A097070 - OEIS

%I #67 Jan 22 2025 08:52:36

%S 1,2,9,40,175,756,3234,13728,57915,243100,1016158,4232592,17577014,

%T 72804200,300874500,1240940160,5109183315,21002455980,86213785350,

%U 353452638000,1447388552610,5920836618840,24197138082780,98801168731200,403095046038750,1643337883690776,6694900194799404

%N 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.

%C Number of compositions of n into n parts, allowing zeros = binomial(2*n-1,n) = A088218 = essentially A001700.

%H Vincenzo Librandi, <a href="/A097070/b097070.txt">Table of n, a(n) for n = 1..1000</a>

%H Paul Barry, <a href="https://arxiv.org/abs/2004.04577">On a Central Transform of Integer Sequences</a>, arXiv:2004.04577 [math.CO], 2020.

%H G.-S. Cheon, H. Kim, and L. W. Shapiro, <a href="http://arxiv.org/abs/1410.1249">Mutation effects in ordered trees</a>, arXiv preprint arXiv:1410.1249 [math.CO], 2014.

%F a(n) = n*binomial(2*n-3, n-1).

%F More generally, total number of k's (k>=0) in all ordered partitions of n into n parts, allowing zeros, is n*binomial(2*n-k-2, n-2) if n >= k, 0 otherwise.

%F Total number of 0's is given by A005430.

%F From _Vladeta Jovovic_, Sep 17 2004: (Start)

%F a(n) = Sum_{k=0..n} k*binomial(n, k)*binomial(n-2, k-2).

%F G.f.: x*(1 -2*x +(1-4*x)^(3/2))/(2*(1-4*x)^(3/2)).

%F E.g.f.: (x/2)*(exp(2*x)*BesselI(0, 2*x)+1). (End)

%F a(n) = A014107(n)*A000108(n-2). - _Philippe Deléham_, Apr 12 2007

%F a(n) = n*A088218(n-1) for n > 0. - _Werner Schulte_, Jan 22 2017

%F From _Bruce J. Nicholson_, Jul 11 2019: (Start)

%F a(n) = A002740(n) + A097613(n).

%F a(n) = A110609(n-1) - A002457(n-2) + A097613(n).

%F a(n) = A005430(n-1) - A000917(n-3) for n > 1.

%F a(n) = A002457(n-1) - A037965(n) - A000917(n-3) for n > 1.

%F a(n) = A037965(n)/2.

%F a(n) = A001700(n-2)*n.

%F a(n) = A001791(n-2)*n + A000984(n-2)*n for n > 1. (End)

%F From _Amiram Eldar_, May 16 2022: (Start)

%F Sum_{n>=1} 1/a(n) = 4*Pi/(3*sqrt(3)) - Pi^2/9.

%F 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)

%F a(n) = 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3) for n > 1. - _Peter Luschny_, Jan 22 2025

%e The compositions for n=2 are 20, 02, 11. There are two 1's in these so a(2) = 2.

%e From _Robert G. Wilson v_, Sep 16 2004: (Start)

%e The case n = 5:

%e A. There are 5 combinations associated with the numbers 50000: 50000, 05000, 00500, 00050, 00005.

%e B. There are 20 combinations associated with the numbers 41000.

%e C. There are 20 combinations associated with 32000.

%e D. There are 30 combinations associated with 31100.

%e E. There are 30 combinations associated with 22100.

%e F. There are 20 combinations associated with 21110.

%e G. There is one combinations associated with 11111.

%e 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.

%e (End)

%p A097070 := n -> ifelse(n=1, 1, 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3)):

%p seq(simplify(A097070(n)), n = 1..28);  # _Peter Luschny_, Jan 22 2025

%t Table[n*Binomial[2n-3, n-1], {n, 30}] (* _Robert G. Wilson v_, Sep 17 2004 *)

%o (PARI) a(n) = n*binomial(2*n-3, n-1); \\ _Joerg Arndt_, Feb 17 2015

%o (Magma) [n*Binomial(2*n-3, n-1): n in [1..30]]; // _Vincenzo Librandi_, Jul 13 2019

%o (Sage) [n*binomial(2*n-3, n-1) for n in (1..30)] # _G. C. Greubel_, Jul 27 2019

%o (GAP) List([1..30], n-> n*Binomial(2*n-3, n-1)); # _G. C. Greubel_, Jul 27 2019

%Y Cf. A037965.

%Y Cf. A000984, A001622, A088218.

%Y cf. A005430, A002740, A002457, A000917, A110609, A097613, A001791, A110609.

%K nonn,easy

%O 1,2

%A _Amy J. Kolan_, Sep 15 2004

%E Formula, more terms and comments from _Vladeta Jovovic_, Sep 15 2004