cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 58 results. Next

A171151 Expansion of (A(x)-1)/(x*A(x)), A(x) the g.f. of A004211.

Original entry on oeis.org

1, 2, 6, 26, 142, 898, 6342, 49114, 412046, 3711746, 35660166, 363484058, 3914162830, 44370673282, 527868672582, 6572992645978, 85461951576974, 1157778354181634, 16310949128381190, 238543136307926810
Offset: 0

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Author

Paul Barry, Dec 04 2009

Keywords

Comments

Hankel transform is A108400.

Formula

G.f.: 1/(1-2x/(1-x/(1-4x/(1-x/(1-6x/(1-x/(1-8x/(1-x/(1-... (continued function).
a(n) = Sum_{k=0..n} A086329(n,k)*2^k. - Philippe Deléham, Dec 05 2009
G.f.: 1/U(0) where U(k) = 1 - x - 2*x*k + x*(2*x*k + 2*x - 1)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Sep 24 2012
G.f.: 1/x - 1/( x*G(0) - 1 ) where G(k) = 1 - (4*x*k-1)/(x - x^4/(x^3 - (4*x*k-1)*(4*x*k+2*x-1)*(4*x*k+4*x-1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 08 2013
G.f.: (1 - G(0))/x where G(k) = 1 - x/(1 - 2*x*(k + 1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Jan 30 2013
G.f.: 1/x + 1 - (2*x+1)/(G(0) + 2*x+1), where G(k)= 2*x*k - x - 1 - 2*(k+1)*x^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 21 2013
G.f.: 1/x - Q(0)/x, where Q(k) = 1 - x*(2*k+1) - (2*k+2)*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 09 2013

A000110 Bell or exponential numbers: number of ways to partition a set of n labeled elements.

Original entry on oeis.org

1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, 1382958545, 10480142147, 82864869804, 682076806159, 5832742205057, 51724158235372, 474869816156751, 4506715738447323, 44152005855084346, 445958869294805289, 4638590332229999353, 49631246523618756274
Offset: 0

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Author

Keywords

Comments

The leading diagonal of its difference table is the sequence shifted, see Bernstein and Sloane (1995). - N. J. A. Sloane, Jul 04 2015
Also the number of equivalence relations that can be defined on a set of n elements. - Federico Arboleda (federico.arboleda(AT)gmail.com), Mar 09 2005
a(n) = number of nonisomorphic colorings of a map consisting of a row of n+1 adjacent regions. Adjacent regions cannot have the same color. - David W. Wilson, Feb 22 2005
If an integer is squarefree and has n distinct prime factors then a(n) is the number of ways of writing it as a product of its divisors. - Amarnath Murthy, Apr 23 2001
Consider rooted trees of height at most 2. Letting each tree 'grow' into the next generation of n means we produce a new tree for every node which is either the root or at height 1, which gives the Bell numbers. - Jon Perry, Jul 23 2003
Begin with [1,1] and follow the rule that [1,k] -> [1,k+1] and [1,k] k times, e.g., [1,3] is transformed to [1,4], [1,3], [1,3], [1,3]. Then a(n) is the sum of all components: [1,1] = 2; [1,2], [1,1] = 5; [1,3], [1,2], [1,2], [1,2], [1,1] = 15; etc. - Jon Perry, Mar 05 2004
Number of distinct rhyme schemes for a poem of n lines: a rhyme scheme is a string of letters (e.g., 'abba') such that the leftmost letter is always 'a' and no letter may be greater than one more than the greatest letter to its left. Thus 'aac' is not valid since 'c' is more than one greater than 'a'. For example, a(3)=5 because there are 5 rhyme schemes: aaa, aab, aba, abb, abc; also see example by Neven Juric. - Bill Blewett, Mar 23 2004
In other words, number of length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(0)=0 and s(k) <= 1 + max(prefix) for k >= 1, see example (cf. A080337 and A189845). - Joerg Arndt, Apr 30 2011
Number of partitions of {1, ..., n+1} into subsets of nonconsecutive integers, including the partition 1|2|...|n+1. E.g., a(3)=5: there are 5 partitions of {1,2,3,4} into subsets of nonconsecutive integers, namely, 13|24, 13|2|4, 14|2|3, 1|24|3, 1|2|3|4. - Augustine O. Munagi, Mar 20 2005
Triangle (addition) scheme to produce terms, derived from the recurrence, from Oscar Arevalo (loarevalo(AT)sbcglobal.net), May 11 2005:
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
... [This is Aitken's array A011971]
With P(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j,i) = the j-th part of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i=1..P(n)} (n!/(Product_{j=1..p(i)} p(i,j)!)) * (1/(Product_{j=1..d(i)} m(i,j)!)). - Thomas Wieder, May 18 2005
a(n+1) is the number of binary relations on an n-element set that are both symmetric and transitive. - Justin Witt (justinmwitt(AT)gmail.com), Jul 12 2005
If the rule from Jon Perry, Mar 05 2004, is used, then a(n-1) = [number of components used to form a(n)] / 2. - Daniel Kuan (dkcm(AT)yahoo.com), Feb 19 2006
a(n) is the number of functions f from {1,...,n} to {1,...,n,n+1} that satisfy the following two conditions for all x in the domain: (1) f(x) > x; (2) f(x)=n+1 or f(f(x))=n+1. E.g., a(3)=5 because there are exactly five functions that satisfy the two conditions: f1={(1,4),(2,4),(3,4)}, f2={(1,4),(2,3),(3,4)}, f3={(1,3),(2,4),(3,4)}, f4={(1,2),(2,4),(3,4)} and f5={(1,3),(2,3),(3,4)}. - Dennis P. Walsh, Feb 20 2006
Number of asynchronic siteswap patterns of length n which have no zero-throws (i.e., contain no 0's) and whose number of orbits (in the sense given by Allen Knutson) is equal to the number of balls. E.g., for n=4, the condition is satisfied by the following 15 siteswaps: 4444, 4413, 4242, 4134, 4112, 3441, 2424, 1344, 2411, 1313, 1241, 2222, 3131, 1124, 1111. Also number of ways to choose n permutations from identity and cyclic permutations (1 2), (1 2 3), ..., (1 2 3 ... n) so that their composition is identity. For n=3 we get the following five: id o id o id, id o (1 2) o (1 2), (1 2) o id o (1 2), (1 2) o (1 2) o id, (1 2 3) o (1 2 3) o (1 2 3). (To see the bijection, look at Ehrenborg and Readdy paper.) - Antti Karttunen, May 01 2006
a(n) is the number of permutations on [n] in which a 3-2-1 (scattered) pattern occurs only as part of a 3-2-4-1 pattern. Example: a(3) = 5 counts all permutations on [3] except 321. See "Eigensequence for Composition" reference a(n) = number of permutation tableaux of size n (A000142) whose first row contains no 0's. Example: a(3)=5 counts {{}, {}, {}}, {{1}, {}}, {{1}, {0}}, {{1}, {1}}, {{1, 1}}. - David Callan, Oct 07 2006
From Gottfried Helms, Mar 30 2007: (Start)
This sequence is also the first column in the matrix-exponential of the (lower triangular) Pascal-matrix, scaled by exp(-1): PE = exp(P) / exp(1) =
1
1 1
2 2 1
5 6 3 1
15 20 12 4 1
52 75 50 20 5 1
203 312 225 100 30 6 1
877 1421 1092 525 175 42 7 1
First 4 columns are A000110, A033306, A105479, A105480. The general case is mentioned in the two latter entries. PE is also the Hadamard-product Toeplitz(A000110) (X) P:
1
1 1
2 1 1
5 2 1 1
15 5 2 1 1 (X) P
52 15 5 2 1 1
203 52 15 5 2 1 1
877 203 52 15 5 2 1 1
(End)
The terms can also be computed with finite steps and precise integer arithmetic. Instead of exp(P)/exp(1) one can compute A = exp(P - I) where I is the identity-matrix of appropriate dimension since (P-I) is nilpotent to the order of its dimension. Then a(n)=A[n,1] where n is the row-index starting at 1. - Gottfried Helms, Apr 10 2007
When n is prime, a(n) == 2 (mod n), but the converse is not always true. Define a Bell pseudoprime to be a composite number n such that a(n) == 2 (mod n). W. F. Lunnon recently found the Bell pseudoprimes 21361 = 41*521 and C46 = 3*23*16218646893090134590535390526854205539989357 and conjectured that Bell pseudoprimes are extremely scarce. So the second Bell pseudoprime is unlikely to be known with certainty in the near future. I confirmed that 21361 is the first. - David W. Wilson, Aug 04 2007 and Sep 24 2007
This sequence and A000587 form a reciprocal pair under the list partition transform described in A133314. - Tom Copeland, Oct 21 2007
Starting (1, 2, 5, 15, 52, ...), equals row sums and right border of triangle A136789. Also row sums of triangle A136790. - Gary W. Adamson, Jan 21 2008
This is the exponential transform of A000012. - Thomas Wieder, Sep 09 2008
From Abdullahi Umar, Oct 12 2008: (Start)
a(n) is also the number of idempotent order-decreasing full transformations (of an n-chain).
a(n) is also the number of nilpotent partial one-one order-decreasing transformations (of an n-chain).
a(n+1) is also the number of partial one-one order-decreasing transformations (of an n-chain). (End)
From Peter Bala, Oct 19 2008: (Start)
Bell(n) is the number of n-pattern sequences [Cooper & Kennedy]. An n-pattern sequence is a sequence of integers (a_1,...,a_n) such that a_i = i or a_i = a_j for some j < i. For example, Bell(3) = 5 since the 3-pattern sequences are (1,1,1), (1,1,3), (1,2,1), (1,2,2) and (1,2,3).
Bell(n) is the number of sequences of positive integers (N_1,...,N_n) of length n such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (see the comment by B. Blewett above). It is interesting to note that if we strengthen the latter condition to N_(i+1) <= 1 + N_i we get the Catalan numbers A000108 instead of the Bell numbers.
(End)
Equals the eigensequence of Pascal's triangle, A007318; and starting with offset 1, = row sums of triangles A074664 and A152431. - Gary W. Adamson, Dec 04 2008
The entries f(i, j) in the exponential of the infinite lower-triangular matrix of binomial coefficients b(i, j) are f(i, j) = b(i, j) e a(i - j). - David Pasino, Dec 04 2008
Equals lim_{k->oo} A071919^k. - Gary W. Adamson, Jan 02 2009
Equals A154107 convolved with A014182, where A014182 = expansion of exp(1-x-exp(-x)), the eigensequence of A007318^(-1). Starting with offset 1 = A154108 convolved with (1,2,3,...) = row sums of triangle A154109. - Gary W. Adamson, Jan 04 2009
Repeated iterates of (binomial transform of [1,0,0,0,...]) will converge upon (1, 2, 5, 15, 52, ...) when each result is prefaced with a "1"; such that the final result is the fixed limit: (binomial transform of [1,1,2,5,15,...]) = (1,2,5,15,52,...). - Gary W. Adamson, Jan 14 2009
From Karol A. Penson, May 03 2009: (Start)
Relation between the Bell numbers B(n) and the n-th derivative of 1/Gamma(1+x) evaluated at x=1:
a) produce a number of such derivatives through seq(subs(x=1, simplify((d^n/dx^n)GAMMA(1+x)^(-1))), n=1..5);
b) leave them expressed in terms of digamma (Psi(k)) and polygamma (Psi(k,n)) functions and unevaluated;
Examples of such expressions, for n=1..5, are:
n=1: -Psi(1),
n=2: -(-Psi(1)^2 + Psi(1,1)),
n=3: -Psi(1)^3 + 3*Psi(1)*Psi(1,1) - Psi(2,1),
n=4: -(-Psi(1)^4 + 6*Psi(1)^2*Psi(1,1) - 3*Psi(1,1)^2 - 4*Psi(1)*Psi(2,1) + Psi(3, 1)),
n=5: -Psi(1)^5 + 10*Psi(1)^3*Psi(1,1) - 15*Psi(1)*Psi(1,1)^2 - 10*Psi(1)^2*Psi(2,1) + 10*Psi(1,1)*Psi(2,1) + 5*Psi(1)*Psi(3,1) - Psi(4,1);
c) for a given n, read off the sum of absolute values of coefficients of every term involving digamma or polygamma functions.
This sum is equal to B(n). Examples: B(1)=1, B(2)=1+1=2, B(3)=1+3+1=5, B(4)=1+6+3+4+1=15, B(5)=1+10+15+10+10+5+1=52;
d) Observe that this decomposition of the Bell number B(n) apparently does not involve the Stirling numbers of the second kind explicitly.
(End)
The numbers given above by Penson lead to the multinomial coefficients A036040. - Johannes W. Meijer, Aug 14 2009
Column 1 of A162663. - Franklin T. Adams-Watters, Jul 09 2009
Asymptotic expansions (0!+1!+2!+...+(n-1)!)/(n-1)! = a(0) + a(1)/n + a(2)/n^2 + ... and (0!+1!+2!+...+n!)/n! = 1 + a(0)/n + a(1)/n^2 + a(2)/n^3 + .... - Michael Somos, Jun 28 2009
Starting with offset 1 = row sums of triangle A165194. - Gary W. Adamson, Sep 06 2009
a(n+1) = A165196(2^n); where A165196 begins: (1, 2, 4, 5, 7, 12, 14, 15, ...). such that A165196(2^3) = 15 = A000110(4). - Gary W. Adamson, Sep 06 2009
The divergent series g(x=1,m) = 1^m*1! - 2^m*2! + 3^m*3! - 4^m*4! + ..., m >= -1, which for m=-1 dates back to Euler, is related to the Bell numbers. We discovered that g(x=1,m) = (-1)^m * (A040027(m) - A000110(m+1) * A073003). We observe that A073003 is Gompertz's constant and that A040027 was published by Gould, see for more information A163940. - Johannes W. Meijer, Oct 16 2009
a(n) = E(X^n), i.e., the n-th moment about the origin of a random variable X that has a Poisson distribution with (rate) parameter, lambda = 1. - Geoffrey Critzer, Nov 30 2009
Let A000110 = S(x), then S(x) = A(x)/A(x^2) when A(x) = A173110; or (1, 1, 2, 5, 15, 52, ...) = (1, 1, 3, 6, 20, 60, ...) / (1, 0, 1, 0, 3, 0, 6, 0, 20, ...). - Gary W. Adamson, Feb 09 2010
The Bell numbers serve as the upper limit for the number of distinct homomorphic images from any given finite universal algebra. Every algebra homomorphism is determined by its kernel, which must be a congruence relation. As the number of possible congruence relations with respect to a finite universal algebra must be a subset of its possible equivalence classes (given by the Bell numbers), it follows naturally. - Max Sills, Jun 01 2010
For a proof of the o.g.f. given in the R. Stephan comment see, e.g., the W. Lang link under A071919. - Wolfdieter Lang, Jun 23 2010
Let B(x) = (1 + x + 2x^2 + 5x^3 + ...). Then B(x) is satisfied by A(x)/A(x^2) where A(x) = polcoeff A173110: (1 + x + 3x^2 + 6x^3 + 20x^4 + 60x^5 + ...) = B(x) * B(x^2) * B(x^4) * B(x^8) * .... - Gary W. Adamson, Jul 08 2010
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 without choosing any two colors the same positive number of times. (See related comments for A000108, A008277, A016098.) - Matthew Vandermast, Nov 22 2010
A binary counter with faulty bits starts at value 0 and attempts to increment by 1 at each step. Each bit that should toggle may or may not do so. a(n) is the number of ways that the counter can have the value 0 after n steps. E.g., for n=3, the 5 trajectories are 0,0,0,0; 0,1,0,0; 0,1,1,0; 0,0,1,0; 0,1,3,0. - David Scambler, Jan 24 2011
No Bell number is divisible by 8, and no Bell number is congruent to 6 modulo 8; see Theorem 6.4 and Table 1.7 in Lunnon, Pleasants and Stephens. - Jon Perry, Feb 07 2011, clarified by Eric Rowland, Mar 26 2014
a(n+1) is the number of (symmetric) positive semidefinite n X n 0-1 matrices. These correspond to equivalence relations on {1,...,n+1}, where matrix element M[i,j] = 1 if and only if i and j are equivalent to each other but not to n+1. - Robert Israel, Mar 16 2011
a(n) is the number of monotonic-labeled forests on n vertices with rooted trees of height less than 2. We note that a labeled rooted tree is monotonic-labeled if the label of any parent vertex is greater than the label of any offspring vertex. See link "Counting forests with Stirling and Bell numbers". - Dennis P. Walsh, Nov 11 2011
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A000772 and A094198. - Peter Bala, Nov 25 2011
B(n) counts the length n+1 rhyme schemes without repetitions. E.g., for n=2 there are 5 rhyme schemes of length 3 (aaa, aab, aba, abb, abc), and the 2 without repetitions are aba, abc. This is basically O. Munagi's result that the Bell numbers count partitions into subsets of nonconsecutive integers (see comment above dated Mar 20 2005). - Eric Bach, Jan 13 2012
Right and left borders and row sums of A212431 = A000110 or a shifted variant. - Gary W. Adamson, Jun 21 2012
Number of maps f: [n] -> [n] where f(x) <= x and f(f(x)) = f(x) (projections). - Joerg Arndt, Jan 04 2013
Permutations of [n] avoiding any given one of the 8 dashed patterns in the equivalence classes (i) 1-23, 3-21, 12-3, 32-1, and (ii) 1-32, 3-12, 21-3, 23-1. (See Claesson 2001 reference.) - David Callan, Oct 03 2013
Conjecture: No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
Sum_{n>=0} a(n)/n! = e^(e-1) = 5.57494152476..., see A234473. - Richard R. Forberg, Dec 26 2013 (This is the e.g.f. for x=1. - Wolfdieter Lang, Feb 02 2015)
Sum_{j=0..n} binomial(n,j)*a(j) = (1/e)*Sum_{k>=0} (k+1)^n/k! = (1/e) Sum_{k=1..oo} k^(n+1)/k! = a(n+1), n >= 0, using the Dobinski formula. See the comment by Gary W. Adamson, Dec 04 2008 on the Pascal eigensequence. - Wolfdieter Lang, Feb 02 2015
In fact it is not really an eigensequence of the Pascal matrix; rather the Pascal matrix acts on the sequence as a shift. It is an eigensequence (the unique eigensequence with eigenvalue 1) of the matrix derived from the Pascal matrix by adding at the top the row [1, 0, 0, 0 ...]. The binomial sum formula may be derived from the definition in terms of partitions: label any element X of a set S of N elements, and let X(k) be the number of subsets of S containing X with k elements. Since each subset has a unique coset, the number of partitions p(N) of S is given by p(N) = Sum_{k=1..N} (X(k) p(N-k)); trivially X(k) = N-1 choose k-1. - Mason Bogue, Mar 20 2015
a(n) is the number of ways to nest n matryoshkas (Russian nesting dolls): we may identify {1, 2, ..., n} with dolls of ascending sizes and the sets of a set partition with stacks of dolls. - Carlo Sanna, Oct 17 2015
Number of permutations of [n] where the initial elements of consecutive runs of increasing elements are in decreasing order. a(4) = 15: `1234, `2`134, `23`14, `234`1, `24`13, `3`124, `3`2`14, `3`24`1, `34`12, `34`2`1, `4`123, `4`2`13, `4`23`1, `4`3`12, `4`3`2`1. - Alois P. Heinz, Apr 27 2016
Taking with alternating signs, the Bell numbers are the coefficients in the asymptotic expansion (Ramanujan): (-1)^n*(A000166(n) - n!/exp(1)) ~ 1/n - 2/n^2 + 5/n^3 - 15/n^4 + 52/n^5 - 203/n^6 + O(1/n^7). - Vladimir Reshetnikov, Nov 10 2016
Number of treeshelves avoiding pattern T231. See A278677 for definitions and examples. - Sergey Kirgizov, Dec 24 2016
Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear. - N. J. A. Sloane, Feb 09 2017
a(n) = Sum(# of standard immaculate tableaux of shape m, m is a composition of n), where this sum is over all integer compositions m of n > 0. This formula is easily seen to hold by identifying standard immaculate tableaux of size n with set partitions of { 1, 2, ..., n }. For example, if we sum over integer compositions of 4 lexicographically, we see that 1+1+2+1+3+3+3+1 = 15 = A000110(4). - John M. Campbell, Jul 17 2017
a(n) is also the number of independent vertex sets (and vertex covers) in the (n-1)-triangular honeycomb bishop graph. - Eric W. Weisstein, Aug 10 2017
Even-numbered entries represent the numbers of configurations of identity and non-identity for alleles of a gene in n diploid individuals with distinguishable maternal and paternal alleles. - Noah A Rosenberg, Jan 28 2019
Number of partial equivalence relations (PERs) on a set with n elements (offset=1), i.e., number of symmetric, transitive (not necessarily reflexive) relations. The idea is to add a dummy element D to the set, and then take equivalence relations on the result; anything equivalent to D is then removed for the partial equivalence relation. - David Spivak, Feb 06 2019
Number of words of length n+1 with no repeated letters, when letters are unlabeled. - Thomas Anton, Mar 14 2019
Named by Becker and Riordan (1948) after the Scottish-American mathematician and writer Eric Temple Bell (1883 - 1960). - Amiram Eldar, Dec 04 2020
Also the number of partitions of {1,2,...,n+1} with at most one n+1 singleton. E.g., a(3)=5: {13|24, 12|34, 123|4, 14|23, 1234}. - Yuchun Ji, Dec 21 2020
a(n) is the number of sigma algebras on the set of n elements. Note that each sigma algebra is generated by a partition of the set. For example, the sigma algebra generated by the partition {{1}, {2}, {3,4}} is {{}, {1}, {2}, {1,2}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. - Jianing Song, Apr 01 2021
a(n) is the number of P_3-free graphs on n labeled nodes. - M. Eren Kesim, Jun 04 2021
a(n) is the number of functions X:([n] choose 2) -> {+,-} such that for any ordered 3-tuple abc we have X(ab)X(ac)X(bc) not in {+-+,++-,-++}. - Robert Lauff, Dec 09 2022
From Manfred Boergens, Mar 11 2025: (Start)
The partitions in the definition can be described as disjoint covers of the set. "Covers" in general give rise to the following amendments:
For disjoint covers which may include one empty set see A186021.
For arbitrary (including non-disjoint) covers see A003465.
For arbitrary (including non-disjoint) covers which may include one empty set see A000371. (End)

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 52*x^5 + 203*x^6 + 877*x^7 + 4140*x^8 + ...
From Neven Juric, Oct 19 2009: (Start)
The a(4)=15 rhyme schemes for n=4 are
  aaaa, aaab, aaba, aabb, aabc, abaa, abab, abac, abba, abbb, abbc, abca, abcb, abcc, abcd
The a(5)=52 rhyme schemes for n=5 are
  aaaaa, aaaab, aaaba, aaabb, aaabc, aabaa, aabab, aabac, aabba, aabbb, aabbc, aabca, aabcb, aabcc, aabcd, abaaa, abaab, abaac, ababa, ababb, ababc, abaca, abacb, abacc, abacd, abbaa, abbab, abbac, abbba, abbbb, abbbc, abbca, abbcb, abbcc, abbcd, abcaa, abcab, abcac, abcad, abcba, abcbb, abcbc, abcbd, abcca, abccb, abccc, abccd, abcda, abcdb, abcdc, abcdd, abcde
(End)
From _Joerg Arndt_, Apr 30 2011: (Start)
Restricted growth strings (RGS):
For n=0 there is one empty string;
for n=1 there is one string [0];
for n=2 there are 2 strings [00], [01];
for n=3 there are a(3)=5 strings [000], [001], [010], [011], and [012];
for n=4 there are a(4)=15 strings
1: [0000], 2: [0001], 3: [0010], 4: [0011], 5: [0012], 6: [0100], 7: [0101], 8: [0102], 9: [0110], 10: [0111], 11: [0112], 12: [0120], 13: [0121], 14: [0122], 15: [0123].
These are one-to-one with the rhyme schemes (identify a=0, b=1, c=2, etc.).
(End)
Consider the set S = {1, 2, 3, 4}. The a(4) = 1 + 3 + 6 + 4 + 1 = 15 partitions are: P1 = {{1}, {2}, {3}, {4}}; P21 .. P23 = {{a,4}, S\{a,4}} with a = 1, 2, 3; P24 .. P29 = {{a}, {b}, S\{a,b}} with 1 <= a < b <= 4;  P31 .. P34 = {S\{a}, {a}} with a = 1 .. 4; P4 = {S}. See the Bottomley link for a graphical illustration. - _M. F. Hasler_, Oct 26 2017
		

References

  • Stefano Aguzzoli, Brunella Gerla and Corrado Manara, Poset Representation for Goedel and Nilpotent Minimum Logics, in Symbolic and Quantitative Approaches to Reasoning with Uncertainty, Lecture Notes in Computer Science, Volume 3571/2005, Springer-Verlag. [Added by N. J. A. Sloane, Jul 08 2009]
  • S. Ainley, Problem 19, QARCH, No. IV, Nov 03, 1980.
  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 205.
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  • J. Balogh, B. Bollobas and D. Weinreich, A jump to the Bell numbers for hereditary graph properties, J. Combin. Theory Ser. B 95 (2005), no. 1, 29-48.
  • R. E. Beard, On the coefficients in the expansion of exp(exp(t)) and exp(-exp(t)), J. Institute Actuaries, 76 (1951), 152-163.
  • H. W. Becker, Abstracts of two papers related to Bell numbers, Bull. Amer. Math. Soc., 52 (1946), p. 415.
  • E. T. Bell, The iterated exponential numbers, Ann. Math., 39 (1938), 539-557.
  • C. M. Bender, D. C. Brody and B. K. Meister, Quantum Field Theory of Partitions, J. Math. Phys., 40,7 (1999), 3239-45.
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  • G. Birkhoff, Lattice Theory, Amer. Math. Soc., Revised Ed., 1961, p. 108, Ex. 1.
  • M. T. L. Bizley, On the coefficients in the expansion of exp(lambda exp(t)), J. Inst. Actuaries, 77 (1952), p. 122.
  • J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 41.
  • Carlier, Jacques; and Lucet, Corinne; A decomposition algorithm for network reliability evaluation. In First International Colloquium on Graphs and Optimization (GOI), 1992 (Grimentz). Discrete Appl. Math. 65 (1996), 141-156.
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  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 210.
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  • John H. Conway et al., The Symmetries of Things, Peters, 2008, p. 207.
  • Colin Defant, Highly sorted permutations and Bell numbers, ECA 1:1 (2021) Article S2R6.
  • De Angelis, Valerio, and Dominic Marcello. "Wilf's Conjecture." The American Mathematical Monthly 123.6 (2016): 557-573.
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  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 52, p. 19, Ellipses, Paris 2008.
  • G. Dobinski, Summierung der Reihe Sum(n^m/n!) für m = 1, 2, 3, 4, 5, ..., Grunert Archiv (Arch. f. Math. und Physik), 61 (1877) 333-336.
  • L. F. Epstein, A function related to the series for exp(exp(z)), J. Math. and Phys., 18 (1939), 153-173.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.8, p. 321.
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  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, 2nd ed., p. 493.
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  • Levinson, H.; Silverman, R. Topologies on finite sets. II. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 699--712, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561090 (81c:54006)
  • S. Linusson, The number of M-sequences and f-vectors, Combinatorica, 19 (1999), 255-266.
  • L. Lovasz, Combinatorial Problems and Exercises, North-Holland, 1993, pp. 14-15.
  • M. Meier, On the number of partitions of a given set, Amer. Math. Monthly, 114 (2007), p. 450.
  • Merris, Russell, and Stephen Pierce. "The Bell numbers and r-fold transitivity." Journal of Combinatorial Theory, Series A 12.1 (1972): 155-157.
  • Moser, Leo, and Max Wyman. An asymptotic formula for the Bell numbers. Trans. Royal Soc. Canada, 49 (1955), 49-53.
  • A. Murthy, Generalization of partition function, introducing Smarandache factor partition, Smarandache Notions Journal, Vol. 11, No. 1-2-3, Spring 2000.
  • Amarnath Murthy and Charles Ashbacher, Generalized Partitions and Some New Ideas on Number Theory and Smarandache Sequences, Hexis, Phoenix; USA 2005. See Section 1.4,1.8.
  • P. Peart, Hankel determinants via Stieltjes matrices. Proceedings of the Thirty-first Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2000). Congr. Numer. 144 (2000), 153-159.
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  • 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).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge; see Section 1.4 and Example 5.2.4.
  • Abdullahi Umar, On the semigroups of order-decreasing finite full transformations, Proc. Roy. Soc. Edinburgh 120A (1992), 129-142.
  • Abdullahi Umar, On the semigroups of partial one-to-one order-decreasing finite transformations, Proc. Roy. Soc. Edinburgh 123A (1993), 355-363.

Crossrefs

Equals row sums of triangle A008277 (Stirling subset numbers).
Partial sums give A005001. a(n) = A123158(n, 0).
See A061462 for powers of 2 dividing a(n).
Rightmost diagonal of triangle A121207. A144293 gives largest prime factor.
Equals row sums of triangle A152432.
Row sums, right and left borders of A212431.
A diagonal of A011971. - N. J. A. Sloane, Jul 31 2012
Diagonal of A102661. - Manfred Boergens, Mar 11 2025
Cf. A054767 (period of this sequence mod n).
Row sums are A048993. - Wolfdieter Lang, Oct 16 2014
Sequences in the Erné (1974) paper: A000110, A000798, A001035, A001927, A001929, A006056, A006057, A006058, A006059.
Bell polynomials B(n,x): A001861 (x=2), A027710 (x=3), A078944 (x=4), A144180 (x=5), A144223 (x=6), A144263 (x=7), A221159 (x=8).
Cf. A243991 (sum of reciprocals), A085686 (inv. Euler Transf.).

Programs

  • Haskell
    type N = Integer
    n_partitioned_k :: N -> N -> N
    1 `n_partitioned_k` 1 = 1
    1 `n_partitioned_k` _ = 0
    n `n_partitioned_k` k = k * (pred n `n_partitioned_k` k) + (pred n `n_partitioned_k` pred k)
    n_partitioned :: N -> N
    n_partitioned 0 = 1
    n_partitioned n = sum $ map (\k -> n `n_partitioned_k` k) $ [1 .. n]
    -- Felix Denis, Oct 16 2012
    
  • Haskell
    a000110 = sum . a048993_row -- Reinhard Zumkeller, Jun 30 2013
    
  • Julia
    function a(n)
        t = [zeros(BigInt, n+1) for _ in 1:n+1]
        t[1][1] = 1
        for i in 2:n+1
            t[i][1] = t[i-1][i-1]
            for j in 2:i
                t[i][j] = t[i-1][j-1] + t[i][j-1]
            end
        end
        return [t[i][1] for i in 1:n+1]
    end
    print(a(28)) # Paul Muljadi, May 07 2024
    
  • Magma
    [Bell(n): n in [0..40]]; // Vincenzo Librandi, Feb 07 2011
    
  • Maple
    A000110 := proc(n) option remember; if n <= 1 then 1 else add( binomial(n-1,i)*A000110(n-1-i),i=0..n-1); fi; end: # version 1
    A := series(exp(exp(x)-1),x,60): A000110 := n->n!*coeff(A,x,n): # version 2
    A000110:= n-> add(Stirling2(n, k), k=0..n): seq(A000110(n), n=0..22); # version 3, from Zerinvary Lajos, Jun 28 2007
    A000110 := n -> combinat[bell](n): # version 4, from Peter Luschny, Mar 30 2011
    spec:= [S, {S=Set(U, card >= 1), U=Set(Z, card >= 1)}, labeled]: G:={P=Set(Set(Atom, card>0))}: combstruct[gfsolve](G, unlabeled, x): seq(combstruct[count]([P, G, labeled], size=i), i=0..22);  # version 5, Zerinvary Lajos, Dec 16 2007
    BellList := proc(m) local A, P, n; A := [1, 1]; P := [1]; for n from 1 to m - 2 do
    P := ListTools:-PartialSums([A[-1], op(P)]); A := [op(A), P[-1]] od; A end: BellList(29); # Peter Luschny, Mar 24 2022
  • Mathematica
    f[n_] := Sum[ StirlingS2[n, k], {k, 0, n}]; Table[ f[n], {n, 0, 40}] (* Robert G. Wilson v *)
    Table[BellB[n], {n, 0, 40}] (* Harvey P. Dale, Mar 01 2011 *)
    B[0] = 1; B[n_] := 1/E Sum[k^(n - 1)/(k-1)!, {k, 1, Infinity}] (* Dimitri Papadopoulos, Mar 10 2015, edited by M. F. Hasler, Nov 30 2018 *)
    BellB[Range[0,40]] (* Eric W. Weisstein, Aug 10 2017 *)
    b[1] = 1; k = 1; Flatten[{1, Table[Do[j = k; k += b[m]; b[m] = j;, {m, 1, n-1}]; b[n] = k, {n, 1, 40}]}] (* Vaclav Kotesovec, Sep 07 2019 *)
    Table[j! Coefficient[Series[Exp[Exp[x] - 1], {x, 0, 20}], x, j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 01 2023 *)
    Table[(D[Exp[Exp[x]], {x, n}] /. x -> 0)/E, {n, 0, 20}] (* Joan Ludevid, Nov 05 2024 *)
  • Maxima
    makelist(belln(n),n,0,40); /* Emanuele Munarini, Jul 04 2011 */
    
  • PARI
    {a(n) = my(m); if( n<0, 0, m = contfracpnqn( matrix(2, n\2, i, k, if( i==1, -k*x^2, 1 - (k+1)*x))); polcoeff(1 / (1 - x + m[2,1] / m[1,1]) + x * O(x^n), n))}; /* Michael Somos */
    
  • PARI
    {a(n) = polcoeff( sum( k=0, n, prod( i=1, k, x / (1 - i*x)), x^n * O(x)), n)}; /* Michael Somos, Aug 22 2004 */
    
  • PARI
    a(n)=round(exp(-1)*suminf(k=0,1.0*k^n/k!)) \\ Gottfried Helms, Mar 30 2007 - WARNING! For illustration only: Gives silently a wrong result for n = 42 and an error for n > 42, with standard precision of 38 digits. - M. F. Hasler, Nov 30 2018
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( exp( x + x * O(x^n)) - 1), n))}; /* Michael Somos, Jun 28 2009 */
    
  • PARI
    Vec(serlaplace(exp(exp('x+O('x^66))-1))) \\ Joerg Arndt, May 26 2012
    
  • PARI
    A000110(n)=sum(k=0,n,stirling(n,k,2)) \\ M. F. Hasler, Nov 30 2018
    
  • Perl
    use bignum;sub a{my($n)=@;my@t=map{[(0)x($n+1)]}0..$n;$t[0][0]=1;for my$i(1..$n){$t[$i][0]=$t[$i-1][$i-1];for my$j(1..$i){$t[$i][$j]=$t[$i-1][$j-1]+$t[$i][$j-1]}}return map{$t[$][0]}0..$n-1}print join(", ",a(28)),"\n" # Paul Muljadi, May 08 2024
  • Python
    # The objective of this implementation is efficiency.
    # m -> [a(0), a(1), ..., a(m)] for m > 0.
    def A000110_list(m):
        A = [0 for i in range(m)]
        A[0] = 1
        R = [1, 1]
        for n in range(1, m):
            A[n] = A[0]
            for k in range(n, 0, -1):
                A[k-1] += A[k]
            R.append(A[0])
        return R
    A000110_list(40) # Peter Luschny, Jan 18 2011
    
  • Python
    # requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
    from itertools import accumulate
    A000110, blist, b = [1,1], [1], 1
    for _ in range(20):
        blist = list(accumulate([b]+blist))
        b = blist[-1]
        A000110.append(b) # Chai Wah Wu, Sep 02 2014, updated Chai Wah Wu, Sep 19 2014
    
  • Python
    from sympy import bell
    print([bell(n) for n in range(27)]) # Michael S. Branicky, Dec 15 2021
    
  • Python
    from functools import cache
    @cache
    def a(n, k=0): return int(n < 1) or k*a(n-1, k) + a(n-1, k+1)
    print([a(n) for n in range(27)])  # Peter Luschny, Jun 14 2022
    
  • Sage
    from sage.combinat.expnums import expnums2; expnums2(30, 1) # Zerinvary Lajos, Jun 26 2008
    
  • Sage
    [bell_number(n) for n in (0..40)] # G. C. Greubel, Jun 13 2019
    

Formula

E.g.f.: exp(exp(x) - 1).
Recurrence: a(n+1) = Sum_{k=0..n} a(k)*binomial(n, k).
a(n) = Sum_{k=0..n} Stirling2(n, k).
a(n) = Sum_{j=0..n-1} (1/(n-1)!)*A000166(j)*binomial(n-1, j)*(n-j)^(n-1). - André F. Labossière, Dec 01 2004
G.f.: (Sum_{k>=0} 1/((1-k*x)*k!))/exp(1) = hypergeom([-1/x], [(x-1)/x], 1)/exp(1) = ((1-2*x)+LaguerreL(1/x, (x-1)/x, 1)+x*LaguerreL(1/x, (2*x-1)/x, 1))*Pi/(x^2*sin(Pi*(2*x-1)/x)), where LaguerreL(mu, nu, z) = (gamma(mu+nu+1)/(gamma(mu+1)*gamma(nu+1)))* hypergeom([-mu], [nu+1], z) is the Laguerre function, the analytic extension of the Laguerre polynomials, for mu not equal to a nonnegative integer. This generating function has an infinite number of poles accumulating in the neighborhood of x=0. - Karol A. Penson, Mar 25 2002
a(n) = exp(-1)*Sum_{k >= 0} k^n/k! [Dobinski]. - Benoit Cloitre, May 19 2002
a(n) is asymptotic to n!*(2 Pi r^2 exp(r))^(-1/2) exp(exp(r)-1) / r^n, where r is the positive root of r exp(r) = n. See, e.g., the Odlyzko reference.
a(n) is asymptotic to b^n*exp(b-n-1/2)*sqrt(b/(b+n)) where b satisfies b*log(b) = n - 1/2 (see Graham, Knuth and Patashnik, Concrete Mathematics, 2nd ed., p. 493). - Benoit Cloitre, Oct 23 2002, corrected by Vaclav Kotesovec, Jan 06 2013
Lovasz (Combinatorial Problems and Exercises, North-Holland, 1993, Section 1.14, Problem 9) gives another asymptotic formula, quoted by Mezo and Baricz. - N. J. A. Sloane, Mar 26 2015
G.f.: Sum_{k>=0} x^k/(Product_{j=1..k} (1-j*x)) (see Klazar for a proof). - Ralf Stephan, Apr 18 2004
a(n+1) = exp(-1)*Sum_{k>=0} (k+1)^(n)/k!. - Gerald McGarvey, Jun 03 2004
For n>0, a(n) = Aitken(n-1, n-1) [i.e., a(n-1, n-1) of Aitken's array (A011971)]. - Gerald McGarvey, Jun 26 2004
a(n) = Sum_{k=1..n} (1/k!)*(Sum_{i=1..k} (-1)^(k-i)*binomial(k, i)*i^n + 0^n). - Paul Barry, Apr 18 2005
a(n) = A032347(n) + A040027(n+1). - Jon Perry, Apr 26 2005
a(n) = (2*n!/(Pi*e))*Im( Integral_{x=0..Pi} e^(e^(e^(ix))) sin(nx) dx ) where Im denotes imaginary part [Cesaro]. - David Callan, Sep 03 2005
O.g.f.: 1/(1-x-x^2/(1-2*x-2*x^2/(1-3*x-3*x^2/(.../(1-n*x-n*x^2/(...)))))) (continued fraction due to Ph. Flajolet). - Paul D. Hanna, Jan 17 2006
From Karol A. Penson, Jan 14 2007: (Start)
Representation of Bell numbers B(n), n=1,2,..., as special values of hypergeometric function of type (n-1)F(n-1), in Maple notation: B(n)=exp(-1)*hypergeom([2,2,...,2],[1,1,...,1],1), n=1,2,..., i.e., having n-1 parameters all equal to 2 in the numerator, having n-1 parameters all equal to 1 in the denominator and the value of the argument equal to 1.
Examples:
B(1)=exp(-1)*hypergeom([],[],1)=1
B(2)=exp(-1)*hypergeom([2],[1],1)=2
B(3)=exp(-1)*hypergeom([2,2],[1,1],1)=5
B(4)=exp(-1)*hypergeom([2,2,2],[1,1,1],1)=15
B(5)=exp(-1)*hypergeom([2,2,2,2],[1,1,1,1],1)=52
(Warning: this formula is correct but its application by a computer may not yield exact results, especially with a large number of parameters.)
(End)
a(n+1) = 1 + Sum_{k=0..n-1} Sum_{i=0..k} binomial(k,i)*(2^(k-i))*a(i). - Yalcin Aktar, Feb 27 2007
a(n) = [1,0,0,...,0] T^(n-1) [1,1,1,...,1]', where T is the n X n matrix with main diagonal {1,2,3,...,n}, 1's on the diagonal immediately above and 0's elsewhere. [Meier]
a(n) = ((2*n!)/(Pi * e)) * ImaginaryPart(Integral[from 0 to Pi](e^e^e^(i*theta))*sin(n*theta) dtheta). - Jonathan Vos Post, Aug 27 2007
From Tom Copeland, Oct 10 2007: (Start)
a(n) = T(n,1) = Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * Lag(n,-1,j-n) = Sum_{j=0..n} [ E(n,j)/n! ] * [ n!*Lag(n,-1, j-n) ] where T(n,x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m. Note that E(n,j)/n! = E(n,j) / (Sum_{k=0..n} E(n,k)).
The Eulerian numbers count the permutation ascents and the expression [n!*Lag(n,-1, j-n)] is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to n!*a(n) = Sum_{j=0..n} E(n,j) * [n!*Lag(n,-1, j-n)].
(End)
Define f_1(x), f_2(x), ... such that f_1(x)=e^x and for n=2,3,... f_{n+1}(x) = (d/dx)(x*f_n(x)). Then for Bell numbers B_n we have B_n=1/e*f_n(1). - Milan Janjic, May 30 2008
a(n) = (n-1)! Sum_{k=1..n} a(n-k)/((n-k)! (k-1)!) where a(0)=1. - Thomas Wieder, Sep 09 2008
a(n+k) = Sum_{m=0..n} Stirling2(n,m) Sum_{r=0..k} binomial(k,r) m^r a(k-r). - David Pasino (davepasino(AT)yahoo.com), Jan 25 2009. (Umbrally, this may be written as a(n+k) = Sum_{m=0..n} Stirling2(n,m) (a+m)^k. - N. J. A. Sloane, Feb 07 2009)
Sum_{k=1..n-1} a(n)*binomial(n,k) = Sum_{j=1..n}(j+1)*Stirling2(n,j+1). - [Zhao] - R. J. Mathar, Jun 24 2024
From Thomas Wieder, Feb 25 2009: (Start)
a(n) = Sum_{k_1=0..n+1} Sum_{k_2=0..n}...Sum_{k_i=0..n-i}...Sum_{k_n=0..1}
delta(k_1,k_2,...,k_i,...,k_n)
where delta(k_1,k_2,...,k_i,...,k_n) = 0 if any k_i > k_(i+1) and k_(i+1) <> 0
and delta(k_1,k_2,...,k_i,...,k_n) = 1 otherwise.
(End)
Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]:=binomial(j-1,i-1), (i<=j), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jul 08 2010
G.f. satisfies A(x) = (x/(1-x))*A(x/(1-x)) + 1. - Vladimir Kruchinin, Nov 28 2011
G.f.: 1 / (1 - x / (1 - 1*x / (1 - x / (1 - 2*x / (1 - x / (1 - 3*x / ... )))))). - Michael Somos, May 12 2012
a(n+1) = Sum_{m=0..n} Stirling2(n, m)*(m+1), n >= 0. Compare with the third formula for a(n) above. Here Stirling2 = A048993. - Wolfdieter Lang, Feb 03 2015
G.f.: (-1)^(1/x)*((-1/x)!/e + (!(-1-1/x))/x) where z! and !z are factorial and subfactorial generalized to complex arguments. - Vladimir Reshetnikov, Apr 24 2013
The following formulas were proposed during the period Dec 2011 - Oct 2013 by Sergei N. Gladkovskii: (Start)
E.g.f.: exp(exp(x)-1) = 1 + x/(G(0)-x); G(k) = (k+1)*Bell(k) + x*Bell(k+1) - x*(k+1)*Bell(k)*Bell(k+2)/G(k+1) (continued fraction).
G.f.: W(x) = (1-1/(G(0)+1))/exp(1); G(k) = x*k^2 + (3*x-1)*k - 2 + x - (k+1)*(x*k+x-1)^2/G(k+1); (continued fraction Euler's kind, 1-step).
G.f.: W(x) = (1 + G(0)/(x^2-3*x+2))/exp(1); G(k) = 1 - (x*k+x-1)/( ((k+1)!) - (((k+1)!)^2)*(1-x-k*x+(k+1)!)/( ((k+1)!)*(1-x-k*x+(k+1)!) - (x*k+2*x-1)*(1-2*x-k*x+(k+2)!)/G(k+1))); (continued fraction).
G.f.: A(x) = 1/(1 - x/(1-x/(1 + x/G(0)))); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1); (continued fraction, 1-step).
G.f.: -1/U(0) where U(k) = x*k - 1 + x - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + x/U(0) where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + 1/(U(0) - x) where U(k) = 1 + x - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1 + x/(U(0)-x) where U(k) = 1 - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1/G(0) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/G(k+1) ))); (continued fraction).
G.f.: G(0)/(1+x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k-1) - x*(2*k+1)*(2*k+3)*(2*x*k-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+x-1)/G(k+1) )); (continued fraction).
G.f.: -(1+2*x) * Sum_{k >= 0} x^(2*k)*(4*x*k^2-2*k-2*x-1) / ((2*k+1) * (2*x*k-1)) * A(k) / B(k) where A(k) = Product_{p=0..k} (2*p+1), B(k) = Product_{p=0..k} (2*p-1) * (2*x*p-x-1) * (2*x*p-2*x-1).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: 1 + x*(S-1) where S = Sum_{k>=0} ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k/Product_{i=0..k-1} (1-x-x*i)/(1-x).
G.f.: (G(0) - 2)/(2*x-1) where G(k) = 2 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: -G(0) where G(k) = 1 - (x*k - 2)/(x*k - 1 - x*(x*k - 1)/(x + (x*k - 2)/G(k+1) )); (continued fraction).
G.f.: G(0) where G(k) = 2 - (2*x*k - 1)/(x*k - 1 - x*(x*k - 1)/(x + (2*x*k - 1)/G(k+1) )); (continued fraction).
G.f.: (G(0) - 1)/(1+x) where G(k) = 1 + 1/(1-k*x)/(1-x/(x+1/G(k+1) )); (continued fraction).
G.f.: 1/(x*(1-x)*G(0)) - 1/x where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/G(k+1) )); (continued fraction).
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 + x/( x*k - 1 )/Q(k+1); (continued fraction).
G.f.: 1+x/Q(0), where Q(k) = 1 - x - x/(1 - x*(k+1)/Q(k+1)); (continued fraction).
G.f.: 1/(1-x*Q(0)), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))); (continued fraction).
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x*(k+1))*(1-x*(k+2))/Q(k+1) ); (continued fraction).
(End)
a(n) ~ exp(exp(W(n))-n-1)*n^n/W(n)^(n+1/2), where W(x) is the Lambert W-function. - Vladimir Reshetnikov, Nov 01 2015
a(n) ~ n^n * exp(n/W(n)-1-n) / (sqrt(1+W(n)) * W(n)^n). - Vaclav Kotesovec, Nov 13 2015
From Natalia L. Skirrow, Apr 13 2025: (Start)
By taking logarithmic derivatives of the equivalent to Kotesovec's asymptotic for Bell polynomials at x=1, we obtain properties of the nth row of A008277 as a statistical distribution (where W=W(n),X=W(n)+1)
a(n+1)/a(n) ~ n/W + W/(2*(W+1)^2) is 1 more than the expectation.
(2*a(n+1)+a(n+2))/a(n) - (a(n+1)/a(n))^2 - a(n+2)/a(n+1) ~ n/(W*X)+1/(2*X^2)-3/(2*X^3)+1/X^4 is 1 more than the variance.
(This is a complete asymptotic characterisation, since they converge to normal distributions; see Harper, 1967)
(End)
a(n) are the coefficients in the asymptotic expansion of -exp(-1)*(-1)^x*x*Gamma(-x,0,-1), where Gamma(a,z0,z1) is the generalized incomplete Gamma function. - Vladimir Reshetnikov, Nov 12 2015
a(n) = 1 + floor(exp(-1) * Sum_{k=1..2*n} k^n/k!). - Vladimir Reshetnikov, Nov 13 2015
a(p^m) == m+1 (mod p) when p is prime and m >= 1 (see Lemma 3.1 in the Hurst/Schultz reference). - Seiichi Manyama, Jun 01 2016
a(n) = Sum_{k=0..n} hypergeom([1, -k], [], 1)*Stirling2(n+1, k+1) = Sum_{k=0..n} A182386(k)*Stirling2(n+1, k+1). - Mélika Tebni, Jul 02 2022
For n >= 1, a(n) = Sum_{i=0..n-1} a(i)*A074664(n-i). - Davide Rotondo, Apr 21 2024
a(n) is the n-th derivative of e^e^x divided by e at point x=0. - Joan Ludevid, Nov 05 2024

Extensions

Edited by M. F. Hasler, Nov 30 2018

A028387 a(n) = n + (n+1)^2.

Original entry on oeis.org

1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651
Offset: 0

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Keywords

Comments

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005
Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006
A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007
Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007
Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008
a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009
sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009
When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010
a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010
The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011
Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012
Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012
Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013
a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014
A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015
An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015
Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015
Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017
The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017
From Klaus Purath, Mar 18 2019: (Start)
Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with
x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.
But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)
a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019
a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021
Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021
(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022
The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

Examples

			From _Ilya Gutkovskiy_, Apr 13 2016: (Start)
Illustration of initial terms:
                                        o               o
                        o           o   o o           o o
            o       o   o o       o o   o o o       o o o
    o   o   o o   o o   o o o   o o o   o o o o   o o o o
o   o o o   o o o o o   o o o o o o o   o o o o o o o o o
n=0  n=1       n=2           n=3               n=4
(End)
From _Klaus Purath_, Mar 18 2019: (Start)
Examples:
a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045).
a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032).
a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060).
a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120).
a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130).
a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121).
a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137).
(End)
		

Crossrefs

Complement of A028392. Third column of array A094954.
Cf. A000217, A002522, A062392, A062786, A127701, A135223, A143596, A052905, A162997, A062938 (squares of this sequence).
A110331 and A165900 are signed versions.
Cf. A002327 (primes), A094210.
Frobenius number for k successive numbers: this sequence (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).

Programs

Formula

a(n) = sqrt(A062938(n)). - Floor van Lamoen, Oct 08 2001
a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1. - Gerald McGarvey, Sep 24 2004
a(n) = A105728(n+2, n+1). - Reinhard Zumkeller, Apr 18 2005
a(n) = A109128(n+2, 2). - Reinhard Zumkeller, Jun 20 2005
a(n) = 2*T(n+1) - 1, where T(n) = A000217(n). - Gary W. Adamson, Aug 15 2007
a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum_{k=0..n} a(k). - Reinhard Zumkeller, Aug 20 2007
Binomial transform of [1, 4, 2, 0, 0, 0, ...] = (1, 5, 11, 19, ...). - Gary W. Adamson, Sep 20 2007
G.f.: (1+2*x-x^2)/(1-x)^3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - R. J. Mathar, Jul 11 2009
a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989... Example: a(3) = 19 = (5 + .6180339...) * (3.381966...). Cf. next to leftmost column in A162997 array. - Gary W. Adamson, Jul 23 2009
a(n) = a(n-1) + 2*(n+1), with n > 0, a(0) = 1. - Vincenzo Librandi, Nov 18 2010
For k < n, a(n) = (k+1)*a(n-k) - k*a(n-k-1) + k*(k+1); e.g., a(5) = 41 = 4*11 - 3*5 + 3*4. - Charlie Marion, Jan 13 2011
a(n) = lower right term in M^2, M = the 2 X 2 matrix [1, n; 1, (n+1)]. - Gary W. Adamson, Jun 29 2011
G.f.: (x^2-2*x-1)/(x-1)^3 = G(0) where G(k) = 1 + x*(k+1)*(k+4)/(1 - 1/(1 + (k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 16 2012
Sum_{n>0} 1/a(n) = 1 + Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - Enrique Pérez Herrero, Oct 11 2013
E.g.f.: exp(x) (1+4*x+x^2). - Tom Copeland, Dec 02 2013
a(n) = A005408(A000217(n)). - Tony Foster III, May 31 2016
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2)/6. (End)
a(5*n+1)/5 = A062786(n+1). - Torlach Rush, Jun 05 2024

Extensions

Minor edits by N. J. A. Sloane, Jul 04 2010, following suggestions from the Sequence Fans Mailing List

A034691 Euler transform of powers of 2 [1,2,4,8,16,...].

Original entry on oeis.org

1, 1, 3, 7, 18, 42, 104, 244, 585, 1373, 3233, 7533, 17547, 40591, 93711, 215379, 493735, 1127979, 2570519, 5841443, 13243599, 29953851, 67604035, 152258271, 342253980, 767895424, 1719854346, 3845443858
Offset: 0

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Comments

This is the number of different hierarchical orderings that can be formed from n unlabeled elements: these are divided into groups and the elements in each group are then arranged in an "unlabeled preferential arrangement" or "composition" as in A000079. - Thomas Wieder and N. J. A. Sloane, Jun 10 2003
From Gus Wiseman, Mar 03 2016: (Start)
The original Sloane-Wieder definition, "To obtain a hierarchical ordering we partition the elements into unlabeled nonempty subsets and form a composition of each subset," [arXiv:math/0307064] has two possible meanings. The first possible meaning is that we should (1) choose a set partition pi of {1...n} and (2) for each block of pi choose a composition of the number of elements. In this case the correct number of such structures would evidently be counted by A004211 which differs from a(n) for n > 2.
The other possible meaning is that after we have done (1) and (2) above we (3) "forget" the choice of pi. We will have produced a collection M of multisets of compositions. The span of M (its set of distinct elements) is correctly counted by A034691 and it seems that non-isomorphic hierarchical orderings of unlabeled sets are nothing more than multisets of compositions. This discovery is due to Wieder. (End)
The asymptotic formula in the article by N. J. A. Sloane and Thomas Wieder, "The Number of Hierarchical Orderings" (Theorem 3) is incorrect (a multiplicative factor of 1.397... is missing, see my formula below). - Vaclav Kotesovec, Sep 08 2014
Number of partitions of n into 1 sort of 1, 2 sorts of 2's, 4 sorts of 3's, ..., 2^(k-1) sorts of k's, ... . - Joerg Arndt, Sep 09 2014
Also number of normal multiset partitions of weight n, where a multiset is normal if it spans an initial interval of positive integers. - Gus Wiseman, Mar 03 2016

Examples

			The normal multiset partitions for a(4) = 18: {{1111},{1222},{1122},{1112},{1233},{1223},{1123},{1234},{1,111},{1,122},{1,112},{1,123},{11,11},{11,12},{12,12},{1,1,11},{1,1,12},{1,1,1,1}}
		

Crossrefs

Cf. A034899, A075729, A247003, A004211, A104500 (Euler transform), A290222 (Multiset transform).

Programs

  • Maple
    oo := 101: mul( 1/(1-x^j)^(2^(j-1)),j=1..oo): series(%,x,oo): t1 := seriestolist(%); A034691 := n-> t1[n+1];
    with(combstruct); SetSeqSetU := [T, {T=Set(S), S=Sequence(U,card >= 1), U=Set(Z,card >=1)},unlabeled]; seq(count(SetSeqSetU,size=j),j=1..12);
    # Alternative, uses EulerTransform from A358369:
    a := EulerTransform(BinaryRecurrenceSequence(2, 0)):
    seq(a(n), n = 0..27); # Peter Luschny, Nov 17 2022
  • Mathematica
    nn = 30; b = Table[2^n, {n, 0, nn}]; CoefficientList[Series[Product[1/(1 - x^m)^b[[m]], {m, nn}], {x, 0, nn}],  x] (* T. D. Noe, Nov 21 2011 *)
    Table[SeriesCoefficient[E^(Sum[x^k/(1 - 2*x^k)/k, {k, 1, n}]), {x, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Sep 08 2014 *)
    allnorm[n_Integer]:=Function[s,Array[Count[s,y_/;y<=#]+1&,n]]/@Subsets[Range[n-1]+1];
    allnmsp[0]={};allnmsp[1]={{{1}}};allnmsp[n_Integer]:=allnmsp[n]=Join[allnmsp[n-1],List/@allnorm[n],Join@@Function[ptn,Append[ptn,#]&/@Select[allnorm[n-Length[Join@@ptn]],OrderedQ[{Last[ptn],#}]&]]/@allnmsp[n-1]];
    Apply[SequenceForm,Select[allnmsp[4],Length[Join@@#]===4&],{2}] (* to construct the example *)
    Table[Length[Complement[allnmsp[n],allnmsp[n-1]]],{n,1,8}] (* Gus Wiseman, Mar 03 2016 *)
  • PARI
    A034691(n,l=1+O('x^(n+1)))={polcoeff(1/prod(k=1,n,(l-'x^k)^2^(k-1)),n)} \\ Michael Somos, Nov 21 2011, edited by M. F. Hasler, Jul 24 2017
    
  • SageMath
    # uses[EulerTransform from A166861]
    a = BinaryRecurrenceSequence(2, 0)
    b = EulerTransform(a)
    print([b(n) for n in range(30)]) # Peter Luschny, Nov 11 2020

Formula

G.f.: 1 / Product_{n>=1} (1-x^n)^(2^(n-1)).
Recurrence: a(n) = (1/n) * Sum_{m=1..n} a(n-m)*c(m) where c(m) = A083413(m).
a(n) ~ c * 2^n * exp(sqrt(2*n)) / (sqrt(2*Pi) * exp(1/4) * 2^(3/4) * n^(3/4)), where c = exp( Sum_{k>=2} 1/(k*(2^k-2)) ) = 1.3976490050836502... (see A247003). - Vaclav Kotesovec, Sep 08 2014

A038207 Triangle whose (i,j)-th entry is binomial(i,j)*2^(i-j).

Original entry on oeis.org

1, 2, 1, 4, 4, 1, 8, 12, 6, 1, 16, 32, 24, 8, 1, 32, 80, 80, 40, 10, 1, 64, 192, 240, 160, 60, 12, 1, 128, 448, 672, 560, 280, 84, 14, 1, 256, 1024, 1792, 1792, 1120, 448, 112, 16, 1, 512, 2304, 4608, 5376, 4032, 2016, 672, 144, 18, 1, 1024, 5120, 11520, 15360, 13440, 8064, 3360, 960, 180, 20, 1
Offset: 0

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Keywords

Comments

This infinite matrix is the square of the Pascal matrix (A007318) whose rows are [ 1,0,... ], [ 1,1,0,... ], [ 1,2,1,0,... ], ...
As an upper right triangle, table rows give number of points, edges, faces, cubes,
4D hypercubes etc. in hypercubes of increasing dimension by column. - Henry Bottomley, Apr 14 2000. More precisely, the (i,j)-th entry is the number of j-dimensional subspaces of an i-dimensional hypercube (see the Coxeter reference). - Christof Weber, May 08 2009
Number of different partial sums of 1+[1,1,2]+[2,2,3]+[3,3,4]+[4,4,5]+... with entries that are zero removed. - Jon Perry, Jan 01 2004
Row sums are powers of 3 (A000244), antidiagonal sums are Pell numbers (A000129). - Gerald McGarvey, May 17 2005
Riordan array (1/(1-2x), x/(1-2x)). - Paul Barry, Jul 28 2005
T(n,k) is the number of elements of the Coxeter group B_n with descent set contained in {s_k}, 0<=k<=n-1. For T(n,n), we interpret this as the number of elements of B_n with empty descent set (since s_n does not exist). - Elizabeth Morris (epmorris(AT)math.washington.edu), Mar 01 2006
Let S be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xSy if x is a subset of y. Then T(n,k) = the number of elements (x,y) of S for which y has exactly k more elements than x. - Ross La Haye, Oct 12 2007
T(n,k) is number of paths in the first quadrant going from (0,0) to (n,k) using only steps B=(1,0) colored blue, R=(1,0) colored red and U=(1,1). Example: T(3,2)=6 because we have BUU, RUU, UBU, URU, UUB and UUR. - Emeric Deutsch, Nov 04 2007
T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (0,1), and two kinds of step (1,0). - Joerg Arndt, Jul 01 2011
T(i,j) is the number of i-permutations of {1,2,3} containing j 1's. Example: T(2,1)=4 because we have 12, 13, 21 and 31; T(3,2)=6 because we have 112, 113, 121, 131, 211 and 311. - Zerinvary Lajos, Dec 21 2007
Triangle of coefficients in expansion of (2+x)^n. - N-E. Fahssi, Apr 13 2008
Sum of diagonals are Jacobsthal-numbers: A001045. - Mark Dols, Aug 31 2009
Triangle T(n,k), read by rows, given by [2,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 15 2009
Eigensequence of the triangle = A004211: (1, 3, 11, 49, 257, 1539, ...). - Gary W. Adamson, Feb 07 2010
f-vectors ("face"-vectors) for n-dimensional cubes [see e.g., Hoare]. (This is a restatement of Bottomley's above.) - Tom Copeland, Oct 19 2012
With P = Pascal matrix, the sequence of matrices I, A007318, A038207, A027465, A038231, A038243, A038255, A027466 ... = P^0, P^1, P^2, ... are related by Copeland's formula below to the evolution at integral time steps n= 0, 1, 2, ... of an exponential distribution exp(-x*z) governed by the Fokker-Planck equation as given in the Dattoli et al. ref. below. - Tom Copeland, Oct 26 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Unsigned diagonals of A133156 are rows of this array. - Tom Copeland, Oct 11 2014
Omitting the first row, this is the production matrix for A039683, where an equivalent differential operator can be found. - Tom Copeland, Oct 11 2016
T(n,k) is the number of functions f:[n]->[3] with exactly k elements mapped to 3. Note that there are C(n,k) ways to choose the k elements mapped to 3, and there are 2^(n-k) ways to map the other (n-k) elements to {1,2}. Hence, by summing T(n,k) as k runs from 0 to n, we obtain 3^n = Sum_{k=0..n} T(n,k). - Dennis P. Walsh, Sep 26 2017
Since this array is the square of the Pascal lower triangular matrix, the row polynomials of this array are obtained as the umbral composition of the row polynomials P_n(x) of the Pascal matrix with themselves. E.g., P_3(P.(x)) = 1 P_3(x) + 3 P_2(x) + 3 P_1(x) + 1 = (x^3 + 3 x^2 + 3 x + 1) + 3 (x^2 + 2 x + 1) + 3 (x + 1) + 1 = x^3 + 6 x^2 + 12 x + 8. - Tom Copeland, Nov 12 2018
T(n,k) is the number of 2-compositions of n+1 with some zeros allowed that have k zeros; see the Hopkins & Ouvry reference. - Brian Hopkins, Aug 16 2020
Also the convolution triangle of A000079. - Peter Luschny, Oct 09 2022

Examples

			Triangle begins with T(0,0):
   1;
   2,  1;
   4,  4,  1;
   8, 12,  6,  1;
  16, 32, 24,  8,  1;
  32, 80, 80, 40, 10,  1;
  ... -  corrected by _Clark Kimberling_, Aug 05 2011
Seen as an array read by descending antidiagonals:
[0] 1, 2,  4,   8,    16,    32,    64,     128,     256, ...     [A000079]
[1] 1, 4,  12,  32,   80,    192,   448,    1024,    2304, ...    [A001787]
[2] 1, 6,  24,  80,   240,   672,   1792,   4608,    11520, ...   [A001788]
[3] 1, 8,  40,  160,  560,   1792,  5376,   15360,   42240, ...   [A001789]
[4] 1, 10, 60,  280,  1120,  4032,  13440,  42240,   126720, ...  [A003472]
[5] 1, 12, 84,  448,  2016,  8064,  29568,  101376,  329472, ...  [A054849]
[6] 1, 14, 112, 672,  3360,  14784, 59136,  219648,  768768, ...  [A002409]
[7] 1, 16, 144, 960,  5280,  25344, 109824, 439296,  1647360, ... [A054851]
[8] 1, 18, 180, 1320, 7920,  41184, 192192, 823680,  3294720, ... [A140325]
[9] 1, 20, 220, 1760, 11440, 64064, 320320, 1464320, 6223360, ... [A140354]
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 155.
  • H. S. M. Coxeter, Regular Polytopes, Dover Publications, New York (1973), p. 122.

Crossrefs

Programs

  • GAP
    Flat(List([0..15], n->List([0..n], k->Binomial(n, k)*2^(n-k)))); # Stefano Spezia, Nov 21 2018
  • Haskell
    a038207 n = a038207_list !! n
    a038207_list = concat $ iterate ([2,1] *) [1]
    instance Num a => Num [a] where
       fromInteger k = [fromInteger k]
       (p:ps) + (q:qs) = p + q : ps + qs
       ps + qs         = ps ++ qs
       (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
        *                = []
    -- Reinhard Zumkeller, Apr 02 2011
    
  • Haskell
    a038207' n k = a038207_tabl !! n !! k
    a038207_row n = a038207_tabl !! n
    a038207_tabl = iterate f [1] where
       f row = zipWith (+) ([0] ++ row) (map (* 2) row ++ [0])
    -- Reinhard Zumkeller, Feb 27 2013
    
  • Magma
    /* As triangle */ [[(&+[Binomial(n,i)*Binomial(i,k): i in [k..n]]): k in [0..n]]: n in [0..15]]; // Vincenzo Librandi, Nov 16 2018
    
  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*2^(i-j), j = 0 .. i) end do; # yields sequence in triangular form - Emeric Deutsch, Nov 04 2007
    # Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
    PMatrix(10, n -> 2^(n-1)); # Peter Luschny, Oct 09 2022
  • Mathematica
    Table[CoefficientList[Expand[(y + x + x^2)^n], y] /. x -> 1, {n, 0,10}] // TableForm (* Geoffrey Critzer, Nov 20 2011 *)
    Table[Binomial[n,k]2^(n-k),{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, May 22 2020 *)
  • PARI
    {T(n, k) = polcoeff((x+2)^n, k)}; /* Michael Somos, Apr 27 2000 */
    
  • Sage
    def A038207_triangle(dim):
        M = matrix(ZZ,dim,dim)
        for n in range(dim): M[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                M[n,k] = M[n-1,k-1]+2*M[n-1,k]
        return M
    A038207_triangle(9)  # Peter Luschny, Sep 20 2012
    

Formula

T(n, k) = Sum_{i=0..n} binomial(n,i)*binomial(i,k).
T(n, k) = (-1)^k*A065109(n,k).
G.f.: 1/(1-2*z-t*z). - Emeric Deutsch, Nov 04 2007
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
From the formalism of A133314, the e.g.f. for the row polynomials of A038207 is exp(x*t)*exp(2x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(-2x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*2x). The results generalize for 2 replaced by any number. - Tom Copeland, Aug 18 2008
Sum_{k=0..n} T(n,k)*x^k = (2+x)^n. - Philippe Deléham, Dec 15 2009
n-th row is obtained by taking pairwise sums of triangle A112857 terms starting from the right. - Gary W. Adamson, Feb 06 2012
T(n,n) = 1 and T(n,k) = T(n-1,k-1) + 2*T(n-1,k) for kJon Perry, Oct 11 2012
The e.g.f. for the n-th row is given by umbral composition of the normalized Laguerre polynomials A021009 as p(n,x) = L(n, -L(.,-x))/n! = 2^n L(n, -x/2)/n!. E.g., L(2,x) = 2 -4*x +x^2, so p(2,x)= (1/2)*L(2, -L(.,-x)) = (1/2)*(2*L(0,-x) + 4*L(1,-x) + L(2,-x)) = (1/2)*(2 + 4*(1+x) + (2+4*x+x^2)) = 4 + 4*x + x^2/2. - Tom Copeland, Oct 20 2012
From Tom Copeland, Oct 26 2012: (Start)
From the formalism of A132440 and A218272:
Let P and P^T be the Pascal matrix and its transpose and H= P^2= A038207.
Then with D the derivative operator,
exp(x*z/(1-2*z))/(1-2*z)= exp(2*z D_z z) e^(x*z)= exp(2*D_x (x D_x)) e^(z*x)
= (1 z z^2 z^3 ...) H (1 x x^2/2! x^3/3! ...)^T
= (1 x x^2/2! x^3/3! ...) H^T (1 z z^2 z^3 ...)^T
= Sum_{n>=0} z^n * 2^n Lag_n(-x/2)= exp[z*EF(.,x)], an o.g.f. for the f-vectors (rows) of A038207 where EF(n,x) is an e.g.f. for the n-th f-vector. (Lag_n(x) are the un-normalized Laguerre polynomials.)
Conversely,
exp(z*(2+x))= exp(2D_x) exp(x*z)= exp(2x) exp(x*z)
= (1 x x^2 x^3 ...) H^T (1 z z^2/2! z^3/3! ...)^T
= (1 z z^2/2! z^3/3! ...) H (1 x x^2 x^3 ...)^T
= exp(z*OF(.,x)), an e.g.f for the f-vectors of A038207 where
OF(n,x)= (2+x)^n is an o.g.f. for the n-th f-vector.
(End)
G.f.: R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ (1+y))*x/((2*k+2+ (1+y))*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
A038207 = exp[M*B(.,2)] where M = A238385-I and (B(.,x))^n = B(n,x) are the Bell polynomials (cf. A008277). B(n,2) = A001861(n). - Tom Copeland, Apr 17 2014
T = (A007318)^2 = A112857*|A167374| = |A118801|*|A167374| = |A118801*A167374| = |P*A167374*P^(-1)*A167374| = |P*NpdP*A167374|. Cf. A118801. - Tom Copeland, Nov 17 2016
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial 2^n*Sum_{k = 0..n} binomial(n,k)*x^k/k!. For example, the e.g.f. for the third subdiagonal is exp(x)*(8 + 24*x + 12*x^2 + 4*x^3/3) = 8 + 32*x + 80*x^2/2! + 160*x^3/3! + .... - Peter Bala, Mar 05 2017
T(3*k+2,k) = T(3*k+2,k+1), T(2*k+1,k) = 2*T(2*k+1,k+1). - Yuchun Ji, May 26 2020
From Robert A. Russell, Aug 05 2020: (Start)
G.f. for column k: x^k / (1-2*x)^(k+1).
E.g.f. for column k: exp(2*x) * x^k / k!. (End)
Also the array A(n, k) read by descending antidiagonals, where A(n, k) = (-1)^n*Sum_{j= 0..n+k} binomial(n + k, j)*hypergeom([-n, j+1], [1], 1). - Peter Luschny, Nov 09 2021

A004212 Shifts one place left under 3rd-order binomial transform.

Original entry on oeis.org

1, 1, 4, 19, 109, 742, 5815, 51193, 498118, 5296321, 60987817, 754940848, 9983845261, 140329768789, 2087182244308, 32725315072135, 539118388883449, 9304591246975030, 167804098493079547, 3155000165773280893
Offset: 0

Views

Author

Keywords

Comments

Equals the eigensequence of triangle A027465, the cube of Pascal's triangle. - Gary W. Adamson, Apr 10 2009
Length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(k)<=F(k)+3 where F(0)=0 and F(k+1)=s(k+1) if s(k+1)-s(k)=3, otherwise F(k+1)=F(k); see example and Fxtbook link. - Joerg Arndt, Apr 30 2011

Examples

			Restricted growth strings: a(0)=1 corresponds to the empty string, a(1)=1 to [0],
a(2)=3 to [00], [01], [02], and [03], a(3) = 19 to
        RGS           F
01:  [ 0 0 0 ]    [ 0 0 0 ]
02:  [ 0 0 1 ]    [ 0 0 0 ]
03:  [ 0 0 2 ]    [ 0 0 0 ]
04:  [ 0 0 3 ]    [ 0 0 3 ]
05:  [ 0 1 0 ]    [ 0 0 0 ]
06:  [ 0 1 1 ]    [ 0 0 0 ]
07:  [ 0 1 2 ]    [ 0 0 0 ]
08:  [ 0 1 3 ]    [ 0 0 3 ]
09:  [ 0 2 0 ]    [ 0 0 0 ]
10:  [ 0 2 1 ]    [ 0 0 0 ]
11:  [ 0 2 2 ]    [ 0 0 0 ]
12:  [ 0 2 3 ]    [ 0 0 3 ]
13:  [ 0 3 0 ]    [ 0 3 3 ]
14:  [ 0 3 1 ]    [ 0 3 3 ]
15:  [ 0 3 2 ]    [ 0 3 3 ]
16:  [ 0 3 3 ]    [ 0 3 3 ]
17:  [ 0 3 4 ]    [ 0 3 3 ]
18:  [ 0 3 5 ]    [ 0 3 3 ]
19:  [ 0 3 6 ]    [ 0 3 6 ]
- _Joerg Arndt_, Apr 30 2011
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A075498 (row sums).
Cf. A027465. - Gary W. Adamson, Apr 10 2009
Cf. A004211 (RGS where s(k)<=F(k)+2), A004213 (s(k)<=F(k)+4), A005011 (s(k)<=F(k)+5), A000110 (s(k)<=F(k)+1). - Joerg Arndt, Apr 30 2011
Cf. A009235.

Programs

  • Mathematica
    Table[Sum[StirlingS2[n,k] 3^(-k+n),{k,n}],{n,20}] (* Vincenzo Librandi, May 21 2012 *)
    Table[3^n BellB[n, 1/3], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(3^(n-k)*binomial(n-1,k-1)*a(k-1),k,1,n); /* Vladimir Kruchinin, Nov 28 2011 */
  • PARI
    x='x+O('x^66); /* that many terms */
    egf=exp(intformal(exp(3*x))); /* =  1 + x + 2*x^2 + 19/6*x^3 + 109/24*x^4 + ... */
    /* egf=exp(1/3*(exp(3*x)-1)) */ /* alternative computation */
    Vec(serlaplace(egf)) /* show terms */ /* Joerg Arndt, Apr 30 2011 */
    

Formula

a_n = Sum_{k=0..n} 3^(n-k)*Stirling2(n, k). - Emeric Deutsch, Feb 11 2002
E.g.f.: exp((exp(3*x)-1)/3).
O.g.f. A(x) satisfies A'(x)/A(x) = e^(3*x).
E.g.f.: exp(Integral_{t = 0..x} exp(3*t)). - Joerg Arndt, Apr 30 2011
O.g.f.: Sum_{k>=0} x^k/Product_{j=1..k} (1-3*j*x). - Joerg Arndt, Apr 30 2011
Hankel transform is A000178(n)*3^C(n+1,2). - Paul Barry, Mar 31 2008
Define f_1(x), f_2(x), ... such that f_1(x)=e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n=2,3,.... Then a(n) = e^(-1/2)*3^{n-1}*f_n(1/3). - Milan Janjic, May 30 2008
a(n) = the upper left term in M^n, M = the following infinite square production matrix:
1, 3, 0, 0, 0, ...
1, 1, 3, 0, 0, ...
1, 2, 1, 3, 0, ...
1, 3, 3, 1, 3, ...
... (in which a diagonal of (3,3,3,...) is appended to the right of Pascal's triangle). - Gary W. Adamson, Jul 29 2011
G.f. satisfies A(x) = 1+x/(1-3*x)*A(x/(1-3*x)). a(n) = Sum_{k=1..n} 3^(n-k)*binomial(n-1,k-1)*a(k-1), n > 0, a(0)=1. - Vladimir Kruchinin, Nov 28 2011 [corrected by Ilya Gutkovskiy, May 02 2019]
From Peter Bala, May 16 2012: (Start)
Recurrence equation: a(n+1) = Sum_{k = 0..n} 3^(n-k)*C(n,k)*a(k). Written umbrally this is a(n+1) = (a + 3)^n (expand the binomial and replace a^k with a(k)). More generally, a*f(a) = f(a+3) holds umbrally for any polynomial f(x). An inductive argument then establishes the umbral recurrence a*(a-3)*(a-6)*...*(a-3*(n-1)) = 1 with a(0) = 1. Compare with the Bell numbers B(n) = A000110(n), which satisfy the umbral recurrence B*(B-1)*...*(B-(n-1)) = 1 with B(0) = 1.
Touchard's congruence holds: for prime p not equal to 3, a(p+k) == (a(k) + a(k+1)) (mod p) for k = 0,1,2,... (adapt the proof of Theorem 10.1 in Gessel). In particular, a(p) == 2 (mod p) for prime p <> 3. (End)
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-x*3*k)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 16 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 3*x^2*(k+1)/( 3*x^2*(k+1) - (1-x-3*x*k)*(1-4*x-3*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 19 2013
a(n) ~ 3^n * n^n * exp(n/LambertW(3*n) - 1/3 - n) / (sqrt(1 + LambertW(3*n)) * LambertW(3*n)^n). - Vaclav Kotesovec, Jul 15 2021
a(n) = exp(-1/3)*Sum_{n >= 0} (3*n)^k/(n!*3^n). - Peter Bala, Jun 29 2024

A007405 Dowling numbers: e.g.f.: exp(x + (exp(b*x) - 1)/b) with b=2.

Original entry on oeis.org

1, 2, 6, 24, 116, 648, 4088, 28640, 219920, 1832224, 16430176, 157554048, 1606879040, 17350255744, 197553645440, 2363935624704, 29638547505408, 388328781668864, 5304452565517824, 75381218537805824, 1112348880749130752, 17014743624340539392, 269360902955086379008
Offset: 0

Views

Author

Keywords

Comments

Binomial transform of A004211.
Equals leftmost term in iterates of M^n * [1,1,1,...], where M = a bidiagonal matrix with (1,3,5,7,...) in the main diagonal and (1,1,1,...) in the superdiagonal. - Gary W. Adamson, Apr 13 2009
This is the number of type B set partitions, see R. Suter. - Per W. Alexandersson, Dec 19 2022

Examples

			a(4) = 116 = sum of top row terms of M^3 = (49 + 44 + 18 + 4 + 1).
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000110 (b=1), this sequence (b=2), A003575 (b=3), A003576 (b=4), A003577 (b=5), A003578 (b=6), A003579 (b=7), A003580 (b=8), A003581 (b=9), A003582 (b=10).

Programs

  • Magma
    m:=20; c:=2; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( Exp(x +(Exp(c*x)-1)/c) )); [Factorial(n-1)*b[n]: n in [1..m-1]]; // G. C. Greubel, Feb 24 2019
  • Mathematica
    max = 19; f[x_]:= Exp[x + Exp[2x]/2 -1/2]; CoefficientList[Series[f[x], {x,0,max}], x]*Range[0, max]! (* Jean-François Alcover, Nov 22 2011 *)
    Table[Sum[Binomial[n, k] * 2^k * BellB[k, 1/2], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 17 2020 *)
  • PARI
    x='x+O('x^66); Vec(serlaplace(exp(x+1/2*exp(2*x)-1/2))) \\ Joerg Arndt, May 13 2013
    
  • Sage
    @CachedFunction
    def S(n, k, m):
        if k > n or k < 0 : return 0
        if n == 0 and k == 0: return 1
        return S(n-1, k-1, m) + (m*(k+1)-1)*S(n-1, k, m)
    def A007405(n): return add(S(n, k, 2) for k in (0..n)) # Peter Luschny, May 20 2013
    
  • Sage
    b=2;
    def A007405_list(prec):
        P. = PowerSeriesRing(QQ, prec)
        return P( exp(x +(exp(b*x)-1)/b) ).egf_to_ogf().list()
    A007405_list(30) # G. C. Greubel, Feb 24 2019
    

Formula

E.g.f.: exp(x + (exp(2*x) - 1)/2).
Row sums of triangles A039755, A039756. - Philippe Deléham, Feb 20 2005
a(n) = sum of top row terms of M^n, M = an infinite square production matrix in which a diagonal of 1's is appended to the right of Pascal's triangle squared; as follows:
1, 1, 0, 0, 0, 0, ...
2, 1, 1, 0, 0, 0, ...
4, 4, 1, 1, 0, 0, ...
8, 12, 6, 1, 1, 0, ...
16, 32, 24, 8, 1, 1, ...
... - Gary W. Adamson, Aug 01 2011
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-(2*k+1)*x)/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 16 2013
G.f.: -G(0) where G(k) = 1 - (x*(2*k+1) - 2)/(x*(2*k+1) - 1 - x*(x*(2*k+1) - 1)/(x + (x*(2*k+1) - 2)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 29 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*(k+1)*x - 2*(k+1)*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: 1/Q(0), where Q(k) = 1 - x - x/(1 - x*(2*k+2)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 13 2013
G.f.: 1/(1-x*Q(0)), where Q(k) = 1 + x/(1 - x + 2*x*(k+1)/(x - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 19 2013
Conjecture: Let M_n be an n X n matrix whose elements are m_ij = 1 for i < j - 1, m_ij = -1 for i = j - 1, and m_ij = binomial(n - i,j - i) otherwise. Then a(n - 1) = Det(M_n). - Benedict W. J. Irwin, Apr 19 2017
a(n) = exp(-1/2) * Sum_{k>=0} (2*k + 1)^n / (2^k * k!). - Ilya Gutkovskiy, Apr 16 2020
a(n) = Sum_{k=0..n} binomial(n,k) * A187251(k) * A187251(n-k). - Vaclav Kotesovec, Apr 17 2020
a(n) ~ 2^(n + 1/2) * n^(n + 1/2) * exp(n/LambertW(2*n) - n - 1/2) / (sqrt(1 + LambertW(2*n)) * LambertW(2*n)^(n + 1/2)). - Vaclav Kotesovec, Jun 26 2022

Extensions

Name edited by G. C. Greubel, Feb 24 2019

A004213 Shifts one place left under 4th-order binomial transform.

Original entry on oeis.org

1, 1, 5, 29, 201, 1657, 15821, 170389, 2032785, 26546673, 376085653, 5736591885, 93614616409, 1625661357673, 29905322979421, 580513190237573, 11850869542405409, 253669139947767777, 5678266212792053029, 132607996474971041789, 3224106929536557918697
Offset: 0

Views

Author

Keywords

Comments

Length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(k)<=F(k)+4 where F(0)=0 and F(k+1)=s(k+1) if s(k+1)-s(k)=4, otherwise F(k+1)=F(k); see example and Fxtbook link. - Joerg Arndt, Apr 30 2011

Examples

			Restricted growth strings: a(0)=1 corresponds to the empty string, a(1)=1 to [0],
a(2)=3 to [00], [01], [02], [03], and [04], a(3) = 29 to
       RGS          F
.1:  [ 0 0 0 ]    [ 0 0 0 ]
.2:  [ 0 0 1 ]    [ 0 0 0 ]
.3:  [ 0 0 2 ]    [ 0 0 0 ]
.4:  [ 0 0 3 ]    [ 0 0 0 ]
.5:  [ 0 0 4 ]    [ 0 0 4 ]
.6:  [ 0 1 0 ]    [ 0 0 0 ]
.7:  [ 0 1 1 ]    [ 0 0 0 ]
.8:  [ 0 1 2 ]    [ 0 0 0 ]
.9:  [ 0 1 3 ]    [ 0 0 0 ]
10:  [ 0 1 4 ]    [ 0 0 4 ]
11:  [ 0 2 0 ]    [ 0 0 0 ]
12:  [ 0 2 1 ]    [ 0 0 0 ]
13:  [ 0 2 2 ]    [ 0 0 0 ]
14:  [ 0 2 3 ]    [ 0 0 0 ]
15:  [ 0 2 4 ]    [ 0 0 4 ]
16:  [ 0 3 0 ]    [ 0 0 0 ]
17:  [ 0 3 1 ]    [ 0 0 0 ]
18:  [ 0 3 2 ]    [ 0 0 0 ]
19:  [ 0 3 3 ]    [ 0 0 0 ]
20:  [ 0 3 4 ]    [ 0 0 4 ]
21:  [ 0 4 0 ]    [ 0 4 4 ]
22:  [ 0 4 1 ]    [ 0 4 4 ]
23:  [ 0 4 2 ]    [ 0 4 4 ]
24:  [ 0 4 3 ]    [ 0 4 4 ]
25:  [ 0 4 4 ]    [ 0 4 4 ]
26:  [ 0 4 5 ]    [ 0 4 4 ]
27:  [ 0 4 6 ]    [ 0 4 4 ]
28:  [ 0 4 7 ]    [ 0 4 4 ]
29:  [ 0 4 8 ]    [ 0 4 8 ]
[_Joerg Arndt_, Apr 30 2011]
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A075499 (row sums).
A004211 (RGS where s(k)<=F(k)+2), A004212 (s(k)<=F(k)+3), A005011 (s(k)<=F(k)+5), A000110 (s(k)<=F(k)+1). - Joerg Arndt, Apr 30 2011

Programs

  • Maple
    A004213 := proc(n)
        add(4^(n-m)*combinat[stirling2](n,m),m=0..n) ;
    end proc:
    seq(A004213(n),n=0..30) ; # R. J. Mathar, Aug 20 2022
  • Mathematica
    Table[4^n BellB[n, 1/4], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(4^(n-k)*binomial(n-1, k-1)*a(k-1), k, 1, n); /* Vladimir Kruchinin, Nov 28 2011 */
  • PARI
    x='x+O('x^66);
    egf=exp(intformal(exp(4*x))); /* =  1 + x + 5/2*x^2 + 29/6*x^3 + 67/8*x^4 + ... */
    /* egf=exp(1/4*(exp(4*x)-1)) */ /* alternative computation */
    Vec(serlaplace(egf)) /* Joerg Arndt, Apr 30 2011 */
    

Formula

a(n) = Sum_{m=0..n} 4^(n-m)*Stirling2(n, m).
E.g.f.: exp((exp(4*x)-1)/4).
O.g.f. A(x) satisfies A'(x)/A(x) = e^(4x).
E.g.f.: exp(Integral_{t = 0..x} exp(4*t)). - Joerg Arndt, Apr 30 2011
O.g.f.: Sum_{k>=0} x^k/Product_{j=1..k} (1-4*j*x). - Joerg Arndt, Apr 30 2011
Define f_1(x), f_2(x), ... such that f_1(x) = e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n = 2, 3, .... Then a(n) = e^(-1/4)*4^{n-1}*f_n(1/4). - Milan Janjic, May 30 2008
a(n) = upper left term in M^n, M = an infinite square production matrix in which a diagonal of (4,4,4,...) is appended to the right of Pascal's triangle:
1, 4, 0, 0, 0, ...
1, 1, 4, 0, 0, ...
1, 2, 1, 4, 0, ...
1, 3, 3, 1, 4, ...
... - Gary W. Adamson, Jul 29 2011
G.f. satisfies A(x) = 1 + x/(1 - 4*x)*A(x/(1 - 4*x)). a(n) = Sum_{k = 1..n} 4^(n-k)*binomial(n-1,k-1)*a(k-1), n > 0, a(0) = 1. - Vladimir Kruchinin, Nov 28 2011 [corrected by Ilya Gutkovskiy, May 02 2019]
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-4*k*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 24 2013
G.f.: (G(0) - 1)/(1+x) where G(k) = 1 + 1/(1-4*k*x)/(1-x/(x+1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 31 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 4*x^2*(k+1)/( 4*x^2*(k+1) - (1-x-4*x*k)*(1-5*x-4*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 19 2013
a(n) = exp(-1/4) * Sum_{k>=0} 4^(n-k) * k^n / k!. - Vaclav Kotesovec, Jul 15 2021
a(n) ~ 4^n * n^n * exp(n/LambertW(4*n) - 1/4 - n) / (sqrt(1 + LambertW(4*n)) * LambertW(4*n)^n). - Vaclav Kotesovec, Jul 15 2021
From Peter Bala, Jun 29 2024: (Start)
a(n) = exp(-1/4)*Sum_{n >= 0} (4*n)^k/(n!*4^n).
Touchard's congruence holds: for odd prime p, a(p+k) == (a(k) + a(k+1)) (mod p) for k = 0,1,2,.... In particular, a(p) == 2 (mod p) for odd prime p. See A004211. (End)

A005011 Shifts one place left under 5th-order binomial transform.

Original entry on oeis.org

1, 1, 6, 41, 331, 3176, 35451, 447981, 6282416, 96546231, 1611270851, 28985293526, 558413253581, 11458179765541, 249255304141006, 5725640423174901, 138407987170952351, 3510263847256823056
Offset: 0

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Keywords

Comments

Length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(k)<=F(k)+5 where F(0)=0 and F(k+1)=s(k+1) if s(k+1)-s(k)=5, otherwise F(k+1)=F(k); see examples in A004211, A004212, and A004213, and Fxtbook link. [Joerg Arndt, Apr 30 2011]

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A075500 (row sums).
A004211 (RGS where s(k)<=F(k)+2), A004212 (s(k)<=F(k)+3), A004213 (s(k)<=F(k)+4), A000110 (s(k)<=F(k)+1). - Joerg Arndt, Apr 30 2011

Programs

  • Mathematica
    Table[5^n BellB[n, 1/5], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)
  • PARI
    x='x+O('x^66);
    egf=exp(intformal(exp(5*x))); /* =  1 + x + 3*x^2 + 41/6*x^3 + 331/24*x^4 + ... */
    /* egf=exp(1/5*(exp(5*x)-1)) */ /* alternative computation */
    Vec(serlaplace(egf)) /* Joerg Arndt, Apr 30 2011 */

Formula

a(n) = Sum_{m=0..n} 5^(n-m)*Stirling2(n, m), n >= 0.
E.g.f.: exp((exp(5*x)-1)/5).
O.g.f. A(x) satisfies A'(x)/A(x) = exp(5*x).
E.g.f.: exp(Integral_{t = 0..x} exp(5*t)). - Joerg Arndt, Apr 30 2011
O.g.f.: Sum_{k>=0} x^k/Product_{j=1..k} (1-5*j*x). - Joerg Arndt, Apr 30 2011
Define f_1(x), f_2(x), ... such that f_1(x)=e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n=2,3,.... Then a(n) = e^(-1/5)*5^{n-1}*f_n(1/5). - Milan Janjic, May 30 2008
a(n) = upper left term in M^n, M = an infinite square production matrix in which a diagonal of (5,5,5,...) is appended to the right of Pascal's triangle:
1, 5, 0, 0, 0, ...
1, 1, 5, 0, 0, ...
1, 2, 1, 5, 0, ...
1, 3, 3, 1, 5, ...
... - Gary W. Adamson, Jul 29 2011
G.f.: T(0)/(1-x), where T(k) = 1 - 5*x^2*(k+1)/( 5*x^2*(k+1) - (1-x-5*x*k)*(1-6*x-5*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 25 2013
G.f. A(x) satisfies: A(x) = 1 + x*A(x/(1 - 5*x))/(1 - 5*x). - Ilya Gutkovskiy, May 03 2019
a(n) ~ 5^n * n^n * exp(n/LambertW(5*n) - 1/5 - n) / (sqrt(1 + LambertW(5*n)) * LambertW(5*n)^n). - Vaclav Kotesovec, Jul 15 2021
From Peter Bala, Jun 29 2024: (Start)
a(n) = exp(-1/5)*Sum_{n >= 0} (5*n)^k/(n!*5^n).
Touchard's congruence holds: for all primes p != 5, a(p+k) == (a(k) + a(k+1)) (mod p) for k = 0,1,2,.... In particular, a(p) == 2 (mod p) for all primes p != 5. See A004211. (End)

A005012 Shifts one place left under 6th-order binomial transform.

Original entry on oeis.org

1, 1, 7, 55, 505, 5497, 69823, 1007407, 16157905, 284214097, 5432922775, 112034017735, 2476196276617, 58332035387017, 1457666574447247, 38485034941511935, 1069787864231083297, 31213730550761076769, 953352927192964243879, 30406448846308128741847
Offset: 0

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Author

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A004211.

Programs

  • GAP
    List([0..20],n->Sum([0..n],m->6^(n-m)*Stirling2(n,m))); # Muniru A Asiru, Mar 20 2018
  • Maple
    seq(6^n*BellB(n,1/6), n = 0 .. 50); # Robert Israel, Oct 20 2015
  • Mathematica
    Table[6^n BellB[n, 1/6], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)

Formula

a(n) = sum((6^(n-m))*stirling2(n,m), m=0..n). stirling2(n,m)=A008277(n,m).
E.g.f.: exp((exp(6*x)-1)/6) satisfies A'(x)/A(x) = exp(6*x).
G.f.: T(0)/(x*(1-x)) -1/x, where T(k) = 1 - 6*x^2*(k+1)/( 6*x^2*(k+1) - (1-x-6*x*k)*(1-7*x-6*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 25 2013
a(n) = 6^n * B(n,1/6) where B(n,x) is the Bell polynomial of degree n. - Vladimir Reshetnikov, Oct 20 2015
O.g.f.: Sum_{k>=0} x^k/Product_{j=1..k} (1 - 6*j*x). - Ilya Gutkovskiy, Mar 20 2018
a(n) ~ 6^n * n^n * exp(n/LambertW(6*n) - 1/6 - n) / (sqrt(1 + LambertW(6*n)) * LambertW(6*n)^n). - Vaclav Kotesovec, Jul 15 2021

Extensions

a(0)=1 inserted by Alois P. Heinz, Oct 20 2015
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