Alexandre Wajnberg and Eric Angelini, Sep 07 2005
a(n) is the largest number with shortest addition chain involving n additions. - David W. Wilson, Apr 23 2006
Beginning with a(1) = 0, numbers not equal to the sum of previous distinct natural numbers. - Giovanni Teofilatto, Aug 06 2006
For n >= 1, a(n) is equal to the number of functions f:{1, 2, ..., n} -> {1, 2} such that for a fixed x in {1, 2, ..., n} and a fixed y in {1, 2} we have f(x) != y. - Aleksandar M. Janjic and Milan Janjic, Mar 27 2007
Let P(A) be the power set of an n-element set A. Then a(n) is the number of pairs of elements {x,y} of P(A) for which x = y. - Ross La Haye, Jan 09 2008
a(n) is the number of permutations on [n+1] such that every initial segment is an interval of integers. Example: a(3) counts 1234, 2134, 2314, 2341, 3214, 3241, 3421, 4321. The map "p -> ascents of p" is a bijection from these permutations to subsets of [n]. An ascent of a permutation p is a position i such that p(i) < p(i+1). The permutations shown map to 123, 23, 13, 12, 3, 2, 1 and the empty set respectively. - David Callan, Jul 25 2008
2^(n-1) is the largest number having n divisors (in the sense of A077569); A005179(n) is the smallest. - T. D. Noe, Sep 02 2008
a(n) appears to match the number of divisors of the modified primorials (excluding 2, 3 and 5). Very limited range examined, PARI example shown. - Bill McEachen, Oct 29 2008
Successive k such that phi(k)/k = 1/2, where phi is Euler's totient function. - Artur Jasinski, Nov 07 2008
A classical transform consists (for general a(n)) in swapping a(2n) and a(2n+1); examples for Jacobsthal A001045 and successive differences: A092808, A094359, A140505. a(n) = A000079 leads to 2, 1, 8, 4, 32, 16, ... = A135520. - Paul Curtz, Jan 05 2009
This is also the (L)-sieve transform of {2, 4, 6, 8, ..., 2n, ...} = A005843. (See A152009 for the definition of the (L)-sieve transform.) - John W. Layman, Jan 23 2009
a(n) = a(n-1)-th even natural number (A005843) for n > 1. - Jaroslav Krizek, Apr 25 2009
For n >= 0, a(n) is the number of leaves in a complete binary tree of height n. For n > 0, a(n) is the number of nodes in an n-cube. - K.V.Iyer, May 04 2009
Permutations of n+1 elements where no element is more than one position right of its original place. For example, there are 4 such permutations of three elements: 123, 132, 213, and 312. The 8 such permutations of four elements are 1234, 1243, 1324, 1423, 2134, 2143, 3124, and 4123. - Joerg Arndt, Jun 24 2009
Catalan transform of A099087. - R. J. Mathar, Jun 29 2009
a(n) written in base 2: 1,10,100,1000,10000,..., i.e., (n+1) times 1, n times 0 (A011557(n)). - Jaroslav Krizek, Aug 02 2009
Or, phi(n) is equal to the number of perfect partitions of n. - Juri-Stepan Gerasimov, Oct 10 2009
These are the 2-smooth numbers, positive integers with no prime factors greater than 2. - Michael B. Porter, Oct 04 2009
A064614(a(n)) = A000244(n) and A064614(m) < A000244(n) for m < a(n). - Reinhard Zumkeller, Feb 08 2010
a(n) is the largest number m such that the number of steps of iterations of {r - (largest divisor d < r)} needed to reach 1 starting at r = m is equal to n. Example (a(5) = 32): 32 - 16 = 16; 16 - 8 = 8; 8 - 4 = 4; 4 - 2 = 2; 2 - 1 = 1; number 32 has 5 steps and is the largest such number. See A105017, A064097, A175125. - Jaroslav Krizek, Feb 15 2010
a(n) is the smallest proper multiple of a(n-1). - Dominick Cancilla, Aug 09 2010
The powers-of-2 triangle T(n, k), n >= 0 and 0 <= k <= n, begins with: {1}; {2, 4}; {8, 16, 32}; {64, 128, 256, 512}; ... . The first left hand diagonal T(n, 0) = A006125(n + 1), the first right hand diagonal T(n, n) = A036442(n + 1) and the center diagonal T(2*n, n) = A053765(n + 1). Some triangle sums, see A180662, are: Row1(n) = A122743(n), Row2(n) = A181174(n), Fi1(n) = A181175(n), Fi2(2*n) = A181175(2*n) and Fi2(2*n + 1) = 2*A181175(2*n + 1). - Johannes W. Meijer, Oct 10 2010
Records in the number of prime factors. - Juri-Stepan Gerasimov, Mar 12 2011
Row sums of A152538. - Gary W. Adamson, Dec 10 2008
A078719(a(n)) = 1; A006667(a(n)) = 0. - Reinhard Zumkeller, Oct 08 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>=1, a(n) equals the number of 2-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Equals A001405 convolved with its right-shifted variant: (1 + 2x + 4x^2 + ...) = (1 + x + 2x^2 + 3x^3 + 6x^4 + 10x^5 + ...) * (1 + x + x^2 + 2x^3 + 3x^4 + 6x^5 + ...). - Gary W. Adamson, Nov 23 2011
The number of odd-sized subsets of an n+1-set. For example, there are 2^3 odd-sized subsets of {1, 2, 3, 4}, namely {1}, {2}, {3}, {4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, and {2, 3, 4}. Also, note that 2^n = Sum_{k=1..floor((n+1)/2)} C(n+1, 2k-1). - Dennis P. Walsh, Dec 15 2011
a(n) is the number of 1's in any row of Pascal's triangle (mod 2) whose row number has exactly n 1's in its binary expansion (see A007318 and A047999). (The result of putting together A001316 and A000120.) - Marcus Jaiclin, Jan 31 2012
A204455(k) = 1 if and only if k is in this sequence. - Wolfdieter Lang, Feb 04 2012
For n>=1 apparently the number of distinct finite languages over a unary alphabet, whose minimum regular expression has alphabetic width n (verified up to n=17), see the Gruber/Lee/Shallit link. - Hermann Gruber, May 09 2012
First differences of A000225. - Omar E. Pol, Feb 19 2013
This is the lexicographically earliest sequence which contains no arithmetic progression of length 3. - Daniel E. Frohardt, Apr 03 2013
a(n-2) is the number of bipartitions of {1..n} (i.e., set partitions into two parts) such that 1 and 2 are not in the same subset. - Jon Perry, May 19 2013
Numbers n such that the n-th cyclotomic polynomial has a root mod 2; numbers n such that the n-th cyclotomic polynomial has an even number of odd coefficients. - Eric M. Schmidt, Jul 31 2013
More is known now about non-power-of-2 "Almost Perfect Numbers" as described in Dagal. - Jonathan Vos Post, Sep 01 2013
Number of symmetric Ferrers diagrams that fit into an n X n box. - Graham H. Hawkes, Oct 18 2013
Numbers n such that sigma(2n) = 2n + sigma(n). - Jahangeer Kholdi, Nov 23 2013
a(1), ..., a(floor(n/2)) are all values of permanent on set of square (0,1)-matrices of order n>=2 with row and column sums 2. - Vladimir Shevelev, Nov 26 2013
Numbers whose base-2 expansion has exactly one bit set to 1, and thus has base-2 sum of digits equal to one. - Stanislav Sykora, Nov 29 2013
A072219(a(n)) = 1. - Reinhard Zumkeller, Feb 20 2014
a(n) is the largest number k such that (k^n-2)/(k-2) is an integer (for n > 1); (k^a(n)+1)/(k+1) is never an integer (for k > 1 and n > 0). - Derek Orr, May 22 2014
If x = A083420(n), y = a(n+1) and z = A087289(n), then x^2 + 2*y^2 = z^2. - Vincenzo Librandi, Jun 09 2014
The mini-sequence b(n) = least number k > 0 such that 2^k ends in n identical digits is given by {1, 18, 39}. The repeating digits are {2, 4, 8} respectively. Note that these are consecutive powers of 2 (2^1, 2^2, 2^3), and these are the only powers of 2 (2^k, k > 0) that are only one digit. Further, this sequence is finite. The number of n-digit endings for a power of 2 with n or more digits id 4*5^(n-1). Thus, for b(4) to exist, one only needs to check exponents up to 4*5^3 = 500. Since b(4) does not exist, it is clear that no other number will exist. - Derek Orr, Jun 14 2014
The least number k > 0 such that 2^k ends in n consecutive decreasing digits is a 3-number sequence given by {1, 5, 25}. The consecutive decreasing digits are {2, 32, 432}. There are 100 different 3-digit endings for 2^k. There are no k-values such that 2^k ends in '987', '876', '765', '654', '543', '321', or '210'. The k-values for which 2^k ends in '432' are given by 25 mod 100. For k = 25 + 100*x, the digit immediately before the run of '432' is {4, 6, 8, 0, 2, 4, 6, 8, 0, 2, ...} for x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...}, respectively. Thus, we see the digit before '432' will never be a 5. So, this sequence is complete. - Derek Orr, Jul 03 2014
a(n) is the number of permutations of length n avoiding both 231 and 321 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
Numbers n such that sigma(n) = sigma(2n) - phi(4n). - Farideh Firoozbakht, Aug 14 2014
This is a B_2 sequence: for i < j, differences a(j) - a(i) are all distinct. Here 2*a(n) < a(n+1) + 1, so a(n) - a(0) < a(n+1) - a(n). - Thomas Ordowski, Sep 23 2014
a(n) counts n-walks (closed) on the graph G(1-vertex; 1-loop, 1-loop). - David Neil McGrath, Dec 11 2014
a(n-1) counts walks (closed) on the graph G(1-vertex; 1-loop, 2-loop, 3-loop, 4-loop, ...). - David Neil McGrath, Jan 01 2015
b(0) = 4; b(n+1) is the smallest number not in the sequence such that b(n+1) - Prod_{i=0..n} b(i) divides b(n+1) - Sum_{i=0..n} b(i). Then b(n) = a(n) for n > 2. - Derek Orr, Jan 15 2015
a(n) counts the permutations of length n+2 whose first element is 2 such that the permutation has exactly one descent. - Ran Pan, Apr 17 2015
a(0)-a(30) appear, with a(26)-a(30) in error, in tablet M 08613 (see CDLI link) from the Old Babylonian period (c. 1900-1600 BC). - Charles R Greathouse IV, Sep 03 2015
Subsequence of A028982 (the squares or twice squares sequence). - Timothy L. Tiffin, Jul 18 2016
A000120(a(n)) = 1. A000265(a(n)) = 1. A000593(a(n)) = 1. - Juri-Stepan Gerasimov, Aug 16 2016
Number of monotone maps f : [0..n] -> [0..n] which are order-increasing (i <= f(i)) and idempotent (f(f(i)) = f(i)). In other words, monads on the n-th ordinal (seen as a posetal category). Any monad f determines a subset of [0..n] that contains n, by considering its set of monad algebras = fixed points { i | f(i) = i }. Conversely, any subset S of [0..n] containing n determines a monad on [0..n], by the function i |-> min { j | i <= j, j in S }. - Noam Zeilberger, Dec 11 2016
Consider n points lying on a circle. Then for n>=2 a(n-2) gives the number of ways to connect two adjacent points with nonintersecting chords. - Anton Zakharov, Dec 31 2016
Satisfies Benford's law [Diaconis, 1977; Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017
Also the number of independent vertex sets and vertex covers in the n-empty graph. - Eric W. Weisstein, Sep 21 2017
Also the number of maximum cliques in the n-halved cube graph for n > 4. - Eric W. Weisstein, Dec 04 2017
Number of pairs of compositions of n corresponding to a seaweed algebra of index n-1. - Nick Mayers, Jun 25 2018
The multiplicative group of integers modulo a(n) is cyclic if and only if n = 0, 1, 2. For n >= 3, it is a product of two cyclic groups. - Jianing Song, Jun 27 2018
k^n is the determinant of n X n matrix M_(i, j) = binomial(k + i + j - 2, j) - binomial(i+j-2, j), in this case k=2. - Tony Foster III, May 12 2019
Solutions to the equation Phi(2n + 2*Phi(2n)) = 2n. - M. Farrokhi D. G., Jan 03 2020
a(n-1) is the number of subsets of {1,2,...,n} which have an element that is the size of the set. For example, for n = 4, a(3) = 8 and the subsets are {1}, {1,2}, {2,3}, {2,4}, {1,2,3}, {1,3,4}, {2,3,4}, {1,2,3,4}. - Enrique Navarrete, Nov 21 2020
a(n) is the number of self-inverse (n+1)-order permutations with 231-avoiding. E.g., a(3) = 8: [1234, 1243, 1324, 1432, 2134, 2143, 3214, 4321]. - Yuchun Ji, Feb 26 2021
For any fixed k > 0, a(n) is the number of ways to tile a strip of length n+1 with tiles of length 1, 2, ... k, where the tile of length k can be black or white, with the restriction that the first tile cannot be black. - Greg Dresden and Bora Bursalı, Aug 31 2023

Examples

			There are 2^3 = 8 subsets of a 3-element set {1,2,3}, namely { -, 1, 2, 3, 12, 13, 23, 123 }.
		

References

  • Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 1016.
  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem, Mathematics and Computer Education Journal, Vol. 31, No. 1, pp. 24-28, Winter 1997.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 73, 84.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.5 Logarithms and §8.1 Terminology, pp. 150, 264.
  • Paul J. Nahin, An Imaginary Tale: The Story of sqrt(-1), Princeton University Press, Princeton, NJ. 1998, pp. 69-70.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 273.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 124.
  • 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).
  • V. E. Tarakanov, Combinatorial problems on binary matrices, Combin. Analysis, MSU, 5 (1980), 4-15. (Russian)
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 141.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Crossrefs

This is the Hankel transform (see A001906 for the definition) of A000984, A002426, A026375, A026387, A026569, A026585, A026671 and A032351. - John W. Layman, Jul 31 2000
Euler transform of A001037, A209406 (multisets), inverse binomial transform of A000244, binomial transform of A000012.
Complement of A057716.
Boustrophedon transforms: A000734, A000752.
Range of values of A006519, A007875, A011782, A030001, A034444, A037445, A053644, and A054243.
Cf. A018900, A014311, A014312, A014313, A023688, A023689, A023690, A023691 (sum of 2, ..., 9 distinct powers of 2).
Cf. A090129.
The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).

Programs

  • Haskell
    a000079 = (2 ^)
    a000079_list = iterate (* 2) 1
    -- Reinhard Zumkeller, Jan 22 2014, Mar 05 2012, Dec 29 2011
    
  • Magma
    [2^n: n in [0..40]]; // Vincenzo Librandi, Feb 17 2014
    
  • Magma
    [n le 2 select n else 5*Self(n-1)-6*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Feb 17 2014
    
  • Maple
    A000079 := n->2^n; [ seq(2^n,n=0..50) ];
    isA000079 := proc(n)
        local fs;
        fs := numtheory[factorset](n) ;
        if n = 1 then
            true ;
        elif nops(fs) <> 1 then
            false;
        elif op(1,fs) = 2 then
            true;
        else
            false ;
        end if;
    end proc: # R. J. Mathar, Jan 09 2017
  • Mathematica
    Table[2^n, {n, 0, 50}]
    2^Range[0, 50] (* Wesley Ivan Hurt, Jun 14 2014 *)
    LinearRecurrence[{2}, {2}, {0, 20}] (* Eric W. Weisstein, Sep 21 2017 *)
    CoefficientList[Series[1/(1 - 2 x), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
    NestList[2# &, 1, 40] (* Harvey P. Dale, Oct 07 2019 *)
  • Maxima
    A000079(n):=2^n$ makelist(A000079(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    A000079(n)=2^n \\ Edited by M. F. Hasler, Aug 27 2014
    
  • PARI
    unimodal(n)=local(x,d,um,umc); umc=0; for (c=0,n!-1, x=numtoperm(n,c); d=0; um=1; for (j=2,n,if (x[j]x[j-1] && d==1,um=0); if (um==0,break)); if (um==1,print(x)); umc+=um); umc
    
  • Python
    def a(n): return 1<Michael S. Branicky, Jul 28 2022
    
  • Python
    def is_powerof2(n) -> bool: return n and (n & (n - 1)) == 0  # Peter Luschny, Apr 10 2025
  • Scala
    (List.fill(20)(2: BigInt)).scanLeft(1: BigInt)( * ) // Alonso del Arte, Jan 16 2020
    
  • Scheme
    (define (A000079 n) (expt 2 n)) ;; Antti Karttunen, Mar 21 2017
    

Formula

a(n) = 2^n.
a(0) = 1; a(n) = 2*a(n-1).
G.f.: 1/(1 - 2*x).
E.g.f.: exp(2*x).
a(n)= Sum_{k = 0..n} binomial(n, k).
a(n) is the number of occurrences of n in A000523. a(n) = A001045(n) + A001045(n+1). a(n) = 1 + Sum_{k = 0..(n - 1)} a(k). The Hankel transform of this sequence gives A000007 = [1, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Feb 25 2004
n such that phi(n) = n/2, for n > 1, where phi is Euler's totient (A000010). - Lekraj Beedassy, Sep 07 2004
a(n + 1) = a(n) XOR 3*a(n) where XOR is the binary exclusive OR operator. - Philippe Deléham, Jun 19 2005
a(n) = StirlingS2(n + 1, 2) + 1. - Ross La Haye, Jan 09 2008
a(n+2) = 6a(n+1) - 8a(n), n = 1, 2, 3, ... with a(1) = 1, a(2) = 2. - Yosu Yurramendi, Aug 06 2008
a(n) = ka(n-1) + (4 - 2k)a(n-2) for any integer k and n > 1, with a(0) = 1, a(1) = 2. - Jaume Oliver Lafont, Dec 05 2008
a(n) = Sum_{l_1 = 0..n + 1} Sum_{l_2 = 0..n}...Sum_{l_i = 0..n - i}...Sum_{l_n = 0..1} delta(l_1, l_2, ..., l_i, ..., l_n) where delta(l_1, l_2, ..., l_i, ..., l_n) = 0 if any l_i <= l_(i+1) and l_(i+1) != 0 and delta(l_1, l_2, ..., l_i, ..., l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
a(0) = 1, a(1) = 2; a(n) = a(n-1)^2/a(n-2), n >= 2. - Jaume Oliver Lafont, Sep 22 2009
a(n) = A173786(n, n)/2 = A173787(n + 1, n). - Reinhard Zumkeller, Feb 28 2010
If p[i] = i - 1 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 02 2010
If p[i] = Fibonacci(i-2) 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 >= 2, a(n-2) = det A. - Milan Janjic, May 08 2010
The sum of reciprocals, 1/1 + 1/2 + 1/4 + 1/8 + ... + 1/(2^n) + ... = 2. - Mohammad K. Azarian, Dec 29 2010
a(n) = 2*A001045(n) + A078008(n) = 3*A001045(n) + (-1)^n. - Paul Barry, Feb 20 2003
a(n) = A118654(n, 2).
a(n) = A140740(n+1, 1).
a(n) = A131577(n) + A011782(n) = A024495(n) + A131708(n) + A024493(n) = A000749(n) + A038503(n) + A038504(n) + A038505(n) = A139761(n) + A139748(n) + A139714(n) + A133476(n) + A139398(n). - Paul Curtz, Jul 25 2011
a(n) = row sums of A007318. - Susanne Wienand, Oct 21 2011
a(n) = Hypergeometric([-n], [], -1). - Peter Luschny, Nov 01 2011
G.f.: A(x) = B(x)/x, B(x) satisfies B(B(x)) = x/(1 - x)^2. - Vladimir Kruchinin, Nov 10 2011
a(n) = Sum_{k = 0..n} A201730(n, k)*(-1)^k. - Philippe Deléham, Dec 06 2011
2^n = Sum_{k = 1..floor((n+1)/2)} C(n+1, 2k-1). - Dennis P. Walsh, Dec 15 2011
A209229(a(n)) = 1. - Reinhard Zumkeller, Mar 07 2012
A001227(a(n)) = 1. - Reinhard Zumkeller, May 01 2012
Sum_{n >= 1} mobius(n)/a(n) = 0.1020113348178103647430363939318... - R. J. Mathar, Aug 12 2012
E.g.f.: 1 + 2*x/(U(0) - x) where U(k) = 6*k + 1 + x^2/(6*k+3 + x^2/(6*k + 5 + x^2/U(k+1) )); (continued fraction, 3-step). - Sergei N. Gladkovskii, Dec 04 2012
a(n) = det(|s(i+2,j)|, 1 <= i,j <= n), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 04 2013
a(n) = det(|ps(i+1,j)|, 1 <= i,j <= n), where ps(n,k) are Legendre-Stirling numbers of the first kind (A129467). - Mircea Merca, Apr 06 2013
G.f.: W(0), where W(k) = 1 + 2*x*(k+1)/(1 - 2*x*(k+1)/( 2*x*(k+2) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013
a(n-1) = Sum_{t_1 + 2*t_2 + ... + n*t_n = n} multinomial(t_1 + t_2 + ... + t_n; t_1, t_2, ..., t_n). - Mircea Merca, Dec 06 2013
Construct the power matrix T(n,j) = [A^*j]*[S^*(j-1)] where A(n)=(1,1,1,...) and S(n)=(0,1,0,0,...) (where * is convolution operation). Then a(n-1) = Sum_{j=1..n} T(n,j). - David Neil McGrath, Jan 01 2015
a(n) = A000005(A002110(n)). - Ivan N. Ianakiev, May 23 2016
From Ilya Gutkovskiy, Jul 18 2016: (Start)
Exponential convolution of A000012 with themselves.
a(n) = Sum_{k=0..n} A011782(k).
Sum_{n>=0} a(n)/n! = exp(2) = A072334.
Sum_{n>=0} (-1)^n*a(n)/n! = exp(-2) = A092553. (End)
G.f.: (r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) = A090129(x) = (1 + 2x + 2x^2 + 4x^3 + 8x^4 + ...). - Gary W. Adamson, Sep 13 2016
a(n) = A000045(n + 1) + A000045(n) + Sum_{k = 0..n - 2} A000045(k + 1)*2^(n - 2 - k). - Melvin Peralta, Dec 22 2017
a(n) = 7*A077020(n)^2 + A077021(n)^2, n>=3. - Ralf Steiner, Aug 08 2021
a(n)= n + 1 + Sum_{k=3..n+1} (2*k-5)*J(n+2-k), where Jacobsthal number J(n) = A001045(n). - Michael A. Allen, Jan 12 2022
Integral_{x=0..Pi} cos(x)^n*cos(n*x) dx = Pi/a(n) (see Nahin, pp. 69-70). - Stefano Spezia, May 17 2023

Extensions

Clarified a comment T. D. Noe, Aug 30 2009
Edited by Daniel Forgues, May 12 2010
Incorrect comment deleted by Matthew Vandermast, May 17 2014
Comment corrected to match offset by Geoffrey Critzer, Nov 28 2014
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