Displaying 1-10 of 19 results found.
a(n) = (n!)^2.
(Formerly M3666 N1492)
+10
123
1, 1, 4, 36, 576, 14400, 518400, 25401600, 1625702400, 131681894400, 13168189440000, 1593350922240000, 229442532802560000, 38775788043632640000, 7600054456551997440000, 1710012252724199424000000, 437763136697395052544000000, 126513546505547170185216000000
COMMENTS
Let M_n be the symmetrical n X n matrix M_n(i,j) = 1/Max(i,j); then for n > 0 det(M_n)=1/a(n). - Benoit Cloitre, Apr 27 2002 The n-th entry of the sequence is the value of the permanent of a k X k matrix A defined as follows: k is the n-th odd number; if we concatenate the rows of A to form a vector v of length n^2, v_{i}=1 if i=1 or a multiple of 2. - Simone Severini, Feb 15 2006 a(n) = number of set partitions of {1,2,...,3n-1,3n} into blocks of size 3 in which the entries of each block mod 3 are distinct. For example, a(2) = 4 counts 123-456, 156-234, 126-345, 135-246. - David Callan, Mar 30 2007 Number of permutations of {1,2,...,2n} with no even entry followed by a smaller entry. Example: a(2)=4 because we have 1234, 1324, 3124 and 2314.
Number of permutations of {1,2,...,2n} with n even entries that are followed by a smaller entry. Example: a(2)=4 because we have 2143, 3421, 4213 and 4321.
Number of permutations of {1,2,...,2n-1} with no even entry followed by a smaller entry. Example: a(2)=4 because we have 123, 132, 312 and 231.
Number of permutations of {1,2,...,2n-1} with n-1 odd entries followed by a smaller entry. Example: a(2)=4 because we have 132, 312, 231 and 321.
(End)
G. Leibniz in his "Ars Combinatoria" established the identity P(n)^2 = P(n-1)[P(n+1)-P(n)], where P(n) = n!. (For example, see the Burton reference.) - Mohammad K. Azarian, Mar 28 2008 a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = sigma_2(gcd(i,j)) for 1 <= i,j <= n, and n>0, where sigma_2 is A001157. - Enrique Pérez Herrero, Aug 13 2011 The o.g.f. of 1/a(n) is BesselI(0,2*sqrt(x)). See Abramowitz-Stegun (reference and link under A008277), p. 375, 9.6.10. - Wolfdieter Lang, Jan 09 2012 Number of n x n x n cubes C of zeros and ones such that C(x,y,z) and C(u,v,w) can be nonzero simultaneously only if either x!=u, y!=v, or z!=w. This generalizes permutations which can be considered as n x n squares P of zeros and ones such that P(x,y) and P(u,v) can be nonzero simultaneously only if either x!=u or y!=v. - Joerg Arndt, May 28 2012 a(n) is the number of functions f:[n]->[n(n+1)/2] such that, if round(sqrt(2f(x))) = round(sqrt(2f(y))), then x=y. - Dennis P. Walsh, Nov 26 2012 a(n) is the number of n X n 0-1 matrices whose row sums and column sums are both {1,2,...,n}.
a(n) is the number of linear arrangements of 2n blocks of n different colors, 2 of each color, such that there are an even number of blocks between each pair of blocks of the same color.
(End)
Number of ways to place n instances of a digit inside an n X n X n cube so that no two instances lie on a plane parallel to a face of the cube (see Khovanova link, Lemma 6, p. 22). - Tanya Khovanova and Wayne Zhao, Oct 17 2018 Number of permutations P of length 2n which maximize Sum_{i=1..2n} |P_i - i|. - Fang Lixing, Dec 07 2018
REFERENCES
Archimedeans Problems Drive, Eureka, 22 (1959), 15.
David Burton, "The History of Mathematics", Sixth Edition, Problem 2, p. 433.
J. Dezert, editor, Smarandacheials, Mathematics Magazine, Aurora, Canada, No. 4/2004 (to appear).
S. M. Kerawala, The enumeration of the Latin rectangle of depth three by means of a difference equation, Bull. Calcutta Math. Soc., 33 (1941), 119-127.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
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).
F. Smarandache, Back and Forth Factorials, Arizona State Univ., Special Collections, 1972.
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.62(b).
LINKS
Rob Pratt (Proposer), Problem 11573, Amer. Math. Monthly, 120 (2013), 372. Simone Severini, Title? [dead link]
FORMULA
a(n) = Integral_{x>=0} 2*BesselK(0, 2*sqrt(x))*x^n. This integral represents the n-th moment of a positive function defined on the positive half-axis. - Karol A. Penson, Oct 09 2001 a(n) ~ 2*Pi*n*e^(-2*n)*n^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 07 2002
a(n) = polygorial(n, 4) = A000142(n)/ A000079(n)* A000165(n) = (n!/2^n)*Product_{i=0..n-1} (2*i + 2) = n!*Pochhammer(1, n) = n!^2. - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003 a(n) = !n!_1 = !n! = Product_{i=0, 1, 2, ... .}_{0 < |n-i| <= n}(n-i) = n(n-1)(n-2)...(2)(1)(-1)(-2)...(-n+2)(-n+1)(-n) = [(-1)^n][(n!)^2]. - J. Dezert (Jean.Dezert(AT)onera.fr), Mar 21 2004
A(x) = Sum_{n>=0,N) a(n)*x^n = 1 + x/(U(0;N-2)-x); N >= 4; U(k)= 1 + x*(k+1)^2 - x*(k+2)^2/G(k+1); besides U(0;infinity)=x; (continued fraction).
Let B(x) = Sum_{n>=0} a(n)*x^n/((n!)*(n+s)!), then B(0) = 1/(1-x) for abs(x) < 1 and B(1)= -1/x * log(1-x) for abs(x)< 1.
(End).
G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - (k+1)^2*(1 - x*G(k+1)). - Sergei N. Gladkovskii, Jan 15 2013 a(n) = det(S(i+2,j), 1 <= i,j <= n), where S(n,k) are Stirling numbers of the second kind. - Mircea Merca, Apr 04 2013 a(n) = (2*n+1)!*2^(-4*n)*Sum_{k=0..n} (-1)^k*C(2*n+1,n-k)/(2*k+1). - Mircea Merca, Nov 12 2013 a(n) = a(n-1) + 2*((n-1)^2)*sqrt(a(n-1)*a(n-2)) + ((n-1)^4)*a(n-2), for n > 1.
a(n) = a(n-1) - 2*(n^2 - 1)*sqrt(a(n-1)*a(n-2)) + (n^2 - 1)*a(n-2), for n > 1.
(End).
Sum_{n>=0} (-1)^n/a(n) = BesselJ(0,2) = A091681. (End) Sum_{n>=0} a(n)/(2*n+1)! = 2*Pi/sqrt(27). - Daniel Suteu, Feb 06 2017 a(n) = (2*n+1)! * [x^(2*n+1)] 4*arcsin(x/2)/sqrt(4-x^2). - Ira M. Gessel, Dec 10 2024
EXAMPLE
Consider the square array
1, 2, 3, 4, 5, 6, ...
2, 4, 6, 8, 10, 12, ...
3, 6, 9, 12, 15, 18, ...
4, 8, 12, 16, 20, 24, ...
5, 10, 15, 20, 25, 30, ...
...
a(3) = 36 since there are 36 functions f:[3]->[6] such that, if round(sqrt(2f(x))) = round(sqrt(2f(y))), then x=y. The functions, denoted by <f(1),f(2),f(3)>, are <1,2,4>, <1,2,5>, <1,2,6>, <1,3,4>, <1,3,5>, <1,3,6> and their respective permutations. - Dennis P. Walsh, Nov 26 2012 1 + x + 4*x^2 + 36*x^3 + 576*x^4 + 14400*x^5 + 518400*x^6 + ...
MATHEMATICA
Join[{1}, Table[Det[DiagonalMatrix[Range[n]^2]], {n, 20}]] (* Harvey P. Dale, Mar 31 2020 *)
PROG
(Haskell)
import Data.List (genericIndex)
a001044 n = genericIndex a001044_list n
a001044_list = 1 : zipWith (*) (tail a000290_list) a001044_list
(Python) import math
for n in range(0, 20): print(math.factorial(n)**2, end=', ') # Stefano Spezia, Oct 29 2018
CROSSREFS
Cf. A000142, A000292, A084939, A084940, A084941, A084942, A084943, A084944, A020549, A046032, A048617. First right-hand column of triangle A008955.
a(n) = n*(n-1)*(n-2) (or n!/(n-3)!).
(Formerly M4159)
+10
101
0, 0, 0, 6, 24, 60, 120, 210, 336, 504, 720, 990, 1320, 1716, 2184, 2730, 3360, 4080, 4896, 5814, 6840, 7980, 9240, 10626, 12144, 13800, 15600, 17550, 19656, 21924, 24360, 26970, 29760, 32736, 35904, 39270, 42840, 46620, 50616, 54834, 59280, 63960, 68880
COMMENTS
Ed Pegg Jr conjectures that n^3 - n = k! has a solution if and only if n is 2, 3, 5 or 9 (when k is 3, 4, 5 and 6).
If Y is a 4-subset of an n-set X then, for n >= 6, a(n-4) is the number of (n-5)-subsets of X having exactly two elements in common with Y. - Milan Janjic, Dec 28 2007 Let H be the n X n Hilbert matrix H(i, j) = 1/(i+j-1) for 1 <= i, j <= n. Let B be the inverse matrix of H. The sum of the elements in row 2 of B equals (-1)^n a(n+1). - T. D. Noe, May 01 2011 a(n) equals 2^(n-1) times the coefficient of log(3) in 2F1(n-2, n-2, n, -2). - John M. Campbell, Jul 16 2011 For n > 2 a(n) = 1/(Integral_{x = 0..Pi/2} (sin(x))^5*(cos(x))^(2*n-5)). - Francesco Daddi, Aug 02 2011 a(n) is the number of functions f:[3] -> [n] that are injective since there are n choices for f(1), (n-1) choices for f(2), and (n-2) choices for f(3). Also, a(n+1) is the number of functions f:[3] -> [n] that are width-2 restricted (that is, the pre-image under f of any element in [n] is of size 2 or less). See "Width-restricted finite functions" link below. - Dennis P. Walsh, Mar 01 2012 This sequence is produced by three consecutive triangular numbers t(n-1), t(n-2) and t(n-3) in the expression 2*t(n-1)*(t(n-2)-t(n-3)) for n = 0, 1, 2, ... - J. M. Bergot, May 14, 2012 Number of contact points between equal spheres arranged in a tetrahedron with n - 1 spheres in each edge. - Ignacio Larrosa Cañestro, Jan 07 2013 Also for n >= 3, area of Pythagorean triangle in which one side differs from hypotenuse by two units. Consider any Pythagorean triple (2n, n^2-1, n^2+1) where n > 1. The area of such a Pythagorean triangle is n(n^2-1). For n = 2, 3, 4,.. the areas are 6, 24, 60, .... which are the given terms of the series. - Jayanta Basu, Apr 11 2013 Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices (chromatic polynomial) of the complete graph K_3. - Tom Copeland, Apr 05 2014 Starting with 6, 24, 60, 120, ..., a(n) is the number of permutations of length n>=3 avoiding the partially ordered pattern (POP) {1>2} of length 5. That is, the number of length n permutations having no subsequences of length 5 in which the first element is larger than the second element. - Sergey Kitaev, Dec 11 2020 For integer m and positive integer r >= 2, the polynomial a(n) + a(n + m) + a(n + 2*m) + ... + a(n + r*m) in n has its zeros on the vertical line Re(n) = (2 - r*m)/2 in the complex plane. - Peter Bala, Jun 02 2024
REFERENCES
R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.
L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 40.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(n) = Sum_{i=1..n} polygorial(3,i) where polygorial(3,i) = A028896(i-1). - Daniel Dockery (peritus(AT)gmail.com), Jun 16 2003 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6, n > 2. - Zak Seidov, Feb 09 2006 G.f.: 6*x^2/(1-x)^4.
a(-n) = -a(n+2).
1/6 + 3/24 + 5/60 + ... = Sum_{k>=1} (2*k-1)/(k*(k+1)*(k+2)) = 3/4. [Jolley Eq. 213]
If the first 0 is eliminated, a(n) = floor(n^5/(n^2+1)). - Gary Detlefs, Feb 11 2010 1/6 + 1/24 + 1/60 + ... = Sum_{n>=1} 1/(n*(n+1)*(n+2)) = 1/4. - Mohammad K. Azarian, Dec 29 2010 1/6 + 1/24 + 1/60 + ... + 1/(n*(n+1)*(n+2)) = n*(n+3)/(4*(n+1)*(n+2)). - Christina Steffan, Jul 20 2015 Sum_{n>=3} (-1)^(n+1)/a(n) = 2*log(2) - 5/4. - Amiram Eldar, Jul 02 2020 For n >= 3, (a(n) + (a(n) + (a(n) + ...)^(1/3))^(1/3))^(1/3) = n - 1. - Paolo Xausa, Apr 09 2022
MATHEMATICA
Table[n^3 - 3n^2 + 2n, {n, 0, 42}]
PROG
(PARI) a(n)=n*(n-1)*(n-2)
(Haskell)
(Sage) [n*(n-1)*(n-2) for n in range(40)] # G. C. Greubel, Feb 11 2019
a(n) = (2n)!/2^n.
(Formerly M4287 N1793)
+10
81
1, 1, 6, 90, 2520, 113400, 7484400, 681080400, 81729648000, 12504636144000, 2375880867360000, 548828480360160000, 151476660579404160000, 49229914688306352000000, 18608907752179801056000000, 8094874872198213459360000000, 4015057936610313875842560000000
COMMENTS
Denominators in the expansion of cos(sqrt(2)*x) = 1 - (sqrt(2)*x)^2/2! + (sqrt(2)*x)^4/4! - (sqrt(2)*x)^6/6! + ... = 1 - x^2 + x^4/6 - x^6/90 + ... By Stirling's formula in A000142: a(n) ~ 2^(n+1) * (n/e)^(2n) * sqrt(Pi*n) - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 20 2001 a(n) is also the constant term in the product: Product_{1<=i, j<=n, i!=j} (1 - x_i/x_j)^2. - Sharon Sela (sharonsela(AT)hotmail.com), Feb 12 2002
a(n) is also the number of lattice paths in the n-dimensional lattice [0..2]^n. - T. D. Noe, Jun 06 2002 Representation as the n-th moment of a positive function on the positive half-axis: a(n) = Integral_{x>=0} (x^n*exp(-sqrt(2*x))/sqrt(2*x)), n=0,1,... - Karol A. Penson, Mar 10 2003 Number of permutations of [2n] with no increasing runs of odd length. Example: a(2) = 6 because we have 1234, 13/24, 14/23, 23/14, 24/13 and 34/12 (runs separated by slashes). - Emeric Deutsch, Aug 29 2004 This is also the number of ways of arranging the elements of n distinct pairs, assuming the order of elements is significant and the pairs are distinguishable. When the pairs are not distinguishable, see A001147 and A132101. For example, there are 6 ways of arranging 2 pairs [1,1], [2,2]: {[1122], [1212], [1221], [2211], [2121], [2112]}. - Ross Drewe, Mar 16 2008 n married couples are seated in a row so that every wife is to the left of her husband. The recurrence a(n+1) = a(n)*((2*n + 1) + binomial(2*n+1, 2)) conditions on whether the (n+1)st couple is seated together or separated by at least one other person. - Geoffrey Critzer, Jun 10 2009 a(n) is the number of functions f:[2n]->[n] such that the preimage of {y} has cardinality 2 for every y in [n]. Note that [k] denotes the set {1,2,...,k} and [0] denotes the empty set. - Dennis P. Walsh, Nov 17 2009 a(n) is also the number of n X 2n (0,1)-matrices with row sum 2 and column sum 1. - Shanzhen Gao, Feb 12 2010 Number of ways that 2n people of different heights can be arranged (for a photograph) in two rows of equal length so that every person in the front row is shorter than the person immediately behind them in the back row.
a(n) is the number of functions f:[n]->[n^2] such that, if floor((f(x))^.5) = floor((f(y))^.5), then x = y. For example, with n = 4, the range of f consists of one element from each of the four sets {1,2,3}, {4,5,6,7,8}, {9,10,11,12,13,14,15}, and {16}. Hence there are 1*3*5*7 = 105 ways to choose the range for f, and there are 4! ways to injectively map {1,2,3,4} to the four elements of the range. Thus there are 105*24 = 2520 such functions. Note also that a(n) = n!*(product of the first n odd numbers). - Dennis P. Walsh, Nov 28 2012 a(n) is also the 2*n th difference of n-powers of A000217 (triangular numbers). For example a(2) is the 4th difference of the squares of triangular numbers. - Enric Reverter i Bigas, Jun 24 2013 a(n) is the multinomial coefficient (2*n) over (2, 2, 2, ..., 2) where there are n 2's in the last parenthesis. It is therefore also the number of words of length 2n obtained with n letters, each letter appearing twice. - Robert FERREOL, Jan 14 2018 Number of ways to put socks and shoes on an n-legged animal, if a sock must be put on before a shoe. - Daniel Bishop, Jan 29 2018
REFERENCES
G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, 1998.
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. 283.
A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 112.
Shanzhen Gao and Kenneth Matheis, Closed formulas and integer sequences arising from the enumeration of (0,1)-matrices with row sum two and some constant column sums. In Proceedings of the Forty-First Southeastern International Conference on Combinatorics, Graph Theory and Computing. Congr. Numer. 202 (2010), 45-53.
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).
C. B. Tompkins, Methods of successive restrictions in computational problems involving discrete variables. 1963, Proc. Sympos. Appl. Math., Vol. XV pp. 95-106; Amer. Math. Soc., Providence, R.I.
FORMULA
E.g.f.: 1/(1 - x^2/2) (with interpolating zeros). - Paul Barry, May 26 2003 a(n) = polygorial(n, 6) = ( A000142(n)/ A000079(n))* A001813(n) = (n!/2^n)*Product_{i=0..n-1} (4*i + 2) = (n!/2^n)*4^n*Pochhammer(1/2, n) = gamma(2*n+1)/2^n. - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003 a(n) = A087127(n,2*n) = Sum_{i=0..2*n} (-1)^(2*n-i)*binomial(2*n, i)*binomial(i+2, 2)^n. Let T(n,k,j) = ((n - k + j)*(2*n - 2*k + 1))^n*binomial(2*n, 2*k-j+1) then a(n) = Sum{k=0..n} (T(n,k,1) - T(n,k,0)). For example a(12) = A087127(12,24) = Sum_{k=0..12} (T(12,k,1) - T(12,k,0)) = 24!/2^12. - André F. Labossière, Mar 29 2004 [Corrected by Jianing Song, Jan 08 2019] For even n, a(n) = binomial(2n, n)*(a(n/2))^2. For odd n, a(n) = binomial(2n, n+1)*a((n+1)/2)*a((n-1)/2). For positive n, a(n) = binomial(2n, 2)*a(n-1) with a(0) = 1. - Dennis P. Walsh, Nov 17 2009 a(n) = Product_{i=1..n} binomial(2i, 2).
a(n) = a(n-1)*binomial(2n, 2).
a(n) = Product_{k = 0..n-1} (T(n) - T(k)), where T(n) = n*(n + 1)/2 is the n-th triangular number.
Compare with n! = Product_{k = 0..n-1} (n - k).
Thus we may view a(n) as a generalized factorial function associated with the triangular numbers A000217. Cf. A010050. The corresponding generalized binomial coefficients a(n)/(a(k)*a(n-k)) are triangle A086645. Also cf. A186432. a(n) = n*(n + n-1)*(n + n-1 + n-2)*...*(n + n-1 + n-2 + ... + 1).
For example, a(5) = 5*(5+4)*(5+4+3)*(5+4+3+2)*(5+4+3+2+1) = 113400. (End).
G.f.: 1/U(0) where U(k)= x*(2*k - 1)*k + 1 - x*(2*k + 1)*(k + 1)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Oct 28 2012 a(n) = n!*(product of the first n odd integers). - Dennis P. Walsh, Nov 28 2012 a(0) = 1, a(n) = a(n-1)*T(2*n-1), where T(n) is the n-th triangular number. For example: a(4) = a(3)*T(7) = 90*28 = 2520. - Enric Reverter i Bigas, Jun 24 2013 E.g.f.: 1/(1 - x/(1 - 2*x/(1 - 3*x/(1 - 4*x/(1 - 5*x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, May 10 2017 Sum_{n>=0} 1/a(n) = cosh(sqrt(2)).
Sum_{n>=0} (-1)^n/a(n) = cos(sqrt(2)). (End)
D-finite with recurrence a(n) -n*(2*n-1)*a(n-1)=0. - R. J. Mathar, Jan 28 2022
EXAMPLE
For n = 2, a(2) = 6 since there are 6 functions f:[4]->[2] with size 2 preimages for both {1} and {2}. In this case, there are binomial(4, 2) = 6 ways to choose the 2 elements of [4] f maps to {1} and the 2 elements of [4] that f maps to {2}. - Dennis P. Walsh, Nov 17 2009
MAPLE
a[0]:=1:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]*(2*n-1)*n od: seq(a[n], n=0..16); # Zerinvary Lajos, Mar 08 2008 seq(product(binomial(2*n-2*k, 2), k=0..n-1), n=0..16); # Dennis P. Walsh, Nov 17 2009
MATHEMATICA
Table[Product[Binomial[2 i, 2], {i, 1, n}], {n, 0, 16}]
polygorial[k_, n_] := FullSimplify[ n!/2^n (k -2)^n*Pochhammer[2/(k -2), n]]; Array[ polygorial[6, #] &, 17, 0] (* Robert G. Wilson v, Dec 26 2016 *)
PROG
(PARI) a(n) = (2*n)! / 2^n
CROSSREFS
A diagonal of the triangle in A241171. Even bisection of column k=0 of A097591.
a(n) = n!*(n-1)!/2^(n-1).
(Formerly M3052)
+10
56
1, 1, 3, 18, 180, 2700, 56700, 1587600, 57153600, 2571912000, 141455160000, 9336040560000, 728211163680000, 66267215894880000, 6958057668962400000, 834966920275488000000, 113555501157466368000000, 17373991677092354304000000, 2970952576782792585984000000
COMMENTS
Number of ways of transforming n distinguishable objects into n singletons via a sequence of n-1 refinements. Example: a(3)=3 because we have XYZ->X|YZ->X|Y|Z, XYZ->Y|XZ->X|Y|Z and XYZ->Z|XY->X|Y|Z. - Emeric Deutsch, Jan 23 2005 In other words, a(n) is the number of maximal chains in the lattice of set partitions of {1, ..., n} ordered by refinement. - Gus Wiseman, Jul 22 2018 With offset 0, a(n) = number of unordered increasing full binary trees of 2n edges with leaf set {n,n+1,...,2n}, where full binary means each nonleaf vertex has two children, increasing means the vertices are labeled 0,1,2,...,2n and each child is greater than its parent, unordered might as well mean ordered and each pair of sibling vertices is increasing left to right. For example, a(2)=3 counts the trees with edge lists {01,02,13,14}, {01,03,12,14}, {01,04,12,13}.
PROOF. Given such a tree of size n, to produce a tree of size n+1, two new leaves must be added to the leaf n. Choose any two of the leaf set {n+1,...,2n,2n+1,2n+2} for the new leaves and use the rest to replace the old leaves n+1,...,2n, maintaining relative order. Thus each tree of size n yields (n+2)-choose-2 trees of the next size up. Since the ratio a(n+1)/a(n)=(n+2)-choose-2, the result follows by induction.
Without the condition on the leaves, these trees are counted by the reduced tangent numbers A002105. (End) a(n) = Sum(M(t)N(t)), where summation is over all rooted trees t with n vertices, M(t) is the number of ways to take apart t by sequentially removing terminal edges (see A206494) and N(t) is the number of ways to build up t from the one-vertex tree by adding successively edges to the existing vertices (the Connes-Moscovici weight; see A206496). See Remark on p. 3801 of the Hoffman reference. Example: a(3) = 3; indeed, there are two rooted trees with 3 vertices: t' = the path r-a-b and t" = V; we have M(t')=N(t')=1 and M(t") =1, N(t")=2, leading to M(t')N(t') + M(t")N(t")=3. - Emeric Deutsch, Jul 20 2012 Number of coalescence sequences or labeled histories for n lineages: the number of sequences by which n distinguishable leaves can coalesce to a single sequence. The coalescence process merges pairs of lineages into new lineages, labeling each newly formed lineage l by a subset of the n initial lineages corresponding to the union of all initial lineages that feed into lineage l. - Noah A Rosenberg, Jan 28 2019 Conjecture: For n > 1, n divides 2*a(n-1) + 4 if and only if n is prime. - Werner Schulte, Oct 04 2020 For a proof of the above conjecture see Himane. The list of primes p such that p^2 divides (2*a(p-1) + 4) (analog of A007540 - Wilson primes) begins [239, 24049, ...]. - Peter Bala, Nov 06 2024
REFERENCES
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 148.
László Lovász, Combinatorial Problems and Exercises, North-Holland, 1979, p. 165.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Mike Steel, Phylogeny: Discrete and Random Processes in Evolution, SIAM, 2016, p. 47.
FORMULA
a(n) = polygorial(n, 3) = ( A000142(n)/ A000079(n))* A000142(n+1) = (n!/2^n)*Product_{i=0..n-1} (i+2) = (n!/2^n)*Pochhammer(2, n) = (n!^2/2^n)*(n+1) = polygorial(n, 4)/2^n*(n+1). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003 a(n-1) = (-1)^(n+1)/(n^2*det(M_n)) where M_n is the matrix M_(i, j) = abs(1/i - 1/j). - Benoit Cloitre, Aug 21 2003 a(n) ~ 4*Pi*n^(2*n)/(2^n*exp(2*n)).
Sum_{n>=1} 1/a(n) = BesselI(1,2*sqrt(2))/sqrt(2) = 2.3948330992734... (End)
D-finite with recurrence 2*a(n) -n*(n-1)*a(n-1)=0. - R. J. Mathar, May 02 2022 Sum_{n>=1} (-1)^(n+1)/a(n) = BesselJ(1,2*sqrt(2))/sqrt(2). - Amiram Eldar, Jun 25 2022
EXAMPLE
The (3) = 3 maximal chains in the lattice of set partitions of {1,2,3}:
{{1},{2},{3}} < {{1},{2,3}} < {{1,2,3}}
{{1},{2},{3}} < {{2},{1,3}} < {{1,2,3}}
{{1},{2},{3}} < {{3},{1,2}} < {{1,2,3}}
(End)
MATHEMATICA
FoldList[Times, 1, Accumulate[Range[20]]] (* Harvey P. Dale, Jan 10 2013 *)
PROG
(Magma) [Factorial(n)*Factorial(n-1)/2^(n-1): n in [1..20]]; // Vincenzo Librandi, Aug 23 2018 (Python)
from math import factorial
6 times triangular numbers: a(n) = 3*n*(n+1).
+10
39
0, 6, 18, 36, 60, 90, 126, 168, 216, 270, 330, 396, 468, 546, 630, 720, 816, 918, 1026, 1140, 1260, 1386, 1518, 1656, 1800, 1950, 2106, 2268, 2436, 2610, 2790, 2976, 3168, 3366, 3570, 3780, 3996, 4218, 4446, 4680, 4920, 5166, 5418, 5676
COMMENTS
Write 1,2,3,4,... in a hexagonal spiral around 0; then a(n) is the sequence found by reading the line from 0 in the direction 0, 6, ...
The spiral begins:
85--84--83--82--81--80
/ \
86 56--55--54--53--52 79
/ / \ \
87 57 33--32--31--30 51 78
/ / / \ \ \
88 58 34 16--15--14 29 50 77
/ / / / \ \ \ \
89 59 35 17 5---4 13 28 49 76
/ / / / / \ \ \ \ \
<==90==60==36==18===6===0 3 12 27 48 75
/ / / / / / / / / /
61 37 19 7 1---2 11 26 47 74
\ \ \ \ / / / /
62 38 20 8---9--10 25 46 73
\ \ \ / / /
63 39 21--22--23--24 45 72
\ \ / /
64 40--41--42--43--44 71
\ /
65--66--67--68--69--70
(End)
If Y is a 4-subset of an n-set X then, for n >= 5, a(n-5) is the number of (n-4)-subsets of X having exactly two elements in common with Y. - Milan Janjic, Dec 28 2007 a(n) is the maximal number of points of intersection of n+1 distinct triangles drawn in the plane. For example, two triangles can intersect in at most a(1) = 6 points (as illustrated in the Star of David configuration). - Terry Stickels (Terrystickels(AT)aol.com), Jul 12 2008
Also sequence found by reading the line from 0, in the direction 0, 6, ... and the same line from 0, in the direction 0, 18, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Axis perpendicular to A195143 in the same spiral. - Omar E. Pol, Sep 18 2011 Also the number of 5-cycles in the (n+5)-triangular honeycomb acute knight graph. - Eric W. Weisstein, Jul 27 2017 a(n-4) is the maximum irregularity over all maximal 3-degenerate graphs with n vertices. The extremal graphs are 3-stars (K_3 joined to n-3 independent vertices). (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, May 29 2023
FORMULA
O.g.f.: 6*x/(1 - x)^3.
a(n) = polygorial(3, n+1). - Daniel Dockery (peritus(AT)gmail.com), Jun 16 2003
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2, a(0)=0, a(1)=6, a(2)=18.
a(n) = T(3*n) - T(2*n-2) + T(n-2), where T(n) = A000217(n). In general, T(k)*T(n) = Sum_{i=0..k-1} (-1)^i*T((k-i)*(n-i)). - Charlie Marion, Dec 04 2020 Sum_{n>=1} 1/a(n) = 1/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/3 - 1/3. (End)
Product_{n>=1} (1 - 1/a(n)) = -(3/Pi)*cos(sqrt(7/3)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (3/Pi)*cosh(Pi/(2*sqrt(3))). (End)
PROG
(PARI) first(n) = Vec(6*x/(1 - x)^3 + O(x^n), -n) \\ Iain Fox, Feb 14 2018
CROSSREFS
Cf. A000217, A000567, A003215, A008588, A024966, A028895, A033996, A046092, A049598, A084939, A084940, A084941, A084942, A084943, A084944, A124080. Cf. A002378 (3-cycles in triangular honeycomb acute knight graph), A045943 (4-cycles), A152773 (6-cycles).
AUTHOR
Joe Keane (jgk(AT)jgk.org), Dec 11 1999
Pentagorials: n-th polygorial for k=5.
+10
21
1, 1, 5, 60, 1320, 46200, 2356200, 164934000, 15173928000, 1775349576000, 257425688520000, 45306921179520000, 9514453447699200000, 2350070001581702400000, 674470090453948588800000, 222575129849803034304000000
FORMULA
a(n) = polygorial(n, 5) = ( A000142(n)/ A000079(n))* A008544(n) = (n!/2^n)*Product_{i=0..n-1} (3*i+2) = (n!/2^n)*3^n*Pochhammer(2/3, n) = (n!/2^n)*3^n*GAMMA(n+2/3)/GAMMA(2/3). a(n) ~ Gamma(1/3) * 3^(n+1/2) * n^(2*n+2/3) / (2^n * exp(2*n)). - Vaclav Kotesovec, Jul 17 2015 D-finite with recurrence a(n+1) = ((n+1)*(3*n+2)/2)*a(n) = A000326(n+1)*a(n). - Muniru A Asiru, Apr 05 2016 E.g.f.: hypergeom([2/3, 1], [], (3/2)*x). - Robert Israel, Apr 05 2016
MAPLE
a := n->(n!/2^n)*mul(3*i+2, i=0..n-1); [seq(a(j), j=0..30)];
MATHEMATICA
Table[k! Pochhammer[2/3, k] (3/2)^k, {k, 0, 20}] (* Jan Mangaldan, Mar 20 2013 *) polygorial[k_, n_] := FullSimplify[ n!/2^n (k -2)^n*Pochhammer[2/(k -2), n]]; Array[polygorial[5, #] &, 17, 0] (* Robert G. Wilson v, Dec 17 2016 *)
AUTHOR
Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
Heptagorials: n-th polygorial for k=7.
+10
20
1, 1, 7, 126, 4284, 235620, 19085220, 2137544640, 316356606720, 59791398670080, 14050978687468800, 4018579904616076800, 1374354327378698265600, 553864793933615401036800, 259762588354865623086259200
FORMULA
a(n) = polygorial(n, 7) = ( A000142(n)/ A000079(n))* A047055(n) = (n!/2^n)*Product_{i=0..n-1}(5*i+2) = (n!/2^n)*5^n*Pochhammer(2/5, n) = (n!/2^n)*5^n*GAMMA(n+2/5)*sin(2*Pi/5)*GAMMA(3/5)/Pi. D-finite with recurrence 2*a(n) = n*(5*n-3)*a(n-1). - R. J. Mathar, Mar 12 2019
MAPLE
a := n->n!/2^n*mul(5*i+2, i=0..n-1); [seq(a(j), j=0..30)];
MATHEMATICA
polygorial[k_, n_] := FullSimplify[ n!/2^n (k -2)^n*Pochhammer[2/(k -2), n]]; Array[ polygorial[7, #] &, 16, 0] (* Robert G. Wilson v, Dec 26 2016 *) Join[{1}, FoldList[Times, PolygonalNumber[7, Range[20]]]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 29 2019 *)
AUTHOR
Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
Hendecagorials: n-th polygorial for k=11.
+10
20
1, 1, 11, 330, 19140, 1818300, 256380300, 50250538800, 13065140088000, 4350691649304000, 1805537034461160000, 913601739437346960000, 553642654099032257760000, 395854497680808064298400000
FORMULA
a(n) = polygorial(n, 11) = ( A000142(n)/ A000079(n))* A084949(n) = (n!/2^n)*Product_{i=0..n-1} (9*i+2) = (n!/2^n)*9^n*Pochhammer(2/9, n) = (n!/2^n)*9^n*GAMMA(n+2/9)/GAMMA(2/9). D-finite with recurrence 2*a(n) = n*(9*n-7)*a(n-1). - R. J. Mathar, Mar 12 2019
MAPLE
a := n->n!/2^n*product(9*i+2, i=0..n-1); [seq(a(j), j=0..30)];
MATHEMATICA
polygorial[k_, n_] := FullSimplify[ n!/2^n (k -2)^n*Pochhammer[2/(k - 2), n]]; Array[polygorial[11, #] &, 16, 0] (* Robert G. Wilson v, Dec 13 2016 *)
AUTHOR
Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
Octagorials: n-th polygorial for k=8.
+10
19
1, 1, 8, 168, 6720, 436800, 41932800, 5577062400, 981562982400, 220851671040000, 61838467891200000, 21086917550899200000, 8603462360766873600000, 4138265395528866201600000, 2317428621496165072896000000
FORMULA
a(n) = polygorial(n, 8) = ( A000142(n)/ A000079(n))* A047657(n) = (n!/2^n)*Product_{i=0..n-1} (6*i+2) = (n!/2^n)*6^n*Pochhammer(1/3, n) = (n!/2)*3^n*sqrt(3)*GAMMA(n+1/3)*GAMMA(2/3)/Pi. D-finite with recurrence a(n) = n*(3*n-2)*a(n-1). - R. J. Mathar, Mar 12 2019
MAPLE
a := n->n!/2^n*product(6*i+2, i=0..n-1); [seq(a(j), j=0..30)];
MATHEMATICA
polygorial[k_, n_] := FullSimplify[ n!/2^n (k -2)^n*Pochhammer[2/(k -2), n]]; Array[polygorial[8, #] &, 16, 0] (* Robert G. Wilson v, Dec 26 2016 *)
PROG
(PARI) a(n) = n! / 2^n * prod(i=0, n-1, 6*i+2) \\ Felix Fröhlich, Dec 13 2016
AUTHOR
Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
Enneagorials: n-th polygorial for k=9.
+10
19
1, 1, 9, 216, 9936, 745200, 82717200, 12738448800, 2598643555200, 678245967907200, 220429939569840000, 87290256069656640000, 41375581377017247360000, 23128949989752641274240000
FORMULA
a(n) = polygorial(n, 9) = ( A000142(n)/ A000079(n))* A084947(n) = (n!/2^n)*Product_{i=0..n-1} (7*i+2) = (n!/2^n)*7^n*Pochhammer(2/7, n) = (n!/2^n)*7^n*GAMMA(n+2/7)/GAMMA(2/7). D-finite with recurrence 2*a(n) = n*(7*n-5)*a(n-1). - R. J. Mathar, Mar 12 2019
MAPLE
a := n->n!/2^n*product(7*i+2, i=0..n-1); [seq(a(j), j=0..30)];
MATHEMATICA
polygorial[k_, n_] := FullSimplify[ n!/2^n (k -2)^n*Pochhammer[2/(k -2), n]]; Array[polygorial[9, #] &, 16, 0] (* Robert G. Wilson v, Dec 26 2016 *)
AUTHOR
Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
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