

A097070


Consider all compositions (ordered partitions) of n into n parts, allowing zeros. E.g., for n = 3 we get 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Then a(n) is the total number of 1's.


5



1, 2, 9, 40, 175, 756, 3234, 13728, 57915, 243100, 1016158, 4232592, 17577014, 72804200, 300874500, 1240940160, 5109183315, 21002455980, 86213785350, 353452638000, 1447388552610, 5920836618840, 24197138082780, 98801168731200, 403095046038750, 1643337883690776, 6694900194799404
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OFFSET

1,2


COMMENTS

Number of compositions of n into n parts, allowing zeros = binomial(2*n1,n) = A088218 = essentially A001700.


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
G.S. Cheon, H. Kim, and L. W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014.


FORMULA

a(n) = n*binomial(2*n3, n1).
More generally, total number of k's (k>=0) in all ordered partitions of n into n parts, allowing zeros, is n*binomial(2*nk2, n2) if n >= k, 0 otherwise.
Total number of 0's is given by A005430.
From Vladeta Jovovic, Sep 17 2004: (Start)
a(n) = Sum_{k=0..n} k*binomial(n, k)*binomial(n2, k2).
G.f.: x*(1 2*x +(14*x)^(3/2))/(2*(14*x)^(3/2)).
E.g.f.: (x/2)*(exp(2*x)*BesselI(0, 2*x)+1). (End)
a(n) = A014107(n)*A000108(n2).  Philippe Deléham, Apr 12 2007
a(n) = n*A088218(n1) for n > 0.  Werner Schulte, Jan 22 2017
From Bruce J. Nicholson, Jul 11 2019: (Start)
a(n) = A002740(n) + A097613(n).
a(n) = A110609(n1)  A002457(n2) + A097613(n).
a(n) = A005430(n1)  A000917(n3) for n > 1.
a(n) = A002457(n1)  A037965(n)  A000917(n3) for n > 1.
a(n) = A037965(n)/2.
a(n) = A001700(n2)*n.
a(n) = A001791(n2)*n + A000984(n2)*n for n > 1. (End)
From Amiram Eldar, May 16 2022: (Start)
Sum_{n>=1} 1/a(n) = 4*Pi/(3*sqrt(3))  Pi^2/9.
Sum_{n>=1} (1)^(n+1)/a(n) = 8*log(phi)/sqrt(5)  4*log(phi)^2, where phi is the golden ratio (A001622). (End)


EXAMPLE

The compositions for n=2 are 20, 02, 11. There are two 1's in these so a(2) = 2.
From Robert G. Wilson v, Sep 16 2004: (Start)
The case n = 5:
A. There are 5 combinations associated with the numbers 50000: 50000, 05000, 00500, 00050, 00005.
B. There are 20 combinations associated with the numbers 41000.
C. There are 20 combinations associated with 32000.
D. There are 30 combinations associated with 31100.
E. There are 30 combinations associated with 22100.
F. There are 20 combinations associated with 21110.
G. There is one combinations associated with 11111.
The number of 1's associated with A is 0, with B 20, with C 0, with D 60, with E 30, with F 60 and with G 5. 0 + 20 + 0 + 60 + 30 + 60 + 5 = 175.
(End)


MATHEMATICA

Table[n*Binomial[2n3, n1], {n, 30}] (* Robert G. Wilson v, Sep 17 2004 *)


PROG

(PARI) a(n) = n*binomial(2*n3, n1); \\ Joerg Arndt, Feb 17 2015
(Magma) [n*Binomial(2*n3, n1): n in [1..30]]; // Vincenzo Librandi, Jul 13 2019
(Sage) [n*binomial(2*n3, n1) for n in (1..30)] # G. C. Greubel, Jul 27 2019
(GAP) List([1..30], n> n*Binomial(2*n3, n1)); # G. C. Greubel, Jul 27 2019


CROSSREFS

Cf. A037965.
Cf. A000984, A001622, A088218.
cf. A005430, A002740, A002457, A000917, A110609, A097613, A001791, A110609.
Sequence in context: A317129 A220309 A019066 * A164033 A020728 A107979
Adjacent sequences: A097067 A097068 A097069 * A097071 A097072 A097073


KEYWORD

nonn,easy


AUTHOR

Amy J. Kolan, Sep 15 2004


EXTENSIONS

Formula, more terms and comments from Vladeta Jovovic, Sep 15 2004


STATUS

approved



