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A262977 a(n) = binomial(4*n-1,n).

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%I A262977 #55 Aug 17 2025 11:17:41
%S A262977 1,3,21,165,1365,11628,100947,888030,7888725,70607460,635745396,
%T A262977 5752004349,52251400851,476260169700,4353548972850,39895566894540,
%U A262977 366395202809685,3371363686069236,31074067324187580,286845713747883300,2651487106659130740,24539426037817994160
%N A262977 a(n) = binomial(4*n-1,n).
%C A262977 From _Gus Wiseman_, Sep 28 2022: (Start)
%C A262977 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 A262977   (6,2)  (1,2,5)  (1,1,5,1)  (1,1,1,1,4)
%C A262977          (2,2,4)  (2,1,4,1)  (1,1,2,1,3)
%C A262977          (3,2,3)  (3,1,3,1)  (1,1,3,1,2)
%C A262977          (4,2,2)  (4,1,2,1)  (1,1,4,1,1)
%C A262977          (5,2,1)  (5,1,1,1)  (2,1,1,1,3)
%C A262977                              (2,1,2,1,2)
%C A262977                              (2,1,3,1,1)
%C A262977                              (3,1,1,1,2)
%C A262977                              (3,1,2,1,1)
%C A262977                              (4,1,1,1,1)
%C A262977 The following pertain to this interpretation:
%C A262977 - The case of partitions is A000712, reverse A006330.
%C A262977 - Allowing any alternating sum gives A013777 (compositions of 4n).
%C A262977 - A011782 counts compositions of n.
%C A262977 - A034871 counts compositions of 2n with alternating sum 2k.
%C A262977 - A097805 counts compositions by alternating (or reverse-alternating) sum.
%C A262977 - A103919 counts partitions by sum and alternating sum (reverse: A344612).
%C A262977 - A345197 counts compositions by length and alternating sum.
%C A262977 (End)
%H A262977 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 A262977 G.f.: A(x)=x*B'(x)/B(x), where B(x) if g.f. of A006632.
%F A262977 a(n) = Sum_{k=0..n}(binomial(n-1,n-k)*binomial(3*n,k)).
%F A262977 a(n) = 3*A224274(n), for n > 0. - _Michel Marcus_, Oct 12 2015
%F A262977 From _Peter Bala_, Nov 04 2015: (Start)
%F A262977 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 A262977 a(n) = [x^n] 1/(1 - x)^(3*n). - _Ilya Gutkovskiy_, Oct 03 2017
%F A262977 a(n) = A071919(3n-1,n+1) = A097805(4n,n+1). - _Gus Wiseman_, Sep 28 2022
%F A262977 From _Peter Bala_, Feb 14 2024: (Start)
%F A262977 a(n) = (-1)^n * binomial(-3*n, n).
%F A262977 a(n) = hypergeom([1 - 3*n, -n], [1], 1).
%F A262977 The g.f. A(x) satisfies A(x/(1 + x)^4) = 1/(1 - 3*x). (End)
%F A262977 a(n) = Sum_{k = 0..n} binomial(2*n+k-1, k)*binomial(2*n-k-1, n-k). - _Peter Bala_, Sep 16 2024
%F A262977 G.f.: 1/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - _Seiichi Manyama_, Aug 17 2025
%t A262977 Table[Binomial[4 n - 1, n], {n, 0, 40}] (* _Vincenzo Librandi_, Oct 06 2015 *)
%o A262977 (Maxima)
%o A262977 B(x):=sum(binomial(4*n-1,n-1)*3/(4*n-1)*x^n,n,1,30);
%o A262977 taylor(x*diff(B(x),x,1)/B(x),x,0,20);
%o A262977 (Magma) [Binomial(4*n-1,n): n in [0..20]]; // _Vincenzo Librandi_, Oct 06 2015
%o A262977 (PARI) a(n) = binomial(4*n-1,n); \\ _Michel Marcus_, Oct 06 2015
%Y A262977 Cf. A006632, A002293, A004331, A005810, A052203, A224274, A257633.
%Y A262977 Cf. A000346, A000984, A001700, A002458, A025047, A058622, A081294, A294175.
%Y A262977 Cf. A088218, A163455, A165817.
%K A262977 nonn,easy
%O A262977 0,2
%A A262977 _Vladimir Kruchinin_, Oct 06 2015