A262977 - OEIS

%I #50 Nov 16 2024 06:31:05

%S 1,3,21,165,1365,11628,100947,888030,7888725,70607460,635745396,

%T 5752004349,52251400851,476260169700,4353548972850,39895566894540,

%U 366395202809685,3371363686069236,31074067324187580,286845713747883300,2651487106659130740,24539426037817994160

%N a(n) = binomial(4*n-1,n).

%C From _Gus Wiseman_, Sep 28 2022: (Start)

%C Also the number of integer compositions of 4n with alternating sum 2n, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A348614. The a(12) = 21 compositions are:

%C   (6,2)  (1,2,5)  (1,1,5,1)  (1,1,1,1,4)

%C          (2,2,4)  (2,1,4,1)  (1,1,2,1,3)

%C          (3,2,3)  (3,1,3,1)  (1,1,3,1,2)

%C          (4,2,2)  (4,1,2,1)  (1,1,4,1,1)

%C          (5,2,1)  (5,1,1,1)  (2,1,1,1,3)

%C                              (2,1,2,1,2)

%C                              (2,1,3,1,1)

%C                              (3,1,1,1,2)

%C                              (3,1,2,1,1)

%C                              (4,1,1,1,1)

%C The following pertain to this interpretation:

%C - The case of partitions is A000712, reverse A006330.

%C - Allowing any alternating sum gives A013777 (compositions of 4n).

%C - A011782 counts compositions of n.

%C - A034871 counts compositions of 2n with alternating sum 2k.

%C - A097805 counts compositions by alternating (or reverse-alternating) sum.

%C - A103919 counts partitions by sum and alternating sum (reverse: A344612).

%C - A345197 counts compositions by length and alternating sum.

%C (End)

%H V. V. Kruchinin and D. V. Kruchinin, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Kruchinin/kruch9.html">A Generating Function for the Diagonal T_{2n,n} in Triangles</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

%F G.f.: A(x)=x*B'(x)/B(x), where B(x) if g.f. of A006632.

%F a(n) = Sum_{k=0..n}(binomial(n-1,n-k)*binomial(3*n,k)).

%F a(n) = 3*A224274(n), for n > 0. - _Michel Marcus_, Oct 12 2015

%F From _Peter Bala_, Nov 04 2015: (Start)

%F The o.g.f. equals f(x)/g(x), where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A005810 (k = 0), A052203 (k = 1), A257633 (k = 2), A224274 (k = 3) and A004331 (k = 4). (End)

%F a(n) = [x^n] 1/(1 - x)^(3*n). - _Ilya Gutkovskiy_, Oct 03 2017

%F a(n) = A071919(3n-1,n+1) = A097805(4n,n+1). - _Gus Wiseman_, Sep 28 2022

%F From _Peter Bala_, Feb 14 2024: (Start)

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

%F a(n) = hypergeom([1 - 3*n, -n], [1], 1).

%F The g.f. A(x) satisfies A(x/(1 + x)^4) = 1/(1 - 3*x). (End)

%F a(n) = Sum_{k = 0..n} binomial(2*n+k-1, k)*binomial(2*n-k-1, n-k). - _Peter Bala_, Sep 16 2024

%t Table[Binomial[4 n - 1, n], {n, 0, 40}] (* _Vincenzo Librandi_, Oct 06 2015 *)

%o (Maxima)

%o B(x):=sum(binomial(4*n-1,n-1)*3/(4*n-1)*x^n,n,1,30);

%o taylor(x*diff(B(x),x,1)/B(x),x,0,20);

%o (Magma) [Binomial(4*n-1,n): n in [0..20]]; // _Vincenzo Librandi_, Oct 06 2015

%o (PARI) a(n) = binomial(4*n-1,n); \\ _Michel Marcus_, Oct 06 2015

%Y Cf. A006632, A002293, A004331, A005810, A052203, A224274, A257633.

%Y Cf. A000346, A000984, A001700, A002458, A025047, A058622, A081294, A294175.

%K nonn,easy,changed

%O 0,2

%A _Vladimir Kruchinin_, Oct 06 2015