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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A002162 Decimal expansion of the natural logarithm of 2.

Original entry on oeis.org

6, 9, 3, 1, 4, 7, 1, 8, 0, 5, 5, 9, 9, 4, 5, 3, 0, 9, 4, 1, 7, 2, 3, 2, 1, 2, 1, 4, 5, 8, 1, 7, 6, 5, 6, 8, 0, 7, 5, 5, 0, 0, 1, 3, 4, 3, 6, 0, 2, 5, 5, 2, 5, 4, 1, 2, 0, 6, 8, 0, 0, 0, 9, 4, 9, 3, 3, 9, 3, 6, 2, 1, 9, 6, 9, 6, 9, 4, 7, 1, 5, 6, 0, 5, 8, 6, 3, 3, 2, 6, 9, 9, 6, 4, 1, 8, 6, 8, 7
Offset: 0

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Comments

Newton calculated the first 16 terms of this sequence.
Area bounded by y = tan x, y = cot x, y = 0. - Clark Kimberling, Jun 26 2020
Choose four values independently and uniformly at random from the unit interval [0,1]. Sort them, and label them a,b,c,d from least to greatest (so that a b^2+c^2. - Akiva Weinberger, Dec 02 2024
Define the trihyperboloid to be the intersection of the three solid hyperboloids x^2+y^2-z^2<1, x^2-y^2+z^2<1, and -x^2+y^2+z^2<1. This fits perfectly within the cube [-1,1]^3. Then this is the ratio of the volume of the trihyperboloid to its bounding cube. - Akiva Weinberger, Dec 02 2024

Examples

			0.693147180559945309417232121458176568075500134360255254120680009493393...
		

References

  • G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
  • Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 227.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 250.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3, 2.21, 6.2, and 7.2.
  • 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).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 25 and appendix A, equations 25:14:3 and A:7:3 at pages 232, 670.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 29.

Crossrefs

Cf. A016730 (continued fraction), A002939, A008288, A142979, A142992.

Programs

  • Mathematica
    RealDigits[N[Log[2],200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)
    RealDigits[Log[2],10,120][[1]] (* Harvey P. Dale, Jan 25 2024 *)
  • PARI
    { default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } \\ Harry J. Smith, Apr 21 2009

Formula

log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
log(2) = Integral_{t=0..1} dt/(1+t).
log(2) = (2/3) * (1 + Sum_{k>=1} 2/((4*k)^3-4*k)) (Ramanujan).
log(2) = 4*Sum_{k>=0} (3-2*sqrt(2))^(2*k+1)/(2*k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = 1 - (1/2)*Sum_{k>=1} 1/(k*(2*k+1)). - Jaume Oliver Lafont, Jan 06 2009, Jan 08 2009
log(2) = 4*Sum_{k>=0} 1/((4*k+1)*(4*k+2)*(4*k+3)). - Jaume Oliver Lafont, Jan 08 2009
log(2) = 7/12 + 24*Sum_{k>=1} 1/(A052787(k+4)*A000079(k)). - R. J. Mathar, Jan 23 2009
From Alexander R. Povolotsky, Jul 04 2009: (Start)
log(2) = (1/4)*(3 - Sum_{n>=1} 1/(n*(n+1)*(2*n+1))).
log(2) = (230166911/9240 - Sum_{k>=1} (1/2)^k*(11/k + 10/(k+1) + 9/(k+2) + 8/(k+3) + 7/(k+4) + 6/(k+5) - 6/(k+7) - 7/(k+8) - 8/(k+9) - 9/(k+10) - 10/(k+11)))/35917. (End)
log(2) = A052882/A000670. - Mats Granvik, Aug 10 2009
From log(1-x-x^2) at x=1/2, log(2) = (1/2)*Sum_{k>=1} L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). - Jaume Oliver Lafont, Oct 24 2009
log(2) = Sum_{k>=1} 1/(cos(k*Pi/3)*k*2^k) (cf. A176900). - Jaume Oliver Lafont, Apr 29 2010
log(2) = (Sum_{n>=1} 1/(n^2*(n+1)^2*(2*n+1)) + 11)/16. - Alexander R. Povolotsky, Jan 13 2011
log(2) = ((Sum_{n>=1} (2*n+1)/(Sum_{k=1..n} k^2)^2)+396)/576. - Alexander R. Povolotsky, Jan 14 2011
From Alexander R. Povolotsky, Dec 16 2008: (Start)
log(2) = 105*(319/44100 - Sum_{n>=1} 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7))).
log(2) = 319/420 - (3/2)*Sum_{n>=1} 1/(6*n^2+39*n+63). (End)
log(2) = Sum_{k>=1} A191907(2,k)/k. - Mats Granvik, Jun 19 2011
log(2) = Integral_{x=0..oo} 1/(1 + e^x) dx. - Jean-François Alcover, Mar 21 2013
log(2) = lim_{s->1} zeta(s)*(1-1/2^(s-1)). - Mats Granvik, Jun 18 2013
From Peter Bala, Dec 10 2013: (Start)
log(2) = 2*Sum_{n>=1} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
log(2) = (1/3)*Sum_{n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = (1/12)*Sum_{n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
log(2) = (3/16)*Sum_{n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = (1/12)*Sum_{n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = Sum_{n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
log(2) = lim_{n->oo} Sum_{k=2^n..2^(n+1)-1} 1/k. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 03 2015: (Start)
log(2) = (2/3)*Sum_{k >= 0} 1/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second-order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = (2/3)*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
log(2) = Sum_{n>=1} (Zeta(2*n)-1)/n. - Vaclav Kotesovec, Dec 11 2015
From Peter Bala, Oct 30 2016: (Start)
Asymptotic expansions:
for N even, log(2) - Sum_{k = 1..N/2} (-1)^(k-1)/k ~ (-1)^(N/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1. See Borwein et al., Theorem 1 (b);
for N odd, log(2) - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/k ~ (-1)^((N-1)/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), by Borwein et al., Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then set x = (N - 1)/2, where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. (End)
log(2) = lim_{n->oo} Sum_{k=1..n} sin(1/(n+k)). See Mathematical Reflections link. - Michel Marcus, Jan 07 2017
log(2) = Sum_{n>=1} A006519(n) / ((1 + 2^A006519(n)) * A000265(n) * (1 + A000265(n))). - Nicolas Nagel, Mar 19 2018
From Amiram Eldar, Jul 02 2020: (Start)
Equals Sum_{k>=2} zeta(k)/2^k.
Equals -Sum_{k>=2} log(1 - 1/k^2).
Equals Sum_{k>=1} 1/A002939(k).
Equals Integral_{x=0..Pi/3} tan(x) dx. (End)
log(2) = Integral_{x=0..Pi/2} (sec(x) - tan(x)) dx. - Clark Kimberling, Jul 08 2020
From Peter Bala, Nov 14 2020: (Start)
log(2) = Integral_{x = 0..1} (x - 1)/log(x) dx (Boros and Moll, p. 97).
log(2) = (1/2)*Integral_{x = 0..1} (x + 2)*(x - 1)^2/log(x)^2 dx.
log(2) = (1/4)*Integral_{x = 0..1} (x^2 + 3*x + 4)*(x - 1)^3/log(x)^3 dx. (End)
log(2) = 2*arcsinh(sqrt(2)/4) = 2*sqrt(2)*Sum_{n >= 0} (-1)^n*C(2*n,n)/ ((8*n+4)*32^n) = 3*Sum_{n >= 0} (-1)^n/((8*n+4)*(2^n)*C(2*n,n)). - Peter Bala, Jan 14 2022
log(2) = Integral_{x=0..oo} ( e^(-x) * (1-e^(-2x)) * (1-e^(-4x)) * (1-e^(-6x)) ) / ( x * (1-e^(-14x)) ) dx (see Crux Mathematicorum link). - Bernard Schott, Jul 11 2022
From Peter Bala, Oct 22 2023: (Start)
log(2) = 23/32 + 2!^3/16 * Sum_{n >= 1} (-1)^n * (n + 1)/(n*(n + 1)*(n + 2))^2 = 707/1024 - 4!^3/(16^2 * 2!^2) * Sum_{n >= 1} (-1)^n * (n + 2)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^2 = 42611/61440 + 6!^3/(16^3 * 3!^2) * Sum_{n >= 1} (-1)^n * (n + 3)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6))^2.
More generally, it appears that for k >= 0, log(2) = c(k) + (2*k)!^3/(16^k * k!^2) * Sum_{n >= 1} (-1)^(n+k+1) * (n + k)/(n*(n + 1)*...*(n + 2*k))^2 , where c(k) is a rational approximation to log(2). The first few values of c(k) are [0, 23/32, 707/1024, 42611/61440, 38154331/55050240, 76317139/110100480, 26863086823/38755368960, ...].
Let P(n,k) = n*(n + 1)*...*(n + k).
Conjecture: for k >= 0 and r even with r - 1 <= k, the series Sum_{n >= 1} (-1)^n * (d/dn)^r (P(n,k)) / (P(n,k)^2 = A(r,k)*log(2) + B(r,k), where A(r,k) and B(r,k) are both rational numbers. (End)
From Peter Bala, Nov 13 2023: (Start)
log(2) = 5/8 + (1/8)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)^2 / ( k*(k + 1) )^4
= 257/384 + (3!^5/2^9)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)^2*(2*k + 5) / ( k*(k + 1)*(k + 2)*(k + 3) )^4
= 267515/393216 + (5!^5/2^19)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)*(2*k + 5)^2*(2*k + 7)*(2*k + 9) / ( k*(k + 1)*(k + 2)*(k + 3)*(k + 4)*(k + 5) )^4
log(2) = 3/4 - 1/128 * Sum_{k >= 0} (-1/16)^k * (10*k + 12)*binomial(2*k+2,k+1)/ ((k + 1)*(2*k + 3)). The terms of the series are O(1/(k^(3/2)*4^n)). (End)
log(2) = eta(1) is a period, where eta(x) is the Dirichlet eta function. - Andrea Pinos, Mar 19 2024
log(2) = K_{n>=0} (n^2 + [n=0])/1, where K is the Gauss notation for an infinite continued fraction. In the expanded form, log(2) = 1/(1 + 1/(1 + 4/(1 + 9/1 + 16/(1 + 25/(1 + ... (see Clawson at p. 227). - Stefano Spezia, Jul 01 2024
log(2) = lim_{n->oo} Sum_{k=1..n} 1/(n + k) = lim_{x->0} (2^x - 1)/x = lim_{x->0} (2^x - 2^(-x))/(2*x) (see Finch). - Stefano Spezia, Oct 19 2024
From Colin Linzer, Nov 08 2024: (Start)
log(2) = Integral_{t=0...oo} (1 - tanh(t)) dt.
log(2) = Integral_{t=0...1} arctanh(t) dt.
log(2) = (1/2) * Integral_{t=-1...1} |arctanh(t)| dt. (End)
log(2) = 1 + Sum_{n >= 1} (-1)^n/(n*(4*n^2 - 1)) = 1/2 + (1/2)*Sum_{n >= 1} 1/(n*(4*n^2 - 1)). - Peter Bala, Jan 07 2025
log(2) = Integral_{x=0..1} Integral_{y=0..1} 1/((1 - x*y)*(1 + x)*(1 + y)) dy dx. - Kritsada Moomuang, May 22 2025

A000262 Number of "sets of lists": number of partitions of {1,...,n} into any number of lists, where a list means an ordered subset.

Original entry on oeis.org

1, 1, 3, 13, 73, 501, 4051, 37633, 394353, 4596553, 58941091, 824073141, 12470162233, 202976401213, 3535017524403, 65573803186921, 1290434218669921, 26846616451246353, 588633468315403843, 13564373693588558173, 327697927886085654441, 8281153039765859726341
Offset: 0

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Author

Keywords

Comments

Determinant of n X n matrix M=[m(i,j)] where m(i,i)=i, m(i,j)=1 if i > j, m(i,j)=i-j if j > i. - Vladeta Jovovic, Jan 19 2003
With p(n) = the number of integer partitions of n, d(i) = the number of different parts of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, Sum_{i=1..p(n)} = sum over i and Product_{j=1..d(i)} = product over j, one has: a(n) = Sum_{i=1..p(n)} n!/(Product_{j=1..d(i)} m(i,j)!). - Thomas Wieder, May 18 2005
Consider all n! permutations of the integer sequence [n] = 1,2,3,...,n. The i-th permutation, i=1,2,...,n!, consists of Z(i) permutation cycles. Such a cycle has the length lc(i,j), j=1,...,Z(i). For a given permutation we form the product of all its cycle lengths Product_{j=1..Z(i)} lc(i,j). Furthermore, we sum up all such products for all permutations of [n] which gives Sum_{i=1..n!} Product_{j=1..Z(i)} lc(i,j) = A000262(n). For n=4 we have Sum_{i=1..n!} Product_{j=1..Z(i)} lc(i,j) = 1*1*1*1 + 2*1*1 + 3*1 + 2*1*1 + 3*1 + 2*1 + 3*1 + 4 + 3*1 + 4 + 2*2 + 2*1*1 + 3*1 + 4 + 3*1 + 2*1*1 + 2*2 + 4 + 2*2 + 4 + 3*1 + 2*1*1 + 3*1 + 4 = 73 = A000262(4). - Thomas Wieder, Oct 06 2006
For a finite set S of size n, a chain gang G of S is a partially ordered set (S,<=) consisting only of chains. The number of chain gangs of S is a(n). For example, with S={a, b}and n=2, there are a(2)=3 chain gangs of S, namely, {(a,a),(b,b)}, {(a,a),(a,b),(b,b)} and {(a,a),(b,a),(b,b)}. - Dennis P. Walsh, Feb 22 2007
(-1)*A000262 with the first term set to 1 and A084358 form a reciprocal pair under the list partition transform and associated operations described in A133314. Cf. A133289. - Tom Copeland, Oct 21 2007
Consider the distribution of n unlabeled elements "1" onto n levels where empty levels are allowed. In addition, the empty levels are labeled. Their names are 0_1, 0_2, 0_3, etc. This sequence gives the total number of such distributions. If the empty levels are unlabeled ("0"), then the answer is A001700. Let the colon ":" separate two levels. Then, for example, for n=3 we have a(3)=13 arrangements: 111:0_1:0_2, 0_1:111:0_2, 0_1:0_2:111, 111:0_2:0_1, 0_2:111:0_1, 0_2:0_1:111, 11:1:0, 11:0:1, 0:11:1, 1:11:0, 1:0:11, 0:1:11, 1:1:1. - Thomas Wieder, May 25 2008
Row sums of exponential Riordan array [1,x/(1-x)]. - Paul Barry, Jul 24 2008
a(n) is the number of partitions of [n] (A000110) into lists of noncrossing sets. For example, a(3)=3 counts 12, 1-2, 2-1 and a(4) = 73 counts the 75 partitions of [n] into lists of sets (A000670) except for 13-24, 24-13 which fail to be noncrossing. - David Callan, Jul 25 2008
a(i-j)/(i-j)! gives the value of the non-null element (i, j) of the lower triangular matrix exp(S)/exp(1), where S is the lower triangular matrix - of whatever dimension - having all its (non-null) elements equal to one. - Giuliano Cabrele, Aug 11 2008, Sep 07 2008
a(n) is also the number of nilpotent partial one-one bijections (of an n-element set). Equivalently, it is the number of nilpotents in the symmetric inverse semigroup (monoid). - Abdullahi Umar, Sep 14 2008
A000262 is the exp transform of the factorial numbers A000142. - Thomas Wieder, Sep 10 2008
If n is a positive integer then the infinite continued fraction (1+n)/(1+(2+n)/(2+(3+n)/(3+...))) converges to the rational number A052852(n)/A000262(n). - David Angell (angell(AT)maths.unsw.edu.au), Dec 18 2008
Vladeta Jovovic's formula dated Sep 20 2006 can be restated as follows: a(n) is the n-th ascending factorial moment of the Poisson distribution with parameter (mean) 1. - Shai Covo (green355(AT)netvision.net.il), Jan 25 2010
The umbral exponential generating function for a(n) is (1-x)^{-B}. In other words, write (1-x)^{-B} as a power series in x whose coefficients are polynomials in B, and then replace B^k with the Bell number B_k. We obtain a(0) + a(1)x + a(2)x^2/2! + ... . - Richard Stanley, Jun 07 2010
a(n) is the number of Dyck n-paths (A000108) with its peaks labeled 1,2,...,k in some order, where k is the number of peaks. For example a(2)=3 counts U(1)DU(2)D, U(2)DU(1)D, UU(1)DD where the label at each peak is in parentheses. This is easy to prove using generating functions. - David Callan, Aug 23 2011
a(n) = number of permutations of the multiset {1,1,2,2,...,n,n} such that for 1 <= i <= n, all entries between the two i's exceed i and if any such entries are present, they include n. There are (2n-1)!! permutations satisfying the first condition, and for example: a(3)=13 counts all 5!!=15 of them except 331221 and 122133 which fail the second condition. - David Callan, Aug 27 2014
a(n) is the number of acyclic, injective functions from subsets of [n] to [n]. Let subset D of [n] have size k. The number of acyclic, injective functions from D to [n] is (n-1)!/(n-k-1)! and hence a(n) = Sum_{k=0..n-1} binomial(n,k)*(n-1)!/(n-k-1)!. - Dennis P. Walsh, Nov 05 2015
a(n) is the number of acyclic, injective, labeled directed graphs on n vertices with each vertex having outdegree at most one. - Dennis P. Walsh, Nov 05 2015
For n > 0, a(n) is the number of labeled-rooted skinny-tree forests on n nodes. A skinny tree is a tree in which each vertex has at most one child. Let k denote the number of trees. There are binomial(n,k) ways to choose the roots, binomial(n-1,k-1) ways to choose the number of descendants for each root, and (n-k)! ways to permute those descendants. Summing over k, we obtain a(n) = Sum_{k=1..n} C(n,k)*C(n-1,k-1)*(n-k)!. - Dennis P. Walsh, Nov 10 2015
This is the Sheffer transform of the Bell numbers A000110 with the Sheffer matrix S = |Stirling1| = (1, -log(1-x)) = A132393. See the e.g.f. formula, a Feb 21 2017 comment on A048993, and R. Stanley's Jun 07 2010 comment above. - Wolfdieter Lang, Feb 21 2017
All terms = {1, 3} mod 10. - Muniru A Asiru, Oct 01 2017
We conjecture that for k = 2,3,4,..., the difference a(n+k) - a(n) is divisible by k: if true, then for each k the sequence a(n) taken modulo k is periodic with period dividing k. - Peter Bala, Nov 14 2017
The above conjecture is true - see the Bala link. - Peter Bala, Jan 20 2018
The terms of this sequence can be derived from the numerators of the fractions, produced by the recursion: b(0) = 1, b(n) = 1 + ((n-1) * b(n-1)) / (n-1 + b(n-1)) for n > 0. The denominators give A002720. - Dimitris Valianatos, Aug 01 2018
a(n) is the number of rooted labeled forests on n nodes that avoid the patterns 213, 312, and 123. It is also the number of rooted labeled forests that avoid 312, 213, and 132, as well as the number of rooted labeled forests that avoid 132, 213, and 321. - Kassie Archer, Aug 30 2018
For n > 0, a(n) is the number of partitions of [2n-1] whose nontrivial blocks are of type {a,b}, with a <= n and b > n. In fact, for n > 0, a(n) = A056953(2n-1). - Francesca Aicardi, Nov 03 2022
For n > 0, a(n) is the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select one person from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them). - Enrique Navarrete, Oct 01 2023

Examples

			Illustration of first terms as sets of ordered lists of the first n integers:
  a(1) = 1  : (1)
  a(2) = 3  : (12), (21), (1)(2).
  a(3) = 13 : (123) (6 ways), (12)(3) (2*3 ways) (1)(2)(3) (1 way)
  a(4) = 73 : (1234) (24 ways), (123)(4) (6*4 ways), (12)(34) (2*2*3 ways), (12)(3)(4) (2*6 ways), (1)(2)(3)(4) (1 way).
G.f. = 1 + x + 3*x^2 + 13*x^3 + 73*x^4 + 501*x^4 + 4051*x^5 + 37633*x^6 + 394353*x^7 + ...
		

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 194.
  • 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).

Crossrefs

Row sums of A271703 and for n >= 1 of A008297. Unsigned row sums of A111596.
A002868 is maximal element of the n-th row of A271703 and for n >= 1 of A008297.
Main diagonal of A257740 and of A319501.

Programs

  • GAP
    a:=[1,1];; for n in [3..10^2] do a[n]:=(2*n-3)*a[n-1]-(n-2)*(n-3)*a[n-2]; od; A000262:=a;  # Muniru A Asiru, Oct 01 2017
    
  • Haskell
    a000262 n = a000262_list !! n
    a000262_list = 1 : 1 : zipWith (-)
                   (tail $ zipWith (*) a005408_list a000262_list)
                          (zipWith (*) a002378_list a000262_list)
    -- Reinhard Zumkeller, Mar 06 2014
    
  • Magma
    I:=[1,3]; [1] cat [n le 2 select I[n]  else (2*n-1)*Self(n-1) - (n-1)*(n-2)*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 14 2019
    
  • Magma
    [Factorial(n)*Evaluate(LaguerrePolynomial(n, -1), -1): n in [0..30]]; // G. C. Greubel, Feb 23 2021
    
  • Maple
    A000262 := proc(n) option remember: if n=0 then RETURN(1) fi: if n=1 then RETURN(1) fi: (2*n-1)*procname(n-1) - (n-1)*(n-2)*procname(n-2) end proc:
    for n from 0 to 20 do printf(`%d,`,a(n)) od:
    spec := [S, {S = Set(Prod(Z,Sequence(Z)))}, labeled]; [seq(combstruct[count](spec, size=n), n=0..40)];
    with(combinat):seq(sum(abs(stirling1(n, k))*bell(k), k=0..n), n=0..18); # Zerinvary Lajos, Nov 26 2006
    B:=[S,{S = Set(Sequence(Z,1 <= card),card <=13)},labelled]: seq(combstruct[count](B, size=n), n=0..19);# Zerinvary Lajos, Mar 21 2009
    a := n -> `if`(n=0, 1, n!*hypergeom([1 - n], [2], -1)): seq(simplify(a(n)), n=0..19); # Peter Luschny, Jun 05 2014
  • Mathematica
    Range[0, 19]! CoefficientList[ Series[E^(x/(1-x)), {x, 0, 19}], x] (* Robert G. Wilson v, Apr 04 2005 *)
    a[ n_]:= If[ n<0, 0, n! SeriesCoefficient[ Exp[x/(1-x)], {x, 0, n}]]; (* Michael Somos, Jul 19 2005 *)
    a[n_]:=If[n==0, 1, n! Sum[Binomial[n-1, j]/(j+1)!, {j, 0, n-1}]]; Table[a[n], {n, 0, 30}] (* Wilf *)
    a[0] = 1; a[n_]:= n!*Hypergeometric1F1[n+1, 2, 1]/E; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jun 18 2012, after Shai Covo *)
    Table[Sum[BellY[n, k, Range[n]!], {k, 0, n}], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *)
    a[ n_]:= If[ n<0, 0, n! SeriesCoefficient[ Product[ QPochhammer[x^k]^(-MoebiusMu[k]/k), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 02 2019 *)
    Table[n!*LaguerreL[n, -1, -1], {n, 0, 30}] (* G. C. Greubel, Feb 23 2021 *)
    RecurrenceTable[{a[n] == (2*n - 1)*a[n-1] - (n-1)*(n-2)*a[n-2], a[0] == 1, a[1] == 1}, a, {n, 0, 30}] (* Vaclav Kotesovec, Jul 21 2022 *)
  • Maxima
    makelist(sum(abs(stirling1(n,k))*belln(k),k,0,n),n,0,24); /* Emanuele Munarini, Jul 04 2011 */
    
  • Maxima
    makelist(hypergeometric([-n+1,-n],[],1),n,0,12); /* Emanuele Munarini, Sep 27 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( x / (1 - x) + x * O(x^n)), n))}; /* Michael Somos, Feb 10 2005 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( prod( k=1, n, eta(x^k + x * O(x^n))^( -moebius(k) / k)), n))}; /* Michael Somos, Feb 10 2005 */
    
  • PARI
    {a(n) = s = 1; for(k = 1, n-1, s = 1 + k * s / (k + s)); return( numerator(s))}; /* Dimitris Valianatos, Aug 03 2018 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( prod( k=1, n, (1 - x^k + x * O(x^n))^(-eulerphi(k) / k)), n))}; /* Michael Somos, Jun 02 2019 */
    
  • PARI
    a(n) = if (n==0, 1, (n-1)!*pollaguerre(n-1,1,-1)); \\ Michel Marcus, Feb 23 2021; Jul 13 2024
    
  • Python
    from sympy import hyper, hyperexpand
    def A000262(n): return hyperexpand(hyper((-n+1, -n), [], 1)) # Chai Wah Wu, Jan 14 2022
  • Sage
    A000262 = lambda n: hypergeometric([-n+1, -n], [], 1)
    [simplify(A000262(n)) for n in (0..19)] # Peter Luschny, Apr 08 2015
    

Formula

D-finite with recurrence: a(n) = (2*n-1)*a(n-1) - (n-1)*(n-2)*a(n-2).
E.g.f.: exp( x/(1-x) ).
a(n) = Sum_{k=0..n} |A008275(n,k)| * A000110(k). - Vladeta Jovovic, Feb 01 2003
a(n) = (n-1)!*LaguerreL(n-1,1,-1) for n >= 1. - Vladeta Jovovic, May 10 2003
Representation as n-th moment of a positive function on positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x-1)*BesselI(1, 2*x^(1/2))/x^(1/2) dx, n >= 1. - Karol A. Penson, Dec 04 2003
a(n) = Sum_{k=0..n} A001263(n, k)*k!. - Philippe Deléham, Dec 10 2003
a(n) = n! Sum_{j=0..n-1} binomial(n-1, j)/(j+1)!, for n > 0. - Herbert S. Wilf, Jun 14 2005
Asymptotic expansion for large n: a(n) -> (0.4289*n^(-1/4) + 0.3574*n^(-3/4) - 0.2531*n^(-5/4) + O(n^(-7/4)))*(n^n)*exp(-n + 2*sqrt(n)). - Karol A. Penson, Aug 28 2002
Minor part of this asymptotic expansion is wrong! Right is (in closed form): a(n) ~ n^(n-1/4)*exp(-1/2+2*sqrt(n)-n)/sqrt(2) * (1 - 5/(48*sqrt(n)) - 95/(4608*n)), numerically a(n) ~ (0.42888194248*n^(-1/4) - 0.0446752023417*n^(-3/4) - 0.00884196713*n^(-5/4) + O(n^(-7/4))) *(n^n)*exp(-n+2*sqrt(n)). - Vaclav Kotesovec, Jun 02 2013
a(n) = exp(-1)*Sum_{m>=0} [m]^n/m!, where [m]^n = m*(m+1)*...*(m+n-1) is the rising factorial. - Vladeta Jovovic, Sep 20 2006
Recurrence: D(n,k) = D(n-1,k-1) + (n-1+k) * D(n-1,k) n >= k >= 0; D(n,0)=0. From this, D(n,1) = n! and D(n,n)=1; a(n) = Sum_{i=0..n} D(n,i). - Stephen Dalton (StephenMDalton(AT)gmail.com), Jan 05 2007
Proof: Notice two distinct subsets of the lists for [n]: 1) n is in its own list, then there are D(n-1,k-1); 2) n is in a list with other numbers. Denoting the separation of lists by |, it is not hard to see n has (n-1+k) possible positions, so (n-1+k) * D(n-1,k). - Stephen Dalton (StephenMDalton(AT)gmail.com), Jan 05 2007
Define f_1(x), f_2(x), ... such that f_1(x) = exp(x), f_{n+1}(x) = (d/dx)(x^2*f_n(x)), for n >= 2. Then a(n-1) = exp(-1)*f_n(1). - Milan Janjic, May 30 2008
a(n) = (n-1)! * Sum_{k=1..n} (a(n-k)*k!)/((n-k)!*(k-1)!), where a(0)=1. - Thomas Wieder, Sep 10 2008
a(n) = exp(-1)*n!*M(n+1,2,1), n >= 1, where M (=1F1) is the confluent hypergeometric function of the first kind. - Shai Covo (green355(AT)netvision.net.il), Jan 20 2010
a(n) = n!* A067764(n) / A067653(n). - Gary Detlefs, May 15 2010
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x)^2*d/dx. Cf. A000110, A049118, A049119 and A049120. - Peter Bala, Nov 25 2011
From Sergei N. Gladkovskii, Nov 17 2011, Aug 02 2012, Dec 11 2012, Jan 27 2013, Jul 31 2013, Dec 25 2013: (Start)
Continued fractions:
E.g.f.: Q(0) where Q(k) = 1+x/((1-x)*(2k+1)-x*(1-x)*(2k+1)/(x+(1-x)*(2k+2)/Q(k+1))).
E.g.f.: 1 + x/(G(0)-x) where G(k) = (1-x)*k + 1 - x*(1-x)*(k+1)/G(k+1).
E.g.f.: exp(x/(1-x)) = 4/(2-(x/(1-x))*G(0))-1 where G(k) = 1 - x^2/(x^2 + 4*(1-x)^2*(2*k+1)*(2*k+3)/G(k+1) ).
E.g.f.: 1 + x*(E(0)-1)/(x+1) where E(k) = 1 + 1/(k+1)/(1-x)/(1-x/(x+1/E(k+1) )).
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - x/(x + (k+1)*(1-x)/E(k+1) )).
E.g.f.: E(0)-1, where E(k) = 2 + x/( (2*k+1)*(1-x) - x/E(k+1) ).
(End)
E.g.f.: Product {n >= 1} ( (1 + x^n)/(1 - x^n) )^( phi(2*n)/(2*n) ), where phi(n) = A000010(n) is the Euler totient function. Cf. A088009. - Peter Bala, Jan 01 2014
a(n) = n!*hypergeom([1-n],[2],-1) for n >= 1. - Peter Luschny, Jun 05 2014
a(n) = (-1)^(n-1)*KummerU(1-n,2,-1). - Peter Luschny, Sep 17 2014
a(n) = hypergeom([-n+1, -n], [], 1) for n >= 0. - Peter Luschny, Apr 08 2015
E.g.f.: Product_{k>0} exp(x^k). - Franklin T. Adams-Watters, May 11 2016
0 = a(n)*(18*a(n+2) - 93*a(n+3) + 77*a(n+4) - 17*a(n+5) + a(n+6)) + a(n+1)*(9*a(n+2) - 80*a(n+3) + 51*a(n+4) - 6*a(n+5)) + a(n+2)*(3*a(n+2) - 34*a(n+3) + 15*a(n+4)) + a(n+3)*(-10*a(n+3)) if n >= 0. - Michael Somos, Feb 27 2017
G.f. G(x)=y satisfies a differential equation: (1-x)*y-2*(1-x)*x^2*y'+x^4*y''=1. - Bradley Klee, Aug 13 2018
a(n) = n! * LaguerreL(n, -1, -1) = c_{n}(n-1; -1) where c_{n}(x; a) are the Poisson - Charlier polynomials. - G. C. Greubel, Feb 23 2021
3 divides a(3*n-1); 9 divides a(9*n-1); 11 divides a(11*n-1). - Peter Bala, Mar 26 2022
For n > 0, a(n) = Sum_{k=0..n-1}*k!*C(n-1,k)*C(n,k). - Francesca Aicardi, Nov 03 2022
For n > 0, a(n) = (n-1)! * (Sum_{i=0..n-1} A002720(i) / i!). - Werner Schulte, Mar 29 2024
a(n+1) = numerator of (1 + n/(1 + n/(1 + (n-1)/(1 + (n-1)/(1 + ... + 1/(1 + 1/(1))))))). See A002720 for the denominators. - Peter Bala, Feb 11 2025

A002033 Number of perfect partitions of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 8, 1, 3, 3, 8, 1, 8, 1, 8, 3, 3, 1, 20, 2, 3, 4, 8, 1, 13, 1, 16, 3, 3, 3, 26, 1, 3, 3, 20, 1, 13, 1, 8, 8, 3, 1, 48, 2, 8, 3, 8, 1, 20, 3, 20, 3, 3, 1, 44, 1, 3, 8, 32, 3, 13, 1, 8, 3, 13, 1, 76, 1, 3, 8, 8, 3, 13, 1, 48, 8, 3, 1, 44, 3, 3, 3, 20, 1, 44, 3, 8, 3, 3, 3, 112
Offset: 0

Views

Author

Keywords

Comments

A perfect partition of n is one which contains just one partition of every number less than n when repeated parts are regarded as indistinguishable. Thus 1^n is a perfect partition for every n; and for n = 7, 4 1^3, 4 2 1, 2^3 1 and 1^7 are all perfect partitions. [Riordan]
Also number of ordered factorizations of n+1, see A074206.
Also number of gozinta chains from 1 to n (see A034776). - David W. Wilson
a(n) is the permanent of the n X n matrix with (i,j) entry = 1 if j|i+1 and = 0 otherwise. For n=3 the matrix is {{1, 1, 0}, {1, 0, 1}, {1, 1, 0}} with permanent = 2. - David Callan, Oct 19 2005
Appears to be the number of permutations that contribute to the determinant that gives the Moebius function. Verified up to a(9). - Mats Granvik, Sep 13 2008
Dirichlet inverse of A153881 (assuming offset 1). - Mats Granvik, Jan 03 2009
Equals row sums of triangle A176917. - Gary W. Adamson, Apr 28 2010
A partition is perfect iff it is complete (A126796) and knapsack (A108917). - Gus Wiseman, Jun 22 2016
a(n) is also the number of series-reduced planted achiral trees with n + 1 unlabeled leaves, where a rooted tree is series-reduced if all terminal subtrees have at least two branches, and achiral if all branches directly under any given node are equal. Also Moebius transform of A067824. - Gus Wiseman, Jul 13 2018

Examples

			n=0: 1 (the empty partition)
n=1: 1 (1)
n=2: 1 (11)
n=3: 2 (21, 111)
n=4: 1 (1111)
n=5: 3 (311, 221, 11111)
n=6: 1 (111111)
n=7: 4 (4111, 421, 2221, 1111111)
From _Gus Wiseman_, Jul 13 2018: (Start)
The a(11) = 8 series-reduced planted achiral trees with 12 unlabeled leaves:
  (oooooooooooo)
  ((oooooo)(oooooo))
  ((oooo)(oooo)(oooo))
  ((ooo)(ooo)(ooo)(ooo))
  ((oo)(oo)(oo)(oo)(oo)(oo))
  (((ooo)(ooo))((ooo)(ooo)))
  (((oo)(oo)(oo))((oo)(oo)(oo)))
  (((oo)(oo))((oo)(oo))((oo)(oo)))
(End)
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 126, see #27.
  • R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 141.
  • D. E. Knuth, The Art of Computer Programming, Pre-Fasc. 3b, Sect. 7.2.1.5, no. 67, p. 25.
  • P. A. MacMahon, The theory of perfect partitions and the compositions of multipartite numbers, Messenger Math., 20 (1891), 103-119.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 123-124.
  • 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).

Crossrefs

Same as A074206, up to the offset and initial term there.
Cf. A176917.
For parity see A008966.

Programs

  • Maple
    a := array(1..150): for k from 1 to 150 do a[k] := 0 od: a[1] := 1: for j from 2 to 150 do for m from 1 to j-1 do if j mod m = 0 then a[j] := a[j]+a[m] fi: od: od: for k from 1 to 150 do printf(`%d,`,a[k]) od: # James Sellers, Dec 07 2000
    # alternative
    A002033 := proc(n)
        option remember;
        local a;
        if n <= 2 then
            return 1;
        else
            a := 0 ;
            for i from 0 to n-1 do
                if modp(n+1,i+1) = 0 then
                    a := a+procname(i);
                end if;
            end do:
        end if;
        a ;
    end proc: # R. J. Mathar, May 25 2017
  • Mathematica
    a[0] = 1; a[1] = 1; a[n_] := a[n] = a /@ Most[Divisors[n]] // Total; a /@ Range[96]  (* Jean-François Alcover, Apr 06 2011, updated Sep 23 2014. NOTE: This produces A074206(n) = a(n-1). - M. F. Hasler, Oct 12 2018 *)
  • PARI
    A002033(n) = if(n,sumdiv(n+1,i,if(i<=n,A002033(i-1))),1) \\ Michael B. Porter, Nov 01 2009, corrected by M. F. Hasler, Oct 12 2018
    
  • Python
    from functools import lru_cache
    from sympy import divisors
    @lru_cache(maxsize=None)
    def A002033(n):
        if n <= 1:
            return 1
        return sum(A002033(i-1) for i in divisors(n+1,generator=True) if i <= n) # Chai Wah Wu, Jan 12 2022

Formula

From David Wasserman, Nov 14 2006: (Start)
a(n-1) = Sum_{i|d, i < n} a(i-1).
a(p^k-1) = 2^(k-1).
a(n-1) = A067824(n)/2 for n > 1.
a(A122408(n)-1) = A122408(n)/2. (End)
a(A025487(n)-1) = A050324(n). - R. J. Mathar, May 26 2017
a(n) = (A253249(n+1)+1)/4, n > 0. - Geoffrey Critzer, Aug 19 2020

Extensions

Edited by M. F. Hasler, Oct 12 2018

A074206 Kalmár's [Kalmar's] problem: number of ordered factorizations of n.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 8, 1, 3, 3, 8, 1, 8, 1, 8, 3, 3, 1, 20, 2, 3, 4, 8, 1, 13, 1, 16, 3, 3, 3, 26, 1, 3, 3, 20, 1, 13, 1, 8, 8, 3, 1, 48, 2, 8, 3, 8, 1, 20, 3, 20, 3, 3, 1, 44, 1, 3, 8, 32, 3, 13, 1, 8, 3, 13, 1, 76, 1, 3, 8, 8, 3, 13, 1, 48, 8, 3, 1, 44, 3, 3, 3, 20, 1, 44, 3, 8, 3, 3, 3, 112
Offset: 0

Views

Author

N. J. A. Sloane, Apr 29 2003

Keywords

Comments

a(0)=0, a(1)=1; thereafter a(n) is the number of ordered factorizations of n as a product of integers greater than 1.
Kalmár (1931) seems to be the earliest reference that mentions this sequence (as opposed to A002033). - N. J. A. Sloane, May 05 2016
a(n) is the permanent of the n-1 X n-1 matrix A with (i,j) entry = 1 if j|i+1 and = 0 otherwise. This is because ordered factorizations correspond to nonzero elementary products in the permanent. For example, with n=6, 3*2 -> 1,3,6 [partial products] -> 6,3,1 [reverse list] -> (6,3)(3,1) [partition into pairs with offset 1] -> (5,3)(2,1) [decrement first entry] -> (5,3)(2,1)(1,2)(3,4)(4,5) [append pairs (i,i+1) to get a permutation] -> elementary product A(1,2)A(2,1)A(3,4)A(4,5)A(5,3). - David Callan, Oct 19 2005
This sequence is important in describing the amount of energy in all wave structures in the Universe according to harmonics theory. - Ray Tomes (ray(AT)tomes.biz), Jul 22 2007
a(n) appears to be the number of permutation matrices contributing to the Moebius function. See A008683 for more information. Also a(n) appears to be the Moebius transform of A067824. Furthermore it appears that except for the first term a(n)=A067824(n)*(1/2). Are there other sequences such that when the Moebius transform is applied, the new sequence is also a constant factor times the starting sequence? - Mats Granvik, Jan 01 2009
Numbers divisible by n distinct primes appear to have ordered factorization values that can be found in an n-dimensional summatory Pascal triangle. For example, the ordered factorization values for numbers divisible by two distinct primes can be found in table A059576. - Mats Granvik, Sep 06 2009
Inverse Mobius transform of A174725 and also except for the first term, inverse Mobius transform of A174726. - Mats Granvik, Mar 28 2010
a(n) is a lower bound on the worst-case number of solutions to the probed partial digest problem for n fragments of DNA; see the Newberg & Naor reference, below. - Lee A. Newberg, Aug 02 2011
All integers greater than 1 are perfect numbers over this sequence (for definition of A-perfect numbers, see comment to A175522). - Vladimir Shevelev, Aug 03 2011
If n is squarefree, then a(n) = A000670(A001221(n)) = A000670(A001222(n)). - Vladimir Shevelev and Franklin T. Adams-Watters, Aug 05 2011
A034776 lists the values taken by this sequence. - Robert G. Wilson v, Jun 02 2012
From Gus Wiseman, Aug 25 2020: (Start)
Also the number of strict chains of divisors from n to 1. For example, the a(n) chains for n = 1, 2, 4, 6, 8, 12, 30 are:
1 2/1 4/1 6/1 8/1 12/1 30/1
4/2/1 6/2/1 8/2/1 12/2/1 30/2/1
6/3/1 8/4/1 12/3/1 30/3/1
8/4/2/1 12/4/1 30/5/1
12/6/1 30/6/1
12/4/2/1 30/10/1
12/6/2/1 30/15/1
12/6/3/1 30/6/2/1
30/6/3/1
30/10/2/1
30/10/5/1
30/15/3/1
30/15/5/1
(End)
a(n) is also the number of ways to tile a strip of length log(n) with tiles having lengths {log(k) : k>=2}. - David Bevan, Jan 07 2025

Examples

			G.f. = x + x^2 + x^3 + 2*x^4 + x^5 + 3*x^6 + x^7 + 4*x^8 + 2*x^9 + 3*x^10 + ...
Number of ordered factorizations of 8 is 4: 8 = 2*4 = 4*2 = 2*2*2.
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 126, see #27.
  • R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 141.
  • Kalmár, Laszlo, A "factorisatio numerorum" problemajarol [Hungarian], Matemat. Fizik. Lapok, 38 (1931), 1-15.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 124.

Crossrefs

Apart from initial term, same as A002033.
a(A002110) = A000670, row sums of A251683.
A173382 (and A025523) gives partial sums.
A124433 has these as unsigned row sums.
A334996 has these as row sums.
A001055 counts factorizations.
A001222 counts prime factors with multiplicity.
A008480 counts ordered prime factorizations.
A067824 counts strict chains of divisors starting with n.
A122651 counts strict chains of divisors summing to n.
A253249 counts strict chains of divisors.

Programs

  • Haskell
    a074206 n | n <= 1 = n
    | otherwise = 1 + (sum $ map (a074206 . (div n)) $
    tail $ a027751_row n)
    -- Reinhard Zumkeller, Oct 01 2012
    
  • Maple
    a := array(1..150): for k from 1 to 150 do a[k] := 0 od: a[1] := 1: for j from 2 to 150 do for m from 1 to j-1 do if j mod m = 0 then a[j] := a[j]+a[m] fi: od: od: for k from 1 to 150 do printf(`%d,`,a[k]) od: # James Sellers, Dec 07 2000
    A074206:= proc(n) option remember; if n > 1 then `+`(op(apply(A074206, numtheory[divisors](n)[1..-2]))) else n fi end: # M. F. Hasler, Oct 12 2018
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := a[n] = a /@ Most[Divisors[n]] // Total; a /@ Range[20000] (* N. J. A. Sloane, May 04 2016, based on program in A002033 *)
    ccc[n_]:=Switch[n,0,{},1,{{1}},,Join@@Table[Prepend[#,n]&/@ccc[d],{d,Most[Divisors[n]]}]]; Table[Length[ccc[n]],{n,0,100}] (* _Gus Wiseman, Aug 25 2020 *)
  • PARI
    A=vector(100);A[1]=1; for(n=2,#A,A[n]=1+sumdiv(n,d,A[d])); A/=2; A[1]=1; concat(0,A) \\ Charles R Greathouse IV, Nov 20 2012
    
  • PARI
    {a(n) = if( n<2, n>0, my(A = divisors(n)); sum(k=1, #A-1, a(A[k])))}; /* Michael Somos, Dec 26 2016 */
    
  • PARI
    A074206(n)=if(n>1, sumdiv(n, i, if(iA074206(i))),n) \\ M. F. Hasler, Oct 12 2018
    
  • PARI
    A74206=[1]; A074206(n)={if(#A74206A074206(i)))} \\ Use memoization for computing many values. - M. F. Hasler, Oct 12 2018
    
  • PARI
    first(n) = {my(res = vector(n, i, 1)); for(i = 2, n, for(j = 2, n \ i, res[i*j] += res[i])); concat(0, res)} \\ David A. Corneth, Oct 13 2018
    
  • PARI
    first(n) = {my(res = vector(n, i, 1)); for(i = 2, n, d = divisors(i); res[i] += sum(j = 1, #d-1, res[d[j]])); concat(0, res)} \\ somewhat faster than progs above for finding first terms of n. \\ David A. Corneth, Oct 12 2018
    
  • PARI
    a(n)={if(!n, 0, my(sig=factor(n)[,2], m=vecsum(sig)); sum(k=0, m, prod(i=1, #sig, binomial(sig[i]+k-1, k-1))*sum(r=k, m, binomial(r,k)*(-1)^(r-k))))} \\ Andrew Howroyd, Aug 30 2020
    
  • Python
    from math import prod
    from functools import lru_cache
    from sympy import divisors, factorint, prime
    @lru_cache(maxsize=None)
    def A074206(n): return sum(A074206(d) for d in divisors(prod(prime(i+1)**e for i,e in enumerate(sorted(factorint(n).values(),reverse=True))),generator=True,proper=True)) if n > 1 else n # Chai Wah Wu, Sep 16 2022
  • SageMath
    @cached_function
    def minus_mu(n):
        if n < 2: return n
        return sum(minus_mu(d) for d in divisors(n)[:-1])
    # Note that changing the sign of the sum gives the Möbius function A008683.
    print([minus_mu(n) for n in (0..96)]) # Peter Luschny, Dec 26 2016
    

Formula

With different offset: a(n) = sum of all a(i) such that i divides n and i < n. - Clark Kimberling
a(p^k) = 2^(k-1) if k>0 and p is a prime.
Dirichlet g.f.: 1/(2-zeta(s)). - Herbert S. Wilf, Apr 29 2003
a(n) = A067824(n)/2 for n>1; a(A122408(n)) = A122408(n)/2. - Reinhard Zumkeller, Sep 03 2006
If p,q,r,... are distinct primes, then a(p*q)=3, a(p^2*q)=8, a(p*q*r)=13, a(p^3*q)=20, etc. - Vladimir Shevelev, Aug 03 2011 [corrected by Charles R Greathouse IV, Jun 02 2012]
a(0) = 0, a(1) = 1; a(n) = [x^n] Sum_{k=1..n-1} a(k)*x^k/(1 - x^k). - Ilya Gutkovskiy, Dec 11 2017
a(n) = a(A046523(n)); a(A025487(n)) = A050324(n): a(n) depends only on the nonzero exponents in the prime factorization of n, more precisely prime signature of n, cf. A124010 and A320390. - M. F. Hasler, Oct 12 2018
a(n) = A000670(A001221(n)) for squarefree n. In particular a(A002110(n)) = A000670(n). - Amiram Eldar, May 13 2019
a(n) = A050369(n)/n, for n>=1. - Ridouane Oudra, Aug 31 2019
a(n) = A361665(A181819(n)). - Pontus von Brömssen, Mar 25 2023
From Ridouane Oudra, Nov 02 2023: (Start)
If p,q are distinct primes, and n,m>0 then we have:
a(p^n*q^m) = Sum_{k=0..min(n,m)} 2^(n+m-k-1)*binomial(n,k)*binomial(m,k);
More generally: let tau[k](n) denote the number of ordered factorizations of n as a product of k terms, also named the k-th Piltz function (see A007425), then we have for n>1:
a(n) = Sum_{j=1..bigomega(n)} Sum_{k=1..j} (-1)^(j-k)*binomial(j,k)*tau[k](n), or
a(n) = Sum_{j=1..bigomega(n)} Sum_{k=0..j-1} (-1)^k*binomial(j,k)*tau[j-k](n). (End)

Extensions

Originally this sequence was merged with A002033, the number of perfect partitions. Herbert S. Wilf suggested that it warrants an entry of its own.

A000182 Tangent (or "Zag") numbers: e.g.f. tan(x), also (up to signs) e.g.f. tanh(x).

Original entry on oeis.org

1, 2, 16, 272, 7936, 353792, 22368256, 1903757312, 209865342976, 29088885112832, 4951498053124096, 1015423886506852352, 246921480190207983616, 70251601603943959887872, 23119184187809597841473536, 8713962757125169296170811392, 3729407703720529571097509625856
Offset: 1

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Keywords

Comments

Number of Joyce trees with 2n-1 nodes. Number of tremolo permutations of {0,1,...,2n}. - Ralf Stephan, Mar 28 2003
The Hankel transform of this sequence is A000178(n) for n odd = 1, 12, 34560, ...; example: det([1, 2, 16; 2, 16, 272, 16, 272, 7936]) = 34560. - Philippe Deléham, Mar 07 2004
a(n) is the number of increasing labeled full binary trees with 2n-1 vertices. Full binary means every non-leaf vertex has two children, distinguished as left and right; labeled means the vertices are labeled 1,2,...,2n-1; increasing means every child has a label greater than its parent. - David Callan, Nov 29 2007
From Micha Hofri (hofri(AT)wpi.edu), May 27 2009: (Start)
a(n) was found to be the number of permutations of [2n] which when inserted in order, to form a binary search tree, yield the maximally full possible tree (with only one single-child node).
The e.g.f. is sec^2(x)=1+tan^2(x), and the same coefficients can be manufactured from the tan(x) itself, which is the e.g.f. for the number of trees as above for odd number of nodes. (End)
a(n) is the number of increasing strict binary trees with 2n-1 nodes. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
For relations to alternating permutations, Euler and Bernoulli polynomials, zigzag numbers, trigonometric functions, Fourier transform of a square wave, quantum algebras, and integrals over and in n-dimensional hypercubes and over Green functions, see Hodges and Sukumar. For further discussion on the quantum algebra, see the later Hodges and Sukumar reference and the paper by Hetyei presenting connections to the general combinatorial theory of Viennot on orthogonal polynomials, inverse polynomials, tridiagonal matrices, and lattice paths (thereby related to continued fractions and cumulants). - Tom Copeland, Nov 30 2014
The Zigzag Hankel transform is A000178. That is, A000178(2*n - k) = det( [a(i+j - k)]{i,j = 1..n} ) for n>0 and k=0,1. - _Michael Somos, Mar 12 2015
a(n) is the number of standard Young tableaux of skew shape (n,n,n-1,n-2,...,3,2)/(n-1,n-2,n-3,...,2,1). - Ran Pan, Apr 10 2015
For relations to the Sheffer Appell operator calculus and a Riccati differential equation for generating the Meixner-Pollaczek and Krawtchouk orthogonal polynomials, see page 45 of the Feinsilver link and Rzadkowski. - Tom Copeland, Sep 28 2015
For relations to an elliptic curve, a Weierstrass elliptic function, the Lorentz formal group law, a Lie infinitesimal generator, and the Eulerian numbers A008292, see A155585. - Tom Copeland, Sep 30 2015
Absolute values of the alternating sums of the odd-numbered rows (where the single 1 at the apex of the triangle is counted as row #1) of the Eulerian triangle, A008292. The actual alternating sums alternate in sign, e.g., 1, -2, 16, -272, etc. (Even-numbered rows have alternating sums always 0.) - Gregory Gerard Wojnar, Sep 28 2018
The sequence is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 1]. - Allen Stenger, Aug 03 2020
From Peter Bala, Dec 24 2021: (Start)
Conjectures:
1) The sequence taken modulo any integer k eventually becomes periodic with period dividing phi(k).
2) The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k, except when p = 2, n = 1 and k = 1 or 2.
3) For i >= 1 define a_i(n) = a(n+i). The Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i >= 1 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients. For an example, see A262145.(End)

Examples

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... = x + 1/3*x^3 + 2/15*x^5 + 17/315*x^7 + 62/2835*x^9 + O(x^11).
tanh(x) = x - 1/3*x^3 + 2/15*x^5 - 17/315*x^7 + 62/2835*x^9 - 1382/155925*x^11 + ...
(sec x)^2 = 1 + x^2 + 2/3*x^4 + 17/45*x^6 + ...
a(3)=16 because we have: {1, 3, 2, 5, 4}, {1, 4, 2, 5, 3}, {1, 4, 3, 5, 2},
  {1, 5, 2, 4, 3}, {1, 5, 3, 4, 2}, {2, 3, 1, 5, 4}, {2, 4, 1, 5, 3},
  {2, 4, 3, 5, 1}, {2, 5, 1, 4, 3}, {2, 5, 3, 4, 1}, {3, 4, 1, 5, 2},
  {3, 4, 2, 5, 1}, {3, 5, 1, 4, 2}, {3, 5, 2, 4, 1}, {4, 5, 1, 3, 2},
  {4, 5, 2, 3, 1}. - _Geoffrey Critzer_, May 19 2013
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 111.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
  • L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 148 (the numbers |C^{2n-1}|).
  • J. W. Milnor and J. D. Stasheff, Characteristic Classes, Princeton, 1974, p. 282.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • H. Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1, p. 20.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoullischen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.
  • 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).
  • E. van Fossen Conrad, Some continued fraction expansions of elliptic functions, PhD thesis, The Ohio State University, 2002, p. 28.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, pp. 267-268.

Crossrefs

A350972 is essentially the same sequence.
a(n)=2^(n-1)*A002105(n). Apart from signs, 2^(2n-2)*A001469(n) = n*a(n).
Cf. A001469, A002430, A036279, A000364 (secant numbers), A000111 (secant-tangent numbers), A024283, A009764. First diagonal of A059419 and of A064190.
Equals A002425(n) * 2^A101921(n).
Equals leftmost column of A162005. - Johannes W. Meijer, Jun 27 2009

Programs

  • Maple
    series(tan(x),x,40);
    with(numtheory): a := n-> abs(2^(2*n)*(2^(2*n)-1)*bernoulli(2*n)/(2*n));
    A000182_list := proc(n) local T,k,j; T[1] := 1;
    for k from 2 to n do T[k] := (k-1)*T[k-1] od;
       for k from 2 to n do
           for j from k to n do
               T[j] := (j-k)*T[j-1]+(j-k+2)*T[j] od od;
    seq(T[j], j=1..n)  end:
    A000182_list(15);  # Peter Luschny, Apr 02 2012
  • Mathematica
    Table[ Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}], {n, 0, 7}] (* Victor Adamchik, Oct 05 2005 *)
    v[1] = 2; v[n_] /; n >= 2 := v[n] = Sum[ Binomial[2 n - 3, 2 k - 2] v[k] v[n - k], {k, n - 1}]; Table[ v[n]/2, {n, 15}] (* Zerinvary Lajos, Jul 08 2009 *)
    Rest@ Union[ Range[0, 29]! CoefficientList[ Series[ Tan[x], {x, 0, 30}], x]] (* Harvey P. Dale, Oct 19 2011; modified by Robert G. Wilson v, Apr 02 2012 *)
    t[1, 1] = 1; t[1, 0] = 0; t[n_ /; n > 1, m_] := t[n, m] = m*(m+1)*Sum[t[n-1, k], {k, m-1, n-1}]; a[n_] := t[n, 1]; Table[a[n], {n, 1, 15}]  (* Jean-François Alcover, Jan 02 2013, after A064190 *)
    a[ n_] := If[ n < 1, 0, With[{m = 2 n - 1}, m! SeriesCoefficient[ Tan[x], {x, 0, m}]]]; (* Michael Somos, Mar 12 2015 *)
    a[ n_] := If[ n < 1, 0, ((-16)^n - (-4)^n) Zeta[1 - 2 n]]; (* Michael Somos, Mar 12 2015 *)
    Table[2 PolyGamma[2n - 1, 1/2]/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
    a[ n_] := a[n] = If[ n < 2, Boole[n == 1], Sum[Binomial[2 n - 2, 2 k - 1] a[k] a[n - k], {k, n - 1}]]; (* Michael Somos, Aug 02 2018 *)
    a[n_] := (2^(2*n)*(2^(2*n) - 1)*Abs[BernoulliB[2*n]])/(2*n); a /@  Range[20]  (* Stan Wagon, Nov 21 2022 *)
  • Maxima
    a(n):=sum(sum(binomial(k,r)*sum(sum(binomial(l,j)/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(j-2*i)^(2*n),i,0,floor((j-1)/2))*(-1)^(l-j),j,1,l)*(-1)^l*binomial(r+l-1,r-1),l,1,2*n)*(-1)^(1-r),r,1,k)/k,k,1,2*n); /* Vladimir Kruchinin, Aug 23 2010 */
    
  • Maxima
    a[n]:=if n=1 then 1 else 2*sum(sum(binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j),k,1,j),j,1,n-1);
    makelist(a[n],n,1,30); /* Tani Akinari, Sep 20 2023 */
    
  • PARI
    {a(n) = if( n<1, 0, ((-4)^n - (-16)^n) * bernfrac(2*n) / (2*n))};
    
  • PARI
    {a(n) = my(an); if( n<2, n==1, an = vector(n, m, 1); for( m=2, n, an[m] = sum( k=1, m-1, binomial(2*m - 2, 2*k - 1) * an[k] * an[m-k])); an[n])}; /* Michael Somos */
    
  • PARI
    {a(n) = if( n<1, 0, (2*n - 1)! * polcoeff( tan(x + O(x^(2*n + 2))), 2*n - 1))}; /* Michael Somos */
    
  • PARI
    {a(n) = my(X=x+x*O(x^n),Egf); Egf = x*sum(m=0,n, prod(k=1,m, tanh(2*k*X))); (n-1)!*polcoeff(Egf,n)} /* Paul D. Hanna, May 11 2010 */
    
  • PARI
    /* Continued Fraction for the e.g.f. tan(x), from Paul D. Hanna: */
    {a(n)=local(CF=1+O(x)); for(i=1, n, CF=1/(2*(n-i+1)-1-x^2*CF)); (2*n-1)!*polcoeff(x*CF, 2*n-1)}
    
  • PARI
    /* O.g.f. Sum_{n>=1} a(n)*x^n, from Paul D. Hanna Feb 05 2013: */
    {a(n)=polcoeff( x+2*x*sum(m=1, n, x^m*prod(k=1, m, (2*k-1)^2/(1+(2*k-1)^2*x +x*O(x^n))) ), n)}
    
  • Python
    # The objective of this implementation is efficiency.
    # n -> [0, a(1), a(2), ..., a(n)] for n > 0.
    def A000182_list(n):
        T = [0 for i in range(1, n+2)]
        T[1] = 1
        for k in range(2, n+1):
            T[k] = (k-1)*T[k-1]
        for k in range(2, n+1):
            for j in range(k, n+1):
                T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
        return T
    print(A000182_list(100)) # Peter Luschny, Aug 07 2011
    
  • Python
    from sympy import bernoulli
    def A000182(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023
    
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A000182_list(len) :
        R = []; A = {-1:0, 0:1}; k = 0; e = 1
        for i in (0..2*len-1) :
            Am = 0; A[k + e] = 0; e = -e
            for j in (0..i) : Am += A[k]; A[k] = Am; k += e
            if e > 0 : R.append(A[i//2])
        return R
    A000182_list(15) # Peter Luschny, Mar 31 2012

Formula

E.g.f.: log(sec x) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!.
E.g.f.: tan x = Sum_{n >= 0} a(n+1)*x^(2*n+1)/(2*n+1)!.
E.g.f.: (sec x)^2 = Sum_{n >= 0} a(n+1)*x^(2*n)/(2*n)!.
2/(exp(2x)+1) = 1 + Sum_{n>=1} (-1)^(n+1) a(n) x^(2n-1)/(2n-1)! = 1 - x + x^3/3 - 2*x^5/15 + 17*x^7/315 - 62*x^9/2835 + ...
a(n) = 2^(2*n) (2^(2*n) - 1) |B_(2*n)| / (2*n) where B_n are the Bernoulli numbers (A000367/A002445 or A027641/A027642).
Asymptotics: a(n) ~ 2^(2*n+1)*(2*n-1)!/Pi^(2*n).
Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}]. - Victor Adamchik, Oct 05 2005
a(n) = abs[c(2*n-1)] where c(n)= 2^(n+1) * (1-2^(n+1)) * Ber(n+1)/(n+1) = 2^(n+1) * (1-2^(n+1)) * (-1)^n * Zeta(-n) = [ -(1+EN(.))]^n = 2^n * GN(n+1)/(n+1) = 2^n * EP(n,0) = (-1)^n * E(n,-1) = (-2)^n * n! * Lag[n,-P(.,-1)/2] umbrally = (-2)^n * n! * C{T[.,P(.,-1)/2] + n, n} umbrally for the signed Euler numbers EN(n), the Bernoulli numbers Ber(n), the Genocchi numbers GN(n), the Euler polynomials EP(n,t), the Eulerian polynomials E(n,t), the Touchard / Bell polynomials T(n,t), the binomial function C(x,y) = x!/[(x-y)!*y! ] and the polynomials P(j,t) of A131758. - Tom Copeland, Oct 05 2007
a(1) = A094665(0,0)*A156919(0,0) and a(n) = Sum_{k=1..n-1} 2^(n-k-1)*A094665(n-1, k)*A156919(k,0) for n = 2, 3, .., see A162005. - Johannes W. Meijer, Jun 27 2009
G.f.: 1/(1-1*2*x/(1-2*3*x/(1-3*4*x/(1-4*5*x/(1-5*6*x/(1-... (continued fraction). - Paul Barry, Feb 24 2010
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-2x-12x^2/(1-18x-240x^2/(1-50x-1260x^2/(1-98x-4032x^2/(1-162x-9900x^2/(1-... (continued fraction);
coefficient sequences given by 4*(n+1)^2*(2n+1)*(2n+3) and 2(2n+1)^2 (see Van Fossen Conrad reference). (End)
E.g.f.: x*Sum_{n>=0} Product_{k=1..n} tanh(2*k*x) = Sum_{n>=1} a(n)*x^n/(n-1)!. - Paul D. Hanna, May 11 2010 [corrected by Paul D. Hanna, Sep 28 2023]
a(n) = (-1)^(n+1)*Sum_{j=1..2*n+1} j!*Stirling2(2*n+1,j)*2^(2*n+1-j)*(-1)^j for n >= 0. Vladimir Kruchinin, Aug 23 2010: (Start)
If n is odd such that 2*n-1 is prime, then a(n) == 1 (mod (2*n-1)); if n is even such that 2*n-1 is prime, then a(n) == -1 (mod (2*n-1)). - Vladimir Shevelev, Sep 01 2010
Recursion: a(n) = (-1)^(n-1) + Sum_{i=1..n-1} (-1)^(n-i+1)*C(2*n-1,2*i-1)* a(i). - Vladimir Shevelev, Aug 08 2011
E.g.f.: tan(x) = Sum_{n>=1} a(n)*x^(2*n-1)/(2*n-1)! = x/(1 - x^2/(3 - x^2/(5 - x^2/(7 - x^2/(9 - x^2/(11 - x^2/(13 -...))))))) (continued fraction from J. H. Lambert - 1761). - Paul D. Hanna, Sep 21 2011
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 09 2013: (Start)
Continued fractions:
E.g.f.: (sec(x))^2 = 1+x^2/(x^2+U(0)) where U(k) = (k+1)*(2k+1) - 2x^2 + 2x^2*(k+1)*(2k+1)/U(k+1).
E.g.f.: tan(x) = x*T(0) where T(k) = 1-x^2/(x^2-(2k+1)*(2k+3)/T(k+1)).
E.g.f.: tan(x) = x/(G(0)+x) where G(k) = 2*k+1 - 2*x + x/(1 + x/G(k+1)).
E.g.f.: tanh(x) = x/(G(0)-x) where G(k) = k+1 + 2*x - 2*x*(k+1)/G(k+1).
E.g.f.: tan(x) = 2*x - x/W(0) where W(k) = 1 + x^2*(4*k+5)/((4*k+1)*(4*k+3)*(4*k+5) - 4*x^2*(4*k+3) + x^2*(4*k+1)/W(k+1)).
E.g.f.: tan(x) = x/T(0) where T(k) = 1 - 4*k^2 + x^2*(1 - 4*k^2)/T(k+1).
E.g.f.: tan(x) = -3*x/(T(0)+3*x^2) where T(k)= 64*k^3 + 48*k^2 - 4*k*(2*x^2 + 1) - 2*x^2 - 3 - x^4*(4*k -1)*(4*k+7)/T(k+1).
G.f.: 1/G(0) where G(k) = 1 - 2*x*(2*k+1)^2 - x^2*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1).
G.f.: 2*Q(0) - 1 where Q(k) = 1 + x^2*(4*k + 1)^2/(x + x^2*(4*k + 1)^2 - x^2*(4*k + 3)^2*(x + x^2*(4*k + 1)^2)/(x^2*(4*k + 3)^2 + (x + x^2*(4*k + 3)^2)/Q(k+1) )).
G.f.: (1 - 1/G(0))*sqrt(-x), where G(k) = 1 + sqrt(-x) - x*(k+1)^2/G(k+1).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/Q(k+1)). (End)
O.g.f.: x + 2*x*Sum_{n>=1} x^n * Product_{k=1..n} (2*k-1)^2 / (1 + (2*k-1)^2*x). - Paul D. Hanna, Feb 05 2013
a(n) = (-4)^n*Li_{1-2*n}(-1). - Peter Luschny, Jun 28 2012
a(n) = (-4)^n*(4^n-1)*Zeta(1-2*n). - Jean-François Alcover, Dec 05 2013
Asymptotic expansion: 4*((2*(2*n-1))/(Pi*e))^(2*n-1/2)*exp(1/2+1/(12*(2*n-1))-1/(360*(2*n-1)^3)+1/(1260*(2*n-1)^5)-...). (See Luschny link.) - Peter Luschny, Jul 14 2015
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = tan(x) satisfies the differential equation A''(x) = 2*A(x)*A'(x) with A(0) = 0 and A'(0) = 1, leading to the recurrence a(0) = 0, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0, 1, 0, 2, 0, 16, 0, 272, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 1/2 and a(1) = 1 produces A080635. Cf. A002105, A234797. (End)
a(n) = 2*polygamma(2*n-1, 1/2)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
a(n) = 2^(2n-2)*|p(2n-1,-1/2)|, where p_n(x) are the shifted row polynomials of A019538. E.g., a(2) = 2 = 2^2 * |1 + 6(-1/2) + 6(-1/2)^2|. - Tom Copeland, Oct 19 2016
From Peter Bala, May 05 2017: (Start)
With offset 0, the o.g.f. A(x) = 1 + 2*x + 16*x^2 + 272*x^3 + ... has the property that its 4th binomial transform 1/(1 - 4*x) A(x/(1 - 4*x)) has the S-fraction representation 1/(1 - 6*x/(1 - 2*x/(1 - 20*x/(1 - 12*x/(1 - 42*x/(1 - 30*x/(1 - ...))))))), where the coefficients [6, 2, 20, 12, 42, 30, ...] in the partial numerators of the continued fraction are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. Compare with the S-fraction associated with A(x) given above by Paul Barry.
A(x) = 1/(1 + x - 3*x/(1 - 4*x/(1 + x - 15*x/(1 - 16*x/(1 + x - 35*x/(1 - 36*x/(1 + x - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36,...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,.... (End)
a(n) = Sum_{k = 1..n-1} binomial(2*n-2, 2*k-1) * a(k) * a(n-k), with a(1) = 1. - Michael Somos, Aug 02 2018
a(n) = 2^(2*n-1) * |Euler(2*n-1, 0)|, where Euler(n,x) are the Euler polynomials. - Daniel Suteu, Nov 21 2018 (restatement of one of Copeland's 2007 formulas.)
x - Sum_{n >= 1} (-1)^n*a(n)*x^(2*n)/(2*n)! = x - log(cosh(x)). The series reversion of x - log(cosh(x)) is (1/2)*x - (1/2)*log(2 - exp(x)) = Sum_{n >= 0} A000670(n)*x^(n+1)/(n+1)!. - Peter Bala, Jul 11 2022
For n > 1, a(n) = 2*Sum_{j=1..n-1} Sum_{k=1..j} binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j). - Tani Akinari, Sep 20 2023
a(n) = A110501(n) * 4^(n-1) / n (Han and Liu, 2018). - Amiram Eldar, May 17 2024

A019538 Triangle of numbers T(n,k) = k!*Stirling2(n,k) read by rows (n >= 1, 1 <= k <= n).

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 14, 36, 24, 1, 30, 150, 240, 120, 1, 62, 540, 1560, 1800, 720, 1, 126, 1806, 8400, 16800, 15120, 5040, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880, 1, 1022, 55980, 818520, 5103000, 16435440, 29635200, 30240000, 16329600, 3628800
Offset: 1

Views

Author

N. J. A. Sloane and Manfred Goebel (goebel(AT)informatik.uni-tuebingen.de), Dec 11 1996

Keywords

Comments

Number of ways n labeled objects can be distributed into k nonempty parcels. Also number of special terms in n variables with maximal degree k.
In older terminology these are called differences of 0. - Michael Somos, Oct 08 2003
Number of surjections (onto functions) from an n-element set to a k-element set.
Also coefficients (in ascending order) of so-called ordered Bell polynomials.
(k-1)!*Stirling2(n,k-1) is the number of chain topologies on an n-set having k open sets [Stephen].
Number of set compositions (ordered set partitions) of n items into k parts. Number of k dimensional 'faces' of the n dimensional permutohedron (see Simion, p. 162). - Mitch Harris, Jan 16 2007
Correction of comment before: Number of (n-k)-dimensional 'faces' of the permutohedron of order n (an (n-1)-dimensional polytope). - Tilman Piesk, Oct 29 2014
This array is related to the reciprocal of an e.g.f. as sketched in A133314. For example, the coefficient of the fourth-order term in the Taylor series expansion of 1/(a(0) + a(1) x + a(2) x^2/2! + a(3) x^3/3! + ...) is a(0)^(-5) * {24 a(1)^4 - 36 a(1)^2 a(2) a(0) + [8 a(1) a(3) + 6 a(2)^2] a(0)^2 - a(4) a(0)^3}. The unsigned coefficients characterize the P3 permutohedron depicted on page 10 in the Loday link with 24 vertices (0-D faces), 36 edges (1-D faces), 6 squares (2-D faces), 8 hexagons (2-D faces) and 1 3-D permutohedron. Summing coefficients over like dimensions gives A019538 and A090582. Compare to A133437 for the associahedron. - Tom Copeland, Sep 29 2008, Oct 07 2008
Further to the comments of Tom Copeland above, the permutohedron of type A_3 can be taken as the truncated octahedron. Its dual is the tetrakis hexahedron, a simplicial polyhedron, with f-vector (1,14,36,24) giving the fourth row of this triangle. See the Wikipedia entry and [Fomin and Reading p. 21]. The corresponding h-vectors of permutohedra of type A give the rows of the triangle of Eulerian numbers A008292. See A145901 and A145902 for the array of f-vectors for type B and type D permutohedra respectively. - Peter Bala, Oct 26 2008
Subtriangle of triangle in A131689. - Philippe Deléham, Nov 03 2008
Since T(n,k) counts surjective functions and surjective functions are "consistent", T(n,k) satisfies a binomial identity, namely, T(n,x+y) = Sum_{j=0..n} C(n,j)*T(j,x)*T(n-j,y). For definition of consistent functions and a generalized binomial identity, see "Toy stories and combinatorial identities" in the link section below. - Dennis P. Walsh, Feb 24 2012
T(n,k) is the number of labeled forests on n+k vertices satisfying the following two conditions: (i) each forest consists of exactly k rooted trees with roots labeled 1, 2, ..., k; (ii) every root has at least one child vertex. - Dennis P. Walsh, Feb 24 2012
The triangle is the inverse binomial transform of triangle A028246, deleting the left column and shifting up one row. - Gary W. Adamson, Mar 05 2012
See A074909 for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
E.g.f. for the shifted signed polynomials is G(x,t) = (e^t-1)/[1+(1+x)(e^t-1)] = 1-(1+x)(e^t-1) + (1+x)^2(e^t-1)^2 - ... (see also A008292 and A074909), which has the infinitesimal generator g(x,u)d/du = [(1-x*u)(1-(1+x)u)]d/du, i.e., exp[t*g(x,u)d/du]u eval. at u=0 gives G(x,t), and dG(x,t)/dt = g(x,G(x,t)). The compositional inverse is log((1-xt)/(1-(1+x)t)). G(x,t) is a generating series associated to the generalized Hirzebruch genera. See the G. Rzadowski link for the relation of the derivatives of g(x,u) to solutions of the Riccatt differential equation, soliton solns. to the KdV equation, and the Eulerian and Bernoulli numbers. In addition A145271 connects products of derivatives of g(x,u) and the refined Eulerian numbers to the inverse of G(x,t), which gives the normalized, reverse face polynomials of the simplices (A135278, divided by n+1). See A028246 for the generator g(x,u)d/dx. - Tom Copeland, Nov 21 2014
For connections to toric varieties and Eulerian polynomials, see the Dolgachev and Lunts and the Stembridge links. - Tom Copeland, Dec 31 2015
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra (this entry) and stellahedra. - Tom Copeland, Nov 14 2016
T(n, k) appears in a Worpitzky identity relating monomials to binomials: x^n = Sum_{k=1..n} T(n, k)*binomial(x,k), n >= 1. See eq. (11.) of the Worpitzky link on p. 209. The relation to the Eulerian numbers is given there in eqs. (14.) and (15.). See the formula below relating to A008292. See also Graham et al. eq. (6.10) (relating monomials to falling factorials) on p. 248 (2nd ed. p. 262). The Worpitzky identity given in the Graham et al. reference as eq. (6.37) (2nd ed. p. 269) is eq. (5.), p. 207, of Worpitzky. - Wolfdieter Lang, Mar 10 2017
T(n, m) is also the number of minimum clique coverings and minimum matchings in the complete bipartite graph K_{m,n}. - Eric W. Weisstein, Apr 26 2017
From the Hasan and Franco and Hasan papers: The m-permutohedra for m=1,2,3,4 are the line segment, hexagon, truncated octahedron and omnitruncated 5-cell. The first three are well-known from the study of elliptic models, brane tilings and brane brick models. The m+1 torus can be tiled by a single (m+2)-permutohedron. Relations to toric Calabi-Yau Kahler manifolds are also discussed. - Tom Copeland, May 14 2020
From Manfred Boergens, Jul 25 2021: (Start)
Number of n X k binary matrices with row sums = 1 and no zero columns. These matrices are a subset of the matrices defining A183109.
The distribution into parcels in the leading comment can be regarded as a covering of [n] by tuples (A_1,...,A_k) in P([n])^k with nonempty and disjoint A_j, with P(.) denoting the power set (corrected for clarity by Manfred Boergens, May 26 2024). For the non-disjoint case see A183109 and A218695.
For tuples with "nonempty" dropped see A089072. For tuples with "nonempty and disjoint" dropped see A092477 and A329943 (amendment by Manfred Boergens, Jun 24 2024). (End)

Examples

			The triangle T(n, k) begins:
  n\k 1    2     3      4       5        6        7        8        9      10
  1:  1
  2:  1    2
  3:  1    6     6
  4:  1   14    36     24
  5:  1   30   150    240     120
  6:  1   62   540   1560    1800      720
  7:  1  126  1806   8400   16800    15120     5040
  8:  1  254  5796  40824  126000   191520   141120    40320
  9:  1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014
---------------------------------------------------------------------------
T(4,1) = 1: {1234}. T(4,2) = 14: {1}{234} (4 ways), {12}{34} (6 ways), {123}{4} (4 ways). T(4,3) = 36: {12}{3}{4} (12 ways), {1}{23}{4} (12 ways), {1}{2}{34} (12 ways). T(4,4) = 1: {1}{2}{3}{4} (1 way).
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 89, ex. 1; also p. 210.
  • Miklos Bona, Combinatorics of Permutations, Chapman and Hall,2004, p.12.
  • G. Boole, A Treatise On The Calculus of Finite Differences, Dover Publications, 1960, p. 20.
  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, 1989, p. 155. Also eqs.(6.10) and (6.37).
  • Kiran S. Kedlaya and Andrew V. Sutherland, Computing L -Series of Hyperelliptic Curves in Algorithmic Number Theory Lecture Notes in Computer Science Volume 5011/2008.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.6.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
  • J. F. Steffensen, Interpolation, 2nd ed., Chelsea, NY, 1950, see p. 54.
  • A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.
  • E. Whittaker and G. Robinson, The Calculus of Observations, Blackie, London, 4th ed., 1949; p. 7.

Crossrefs

Row sums give A000670. Maximal terms in rows give A002869. Central terms T(2k-1,k) give A233734.
Diagonal is n! (A000142). 2nd diagonal is A001286. 3rd diagonal is A037960.
Reflected version of A090582. A371568 is another version.
See also the two closely related triangles: A008277(n, k) = T(n, k)/k! (Stirling numbers of second kind) and A028246(n, k) = T(n, k)/k.
Cf. A033282 'faces' of the associahedron.
Cf. A008292, A047969, A145901, A145902. - Peter Bala, Oct 26 2008
Visible in the 3-D array in A249042.
See also A000182.

Programs

  • Haskell
    a019538 n k = a019538_tabl !! (n-1) !! (k-1)
    a019538_row n = a019538_tabl !! (n-1)
    a019538_tabl = iterate f [1] where
       f xs = zipWith (*) [1..] $ zipWith (+) ([0] ++ xs) (xs ++ [0])
    -- Reinhard Zumkeller, Dec 15 2013
    
  • Maple
    with(combinat): A019538 := (n,k)->k!*stirling2(n,k);
  • Mathematica
    Table[k! StirlingS2[n, k], {n, 9}, {k, n}] // Flatten
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, sum(i=0, k, (-1)^i * binomial(k, i) * (k-i)^n))}; /* Michael Somos, Oct 08 2003 */
    
  • Sage
    def T(n, k): return factorial(k)*stirling_number2(n,k) # Danny Rorabaugh, Oct 10 2015

Formula

T(n, k) = k*(T(n-1, k-1)+T(n-1, k)) with T(0, 0) = 1 [or T(1, 1) = 1]. - Henry Bottomley, Mar 02 2001
E.g.f.: (y*(exp(x)-1) - exp(x))/(y*(exp(x)-1) - 1). - Vladeta Jovovic, Jan 30 2003
Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*j^n*binomial(k, j). - Mario Catalani (mario.catalani(AT)unito.it), Nov 28 2003. See Graham et al., eq. (6.19), p. 251. For a proof see Bert Seghers, Jun 29 2013.
Sum_{k=0..n} T(n, k)(-1)^(n-k) = 1, Sum_{k=0..n} T(n, k)(-1)^k = (-1)^n. - Mario Catalani (mario.catalani(AT)unito.it), Dec 11 2003
O.g.f. for n-th row: polylog(-n, x/(1+x))/(x+x^2). - Vladeta Jovovic, Jan 30 2005
E.g.f.: 1 / (1 + t*(1-exp(x))). - Tom Copeland, Oct 13 2008
From Peter Bala, Oct 26 2008: (Start)
O.g.f. as a continued fraction: 1/(1 - x*t/(1 - (x + 1)*t/(1 - 2*x*t/(1 - 2*(x + 1)*t/(1 - ...))))) = 1 + x*t + (x + 2*x^2)*t^2 + (x + 6*x^2 + 6*x^3)*t^3 + ... .
The row polynomials R(n,x), which begin R(1,x) = x, R(2,x) = x + 2*x^2, R(3,x) = x + 6*x^2 + 6*x^3, satisfy the recurrence x*d/dx ((x + 1)*R(n,x)) = R(n+1,x). It follows that the zeros of R(n,x) are real and negative (apply Corollary 1.2 of [Liu and Wang]).
Since this is the triangle of f-vectors of the (simplicial complexes dual to the) type A permutohedra, whose h-vectors form the Eulerian number triangle A008292, the coefficients of the polynomial (x-1)^n*R(n,1/(x-1)) give the n-th row of A008292. For example, from row 3 we have x^2 + 6*x + 6 = 1 + 4*y + y^2, where y = x + 1, producing [1,4,1] as the third row of A008292. The matrix product A008292 * A007318 gives the mirror image of this triangle (see A090582).
For n,k >= 0, T(n+1,k+1) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*[(j+1)^(n+1) - j^(n+1)]. The matrix product of Pascal's triangle A007318 with the current array gives (essentially) A047969. This triangle is also related to triangle A047969 by means of the S-transform of [Hetyei], a linear transformation of polynomials whose value on the basis monomials x^k is given by S(x^k) = binomial(x,k). The S-transform of the shifted n-th row polynomial Q(n,x) := R(n,x)/x is S(Q(n,x)) = (x+1)^n - x^n. For example, from row 3 we obtain S(1 + 6*x + 6*x^2) = 1 + 6*x + 6*x*(x-1)/2 = 1 + 3*x + 3*x^2 = (x+1)^3 - x^3. For fixed k, the values S(Q(n,k)) give the nonzero entries in column (k-1) of the triangle A047969 (the Hilbert transform of the Eulerian numbers). (End)
E.g.f.: (exp(x)-1)^k = sum T(n,k)x^n/n!. - Vladimir Kruchinin, Aug 10 2010
T(n,k) = Sum_{i=1..k} A(n,i)*Binomial(n-i,k-i) where A(n,i) is the number of n-permutations that have i ascending runs, A008292.
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f. A(x,t) = -1 + 1/(1+t*(1-exp(x))), the comp. inverse in x is B(x,t) = log(((1+t)/t) - 1/(t(1+x))).
With h(x,t) = 1/(dB/dx)= (1+x)((1+t)(1+x)-1), the row polynomial P(n,t) is given by (h(x,t)*d/dx)^n x, eval. at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t), with P(0,t)=0.
(A factor of -1/n! was removed by Copeland on Aug 25 2016.) (End)
The term linear in x of [x*h(d/dx,t)]^n 1 gives the n-th row polynomial. (See A134685.) - Tom Copeland, Nov 07 2011
Row polynomials are given by D^n(1/(1-x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. - Peter Bala, Nov 25 2011
T(n,x+y) = Sum_{j=0..n} binomial(n,j)*T(j,x)*T(n-j,y). - Dennis P. Walsh, Feb 24 2012
Let P be a Rota-Baxter operator of weight 1 satisfying the identity P(x)*P(y) = P(P(x)*y) + P(x*P(y)) + P(x*y). Then P(1)^2 = P(1) + 2*P^2(1). More generally, Guo shows that P(1)^n = Sum_{k=1..n} T(n,k)*P^k(1). - Peter Bala, Jun 08 2012
Sum_{i=1..n} (-1)^i*T(n,i)/i = 0, for n > 1. - Leonid Bedratyuk, Aug 09 2012
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(k, j)*(k-j)^n. [M. Catalani's re-indexed formula from Nov 28 2003] Proof: count the surjections of [n] onto [k] with the inclusion-exclusion principle, as an alternating sum of the number of functions from [n] to [k-j]. - Bert Seghers, Jun 29 2013
n-th row polynomial = 1/(1 + x)*( Sum_{k>=0} k^n*(x/(1 + x))^k ), valid for x in the open interval (-1/2, inf). See Tanny link. Cf. A145901. - Peter Bala, Jul 22 2014
T(n,k) = k * A141618(n,k-1) / binomial(n,k-1). - Tom Copeland, Oct 25 2014
Sum_{n>=0} n^k*a^n = Sum_{i=1..k} (a / (1 - a))^i * T(k, i)/(1-a) for |a| < 1. - David A. Corneth, Mar 09 2015
From Peter Bala, May 26 2015: (Start)
The row polynomials R(n,x) satisfy (1 + x)*R(n,x) = (-1)^n*x*R(n,-(1 + x)).
For a fixed integer k, the expansion of the function A(k,z) := exp( Sum_{n >= 1} R(n,k)*z^n/n ) has integer coefficients and satisfies the functional equation A(k,z)^(k + 1) = BINOMIAL(A(k,z))^k, where BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z). Cf. A145901. For cases see A084784 (k = 1), A090352 (k = 2), A090355 (k = 3), A090357 (k = 4), A090362 (k = 5) and A084785 (k = -2 with z -> -z).
A(k,z)^(k + 1) = A(-(k + 1),-z)^k and hence BINOMIAL(A(k,z)) = A(-(k + 1),-z). (End)
From Tom Copeland, Oct 19 2016: (Start)
Let a(1) = 1 + x + B(1) = x + 1/2 and a(n) = B(n) = (B.)^n, where B(n) are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted, signed row polynomials of this array: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... and p_n(x) = n * b(n-1), where b(n) are the partition polynomials of A133314 evaluated with these a(n).
Sum_{n > 0} R(n,-1/2) x^n/n! = 2 * tanh(x/2), where R(n,x) = Sum_{k = 1..n} T(n,k) x^(k-1) are the shifted row polynomials of this entry, so R(n,-1/2) = 4 * (2^(n+1)-1) B(n+1)/(n+1). (Cf. A000182.)
(End)
Also the Bernoulli numbers are given by B(n) = Sum_{k =1..n} (-1)^k T(n,k) / (k+1). - Tom Copeland, Nov 06 2016
G.f. for column k: k! x^k / Product_{i=1..k} (1-i*x). - Robert A. Russell, Sep 25 2018
a(j) <= A183109(j). - Manfred Boergens, Jul 25 2021

A333217 Numbers k such that the k-th composition in standard order covers an initial interval of positive integers.

Original entry on oeis.org

0, 1, 3, 5, 6, 7, 11, 13, 14, 15, 21, 22, 23, 26, 27, 29, 30, 31, 37, 38, 41, 43, 44, 45, 46, 47, 50, 52, 53, 54, 55, 58, 59, 61, 62, 63, 75, 77, 78, 83, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 101, 102, 105, 106, 107, 108, 109, 110, 111, 114, 116, 117, 118
Offset: 1

Views

Author

Gus Wiseman, Mar 15 2020

Keywords

Comments

The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again.

Examples

			The sequence of terms together with the corresponding compositions begins:
    0: ()              37: (3,2,1)           75: (3,2,1,1)
    1: (1)             38: (3,1,2)           77: (3,1,2,1)
    3: (1,1)           41: (2,3,1)           78: (3,1,1,2)
    5: (2,1)           43: (2,2,1,1)         83: (2,3,1,1)
    6: (1,2)           44: (2,1,3)           85: (2,2,2,1)
    7: (1,1,1)         45: (2,1,2,1)         86: (2,2,1,2)
   11: (2,1,1)         46: (2,1,1,2)         87: (2,2,1,1,1)
   13: (1,2,1)         47: (2,1,1,1,1)       89: (2,1,3,1)
   14: (1,1,2)         50: (1,3,2)           90: (2,1,2,2)
   15: (1,1,1,1)       52: (1,2,3)           91: (2,1,2,1,1)
   21: (2,2,1)         53: (1,2,2,1)         92: (2,1,1,3)
   22: (2,1,2)         54: (1,2,1,2)         93: (2,1,1,2,1)
   23: (2,1,1,1)       55: (1,2,1,1,1)       94: (2,1,1,1,2)
   26: (1,2,2)         58: (1,1,2,2)         95: (2,1,1,1,1,1)
   27: (1,2,1,1)       59: (1,1,2,1,1)      101: (1,3,2,1)
   29: (1,1,2,1)       61: (1,1,1,2,1)      102: (1,3,1,2)
   30: (1,1,1,2)       62: (1,1,1,1,2)      105: (1,2,3,1)
   31: (1,1,1,1,1)     63: (1,1,1,1,1,1)    106: (1,2,2,2)
		

Crossrefs

Sequences covering an initial interval are counted by A000670.
Composition in standard order are A066099.
The case of strictly increasing initial intervals is A164894.
The case of strictly decreasing initial intervals is A246534.
The case of permutations is A333218.
The weakly increasing version is A333379.
The weakly decreasing version is A333380.

Programs

  • Mathematica
    normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],normQ[stc[#]]&]

A000629 Number of necklaces of partitions of n+1 labeled beads.

Original entry on oeis.org

1, 2, 6, 26, 150, 1082, 9366, 94586, 1091670, 14174522, 204495126, 3245265146, 56183135190, 1053716696762, 21282685940886, 460566381955706, 10631309363962710, 260741534058271802, 6771069326513690646, 185603174638656822266, 5355375592488768406230
Offset: 0

Views

Author

N. J. A. Sloane, Don Knuth, Nick Singer (nsinger(AT)eos.hitc.com)

Keywords

Comments

Also the number of logically distinct strings of first order quantifiers in which n variables occur (C. S. Peirce, c. 1903). - Stephen Pollard (spollard(AT)truman.edu), Jun 07 2002
Stirling transform of A052849(n) = [2, 4, 12, 48, 240, ...] is a(n) = [2, 6, 26, 150, 1082, ...]. - Michael Somos, Mar 04 2004
Stirling transform of A000142(n-1) = [1, 1, 2, 6, 24, ...] is a(n-1) = [1, 2, 6, 26, ...]. - Michael Somos, Mar 04 2004
Stirling transform of (-1)^n * A024167(n-1) = [0, 1, -1, 5, -14, 94, ...] is a(n-2) = [0, 1, 2, 6, 26, ...]. - Michael Somos, Mar 04 2004
The asymptotic expansion of 2*log(n) - (2^1*log(1) + 2^2*log(2) + ... + 2^n*log(n))/2^n is (a(1)/1)/n + (a(2)/2)/n^2 + (a(3)/3)/n^3 + ... - Michael Somos, Aug 22 2004
This is the sequence of cumulants of the probability distribution of the number of tails before the first head in a sequence of fair coin tosses. - Michael Hardy (hardy(AT)math.umn.edu), May 01 2005
Appears to be row sums of A154921. - Mats Granvik, Jan 18 2009
This is the number of cyclically ordered partitions of n+1 labeled points. The ordered version is A000670. - Michael Somos, Jan 08 2011
A000670(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, Apr 27 2012
Row sums of A154921 as conjectured above by Granvik. a(n) gives the number of outcomes of a race between n horses H1,...,Hn, where if a horse falls it is not ranked. For example, when n = 2 the 6 outcomes are a dead heat, H1 wins H2 second, H2 wins H1 second, H1 wins H2 falls, H2 wins H1 falls or both fall. - Peter Bala, May 15 2012
Also the number of disjoint areas of a Venn diagram for n multisets. - Aurelian Radoaca, Jun 27 2016
Also the number of ways of ordering n nonnegative integers, allowing for the possibility of ties, and also comparing the smallest integers with 0. Each comparison with 0 gives two possibilities, x > 0 or x=0. As such, without comparison with 0, we get A000670, the number of ways of ordering n nonnegative integers, allowing for the possibility of ties, or the number of ways n competitors can rank in a competition, allowing for the possibility of ties. For instance, for 2 nonnegative integers x,y, there are the following 6 ways of ordering them: x = y = 0, x = y > 0, x > y = 0, x > y > 0, y > x = 0, y > x > 0. - Aurelian Radoaca, Jul 09 2016
Also the number of ordered set partitions of subsets of {1,...,n}. Also the number of chains of distinct nonempty subsets of {1,...,n}. - Gus Wiseman, Feb 01 2019
Number of combinations of a Simplex lock having n buttons.
Row sums of the unsigned cumulant expansion polynomials A127671 and logarithmic polynomials A263634. - Tom Copeland, Jun 04 2021
Also the number of vertices in the axis-aligned polytope consisting of all vectors x in R^n where, for all k in {1,...,n}, the k-th smallest coordinate of x lies in the interval [0, k]. - Adam P. Goucher, Jan 18 2023
Number of idempotent Boolean relation matrices whose complement is also idempotent. See Rosenblatt link. - Geoffrey Critzer, Feb 26 2023

Examples

			a(2)=6: the necklace representatives on 1,2,3 are ({123}), ({12},{3}), ({13},{2}), ({23},{1}), ({1},{2},{3}), ({1},{3},{2})
G.f. = 1 + 2*x + 6*x^2 + 26*x^3 + 150*x^4 + 1082*x^5 + 9366*x^6 + 94586*x^7 + ...
From _Gus Wiseman_, Feb 01 2019: (Start)
The a(3) = 26 ordered set partitions of subsets of {1,2,3} are:
  {}  {{1}}  {{2}}  {{3}}  {{12}}    {{13}}    {{23}}    {{123}}
                           {{1}{2}}  {{1}{3}}  {{2}{3}}  {{1}{23}}
                           {{2}{1}}  {{3}{1}}  {{3}{2}}  {{12}{3}}
                                                         {{13}{2}}
                                                         {{2}{13}}
                                                         {{23}{1}}
                                                         {{3}{12}}
                                                         {{1}{2}{3}}
                                                         {{1}{3}{2}}
                                                         {{2}{1}{3}}
                                                         {{2}{3}{1}}
                                                         {{3}{1}{2}}
                                                         {{3}{2}{1}}
(End)
		

References

  • R. Austin, R. K. Guy, and R. Nowakowski, unpublished notes, circa 1987.
  • N. G. de Bruijn, Asymptotic Methods in Analysis, Dover, 1981, p. 36.
  • Eric Hammer, The Calculations of Peirce's 4.453, Transactions of the Charles S. Peirce Society, Vol. 31 (1995), pp. 829-839.
  • D. E. Knuth, personal communication.
  • J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 174.
  • Charles Sanders Peirce, Collected Papers, eds. C. Hartshorne and P. Weiss, Harvard University Press, Cambridge, Vol. 4, 1933, pp. 364-365. (CP 4.453 in the electronic edition of The Collected Papers of Charles Sanders Peirce.)
  • Dawidson Razafimahatolotra, Number of Preorders to Compute Probability of Conflict of an Unstable Effectivity Function, Preprint, Paris School of Economics, University of Paris I, Nov 23 2007.

Crossrefs

Same as A076726 except for a(0). Cf. A008965, A052861, A008277.
Binomial transform of A000670, also double of A000670. - Joe Keane (jgk(AT)jgk.org)
A002050(n) = a(n) - 1.
A000629, A000670, A002050, A052856, A076726 are all more-or-less the same sequence. - N. J. A. Sloane, Jul 04 2012
Row sums of A028246.
A diagonal of the triangular array in A241168.
Row sums of unsigned A127671 and A263634.

Programs

  • Maple
    spec := [ B, {B=Cycle(Set(Z,card>=1))}, labeled ]; [seq(combstruct[count](spec, size=n), n=0..20)];
    a:=n->add(Stirling2(n+1,k)*(k-1)!,k=1..n+1); # Mike Zabrocki, Feb 05 2005
  • Mathematica
    a[ 0 ] = 1; a[ n_ ] := (a[ n ] = 1 + Sum[ Binomial[ n, k ] a[ n-k ], {k, 1, n} ])
    Table[ PolyLog[n, 1/2], {n, 0, -18, -1}] (* Robert G. Wilson v, Aug 05 2010 *)
    a[ n_] := If[ n<0, 0, PolyLog[ -n, 1/2]]; (* Michael Somos, Mar 07 2011 *)
    Table[Sum[(-1)^(n-k) StirlingS2[n,k]k! 2^k,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Oct 21 2011 *)
    Join[{1}, Rest[t=30; Range[0, t]! CoefficientList[Series[2/(2 - Exp[x]), {x, 0, t}], x]]] (* Vincenzo Librandi, Jan 02 2016 *)
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff(subst( (1 + y) / (1 - y), y, exp(x + x * O(x^n)) - 1), n))} /* Michael Somos, Mar 04 2004 */
    
  • PARI
    {a(n)=polcoeff(sum(m=0, n, 2^m*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)} \\ Paul D. Hanna, Jul 20 2011
    
  • Python
    from math import comb
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A000629(n): return 1+sum(comb(n,j)*A000629(j) for j in range(n)) if n else 1 # Chai Wah Wu, Sep 25 2023

Formula

a(n) = 2*A000670(n) - 0^n. - Michael Somos, Jan 08 2011
O.g.f.: Sum_{n>=0} 2^n*n!*x^n / Product_{k=0..n} (1+k*x). - Paul D. Hanna, Jul 20 2011
E.g.f.: exp(x) / (2 - exp(x)) = d/dx log(1 / (2 - exp(x))).
a(n) = Sum_{k>=1} k^n/2^k.
a(n) = 1 + Sum_{j=0..n-1} C(n, j)*a(j).
a(n) = round(n!/log(2)^(n+1)) (just for n <= 15). - Henry Bottomley, Jul 04 2000
a(n) is asymptotic to n!/log(2)^(n+1). - Benoit Cloitre, Oct 20 2002
a(n) = Sum_{k=0..n} (-1)^(n-k)*Stirling2(n, k)*k!*2^k. - Vladeta Jovovic, Sep 29 2003
a(n) = Sum_{k=1..n} A008292(n, k)*2^k; A008292: triangle of Eulerian numbers. - Philippe Deléham, Jun 05 2004
a(1) = 1, a(n) = 2*Sum_{k=1..n-1} k!*A008277(n-1, k) for n>1 or a(n) = Sum_{k=1..n} (k-1)!*A008277(n, k). - Mike Zabrocki, Feb 05 2005
a(n) = Sum_{k=0..n} Stirling2(n+1, k+1)*k!. - Paul Barry, Apr 20 2005
A000629 = binomial transform of this sequence. a(n) = sum of terms in n-th row of A028246. - Gary W. Adamson, May 30 2005
a(n) = 2*(-1)^n * n!*Laguerre(n,P((.),2)), umbrally, where P(j,t) are the polynomials in A131758. - Tom Copeland, Sep 28 2007
a(n) = 2^n*A(n,1/2); A(n,x) the Eulerian polynomials. - Peter Luschny, Aug 03 2010
a(n) = (-1)^n*b(n), where b(n) = -2*Sum_{k=0..n-1} binomial(n,k)*b(k), b(0)=1. - Vladimir Kruchinin, Jan 29 2011
Row sums of A028246. Let f(x) = x+x^2. Then a(n+1) = (f(x)*d/dx)^n f(x) evaluated at x = 1. - Peter Bala, Oct 06 2011
O.g.f.: 1+2*x/(U(0)-2*x) where U(k)=1+3*x+3*x*k-2*x*(k+2)*(1+x+x*k)/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 14 2011
E.g.f.: exp(x)/(2 - exp(x)) = 2/(2-Q(0))-1; Q(k)=1+x/(2*k+1-x*(2*k+1)/(x+(2*k+2)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 14 2011
G.f.: 1 / (1 - 2*x / (1 - 1*x / (1 - 4*x / (1 - 2*x / (1 - 6*x / ...))))). - Michael Somos, Apr 27 2012
PSUM transform of A162509. BINOMIAL transform is A007047. - Michael Somos, Apr 27 2012
G.f.: 1/G(0) where G(k) = 1 - x*(2*k+2)/( 1 - x*(k+1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 23 2013
E.g.f.: 1/E(0) where E(k) = 1 - x/(k+1)/(1 - 1/(1 + 1/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Mar 27 2013
G.f.: T(0)/(1-2*x), where T(k) = 1 - 2*x^2*(k+1)^2/(2*x^2*(k+1)^2 - (1 - 2*x - 3*x*k)*(1 - 5*x - 3*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 29 2013
a(n) = log(2)*integral_{x>=0} (ceiling(x))^n * 2^(-x) dx. - Peter Bala, Feb 06 2015

Extensions

a(19) from Michael Somos, Mar 07 2011

A007489 a(n) = Sum_{k=1..n} k!.

Original entry on oeis.org

0, 1, 3, 9, 33, 153, 873, 5913, 46233, 409113, 4037913, 43954713, 522956313, 6749977113, 93928268313, 1401602636313, 22324392524313, 378011820620313, 6780385526348313, 128425485935180313, 2561327494111820313, 53652269665821260313, 1177652997443428940313
Offset: 0

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Author

Keywords

Comments

Equals row sums of triangle A143122 starting (1, 3, 9, 33, ...). - Gary W. Adamson, Jul 26 2008
a(n) for n>=4 is never a perfect square. - Alexander R. Povolotsky, Oct 16 2008
Number of cycles that can be written in the form (j,j+1,j+2,...), in all permutations of {1,2,...,n}. Example: a(3)=9 because in (1)(2)(3), (1)(23), (12)(3), (13)(2), (123), (132) we have 3+2+2+1+1+0=9 such cycles. - Emeric Deutsch, Jul 14 2009
Conjectured to be the length of the shortest word over {1,...,n} that contains each of the n! permutations as a factor (cf. A180632) [see Johnston]. - N. J. A. Sloane, May 25 2013
The above conjecture has been disproven for n>=6. See A180632 and the Houston 2014 reference. - Dmitry Kamenetsky, Mar 07 2016
a(n) is also the number of compositions of n if cardinal values do not matter but ordinal rankings do. Since cardinal values do not matter, a sequence of k summands summing to n can be represented as (s(1),...,s(k)), where the s's are positive integers and the numbers in parentheses are the initial ordinal rankings. The number of compositions of these summands are equal to k!, with k ranging from 1 to n. - Gregory L. Simay, Jul 31 2016
When the numbers denote finite permutations (as row numbers of A055089) these are the circular shifts to the left. Compare array A211370 for circular shifts to the left in a broader sense. Compare sequence A001563 for circular shifts to the right. - Tilman Piesk, Apr 29 2017
Since a(n) = (1!+2!+3!+...+n!) = 3(1+3!/3+4!/3+...+n!/3) is a multiple of 3 for n>2, the only prime in this sequence is a(2) = 3. - Eric W. Weisstein, Jul 15 2017
Generalization of 2nd comment: a(n) for n>=4 is never a perfect power (A007916) (Chentzov link). - Bernard Schott, Jan 26 2023

Examples

			a(4) = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33. - _Michael B. Porter_, Aug 03 2016
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, 3rd ed., Section B44, Springer 2010.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Equals A003422(n+1) - 1.
Column k=0 of A120695.

Programs

Formula

a(n) = Sum_{k=1..n} P(n, k)/C(n, k). - Ross La Haye, Sep 21 2004
a(n) = 3*A056199(n) for n>=2. - Philippe Deléham, Feb 10 2007
a(n) = !(n+1)-1=A003422(n+1)-1. - Artur Jasinski, Nov 08 2007 [corrected by Werner Schulte, Oct 20 2021]
Starting (1, 3, 9, 33, 153, ...), = row sums of triangle A137593 - Gary W. Adamson, Jan 28 2008
a(n) = a(n-1) + n! for n >= 1. - Jaroslav Krizek, Jun 16 2009
E.g.f. A(x) satisfies to the differential equation A'(x)=A(x)+x/(1-x)^2+1. - Vladimir Kruchinin, Jan 22 2011
a(0)=0, a(1)=1, a(n) = (n+1)*a(n-1)-n*a(n-2). - Sergei N. Gladkovskii, Jul 05 2012
G.f.: W(0)*x/(2-2*x) , where W(k) = 1 + 1/( 1 - x*(k+2)/( x*(k+2) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
G.f.: x /(1-x)/Q(0),m=+2, where Q(k) = 1 - 2*x*(2*k+1) - m*x^2*(k+1)*(2*k+1)/( 1 - 2*x*(2*k+2) - m*x^2*(k+1)*(2*k+3)/Q(k+1) ) ; (continued fraction). - Sergei N. Gladkovskii, Sep 24 2013
E.g.f.: exp(x-1)*(Ei(1) - Ei(1-x)) - exp(x) + 1/(1 - x), where Ei(x) is the exponential integral. - Ilya Gutkovskiy, Nov 27 2016
a(n) = sqrt(a(n-1)*a(n+1)-a(n-2)*n*n!), n >= 2. - Gary Detlefs, Oct 26 2020
a(n) ~ n!. - Ridouane Oudra, Jun 11 2025

A335448 Numbers whose prime indices are inseparable.

Original entry on oeis.org

4, 8, 9, 16, 24, 25, 27, 32, 40, 48, 49, 54, 56, 64, 80, 81, 88, 96, 104, 112, 121, 125, 128, 135, 136, 144, 152, 160, 162, 169, 176, 184, 189, 192, 208, 224, 232, 240, 243, 248, 250, 256, 272, 288, 289, 296, 297, 304, 320, 324, 328, 336, 343, 344, 351, 352
Offset: 1

Views

Author

Gus Wiseman, Jun 21 2020

Keywords

Comments

First differs from A212164 in lacking 72.
First differs from A293243 in lacking 72.
No terms are squarefree.
Also Heinz numbers of inseparable partitions (A325535). The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
These are also numbers that can be written as a product of prime numbers, each different from the last but not necessarily different from those prior to the last.
A multiset is inseparable iff its maximal multiplicity is greater than one plus the sum of its remaining multiplicities.

Examples

			The sequence of terms together with their prime indices begins:
   4: {1,1}
   8: {1,1,1}
   9: {2,2}
  16: {1,1,1,1}
  24: {1,1,1,2}
  25: {3,3}
  27: {2,2,2}
  32: {1,1,1,1,1}
  40: {1,1,1,3}
  48: {1,1,1,1,2}
  49: {4,4}
  54: {1,2,2,2}
  56: {1,1,1,4}
  64: {1,1,1,1,1,1}
  80: {1,1,1,1,3}
  81: {2,2,2,2}
  88: {1,1,1,5}
  96: {1,1,1,1,1,2}
		

Crossrefs

Complement of A335433.
Separations are counted by A003242 and A335452 and ranked by A333489.
Permutations of prime indices are counted by A008480.
Inseparable partitions are counted by A325535.
Strict permutations of prime indices are counted by A335489.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Select[Range[100],Select[Permutations[primeMS[#]],!MatchQ[#,{_,x_,x_,_}]&]=={}&]
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