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A232683 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals (3*n)!/n!^3, which is de Bruijn's sequence S(3,n) (A006480), for n>=0.

%I A232683 #20 Mar 17 2025 15:41:42
%S A232683 1,5,27,191,1732,18690,226300,2964284,41082774,593967362,8873943769,
%T A232683 136095567381,2132329828638,34008171994644,550591656446061,
%U A232683 9029248417359913,149726007326186129,2507013639225903129,42337830100883644650,720436676774318943294,12342627498327879008169
%N A232683 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals (3*n)!/n!^3, which is de Bruijn's sequence S(3,n) (A006480), for n>=0.
%C A232683 Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = (2*n)!/n!^2 = A000984(n).
%C A232683 Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^(2*n) = (3*n)!/(n!*(2*n)!) = A005809(n).
%H A232683 Vaclav Kotesovec, <a href="/A232683/b232683.txt">Table of n, a(n) for n = 0..200</a>
%F A232683 Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = (3*n)!/n!^3 = A000984(n)*A005809(n).
%F A232683 Given g.f. A(x), let G(x) = A(x*G(x)) then (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = Sum_{n>=0} (3*n)!/n!^3 * x^n.
%e A232683 G.f.: A(x) = 1 + 5*x + 27*x^2 + 191*x^3 + 1732*x^4 + 18690*x^5 +...
%e A232683 ILLUSTRATION OF INITIAL TERMS.
%e A232683 If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
%e A232683 A^0: [1], 0,    0,     0,      0,       0,        0,         0, ...;
%e A232683 A^1: [1,  5],  27,   191,   1732,   18690,   226300,   2964284, ...;
%e A232683 A^2: [1, 10,   79],  652,   6103,   65014,   769509,   9862452, ...;
%e A232683 A^3: [1, 15,  156,  1508], 15138,  164232,  1933920,  24464106, ...;
%e A232683 A^4: [1, 20,  258,  2884,  31487], 355104,  4228676,  53345608, ...;
%e A232683 A^5: [1, 25,  385,  4905,  58425,  693015], 8452145, 107398205, ...;
%e A232683 A^6: [1, 30,  537,  7696,  99852, 1253100, 15791920],203842404, ...;
%e A232683 A^7: [1, 35,  714, 11382, 160293, 2133369, 27940444, 368826722], ...; ...
%e A232683 then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals (3*n)!/n!^3 (A006480):
%e A232683 (3*0)!/0!^3 = 1 = 1;
%e A232683 (3*1)!/1!^3 = 1 +  5 = 6;
%e A232683 (3*2)!/2!^3 = 1 + 10 +  79 = 90;
%e A232683 (3*3)!/3!^3 = 1 + 15 + 156 + 1508 = 1680;
%e A232683 (3*4)!/4!^3 = 1 + 20 + 258 + 2884 + 31487 = 34650;
%e A232683 (3*5)!/5!^3 = 1 + 25 + 385 + 4905 + 58425 +  693015 = 756756;
%e A232683 (3*6)!/6!^3 = 1 + 30 + 537 + 7696 + 99852 + 1253100 + 15791920 = 17153136; ...
%e A232683 RELATED SERIES.
%e A232683 From a main diagonal in the above array we can derive the sequence:
%e A232683 [1/1, 10/2, 156/3, 2884/4, 58425/5, 1253100/6, 27940444/7, ...] =
%e A232683 [1, 5, 52, 721, 11685, 208850, 3991492, 80086117, 1667185489, ...];
%e A232683 from which we can form the series G(x) = A(x*G(x)):
%e A232683 G(x) = 1 + 5*x + 52*x^2 + 721*x^3 + 11685*x^4 + 208850*x^5 + 3991492*x^6 +...
%e A232683 such that
%e A232683 (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = 1 + 6*x + 90*x^2 + 1680*x^3 + 34650*x^4 + 756756*x^5 + 17153136*x^6 +...+ A006480(n)*x^n +...
%t A232683 a[0] = 1; a[n_] := Module[{S3, G}, S3 = Sum[((3*k)!/k!^3)*x^k, {k, 0, n + 1}] + x^3*O[x]^n; G = 1 + x*O[x]^n; For[i = 1, i <= n, i++, G = 1 + Integrate[(S3-1)*(G/x) - S3*G^2, x]]; SeriesCoefficient[ x/InverseSeries[ x*G, x], {x, 0, n}]];
%t A232683 Table[a[n], {n, 0, 20}] (* _Jean-François Alcover_, Jan 15 2018, translated from 2nd PARI program *)
%o A232683 (PARI) /* By Definition: */
%o A232683 {a(n)=if(n==0, 1, ((3*n)!/n!^3 - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)}
%o A232683 for(n=0, 20, print1(a(n)*1!, ", "))
%o A232683 (PARI) /* Faster, using series reversion: */
%o A232683 {a(n)=local(S3=sum(k=0, n+1, (3*k)!/k!^3*x^k)+x^3*O(x^n), G=1+x*O(x^n));
%o A232683 for(i=1, n, G = 1 + intformal( (S3-1)*G/x - S3*G^2)); polcoeff(x/serreverse(x*G), n)}
%o A232683 for(n=0, 30, print1(a(n), ", "))
%Y A232683 Cf. A232606, A006480.
%K A232683 nonn
%O A232683 0,2
%A A232683 _Paul D. Hanna_, Nov 27 2013