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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. 9

%I #16 Jan 15 2018 08:53:16

%S 1,5,27,191,1732,18690,226300,2964284,41082774,593967362,8873943769,

%T 136095567381,2132329828638,34008171994644,550591656446061,

%U 9029248417359913,149726007326186129,2507013639225903129,42337830100883644650,720436676774318943294,12342627498327879008169

%N 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 Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = (2*n)!/n!^2 = A000984(n).

%C Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^(2*n) = (3*n)!/(n!*(2*n)!) = A005809(n).

%H Vaclav Kotesovec, <a href="/A232683/b232683.txt">Table of n, a(n) for n = 0..200</a>

%F Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = (3*n)!/n!^3 = A000984(n)*A005809(n).

%F 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 G.f.: A(x) = 1 + 5*x + 27*x^2 + 191*x^3 + 1732*x^4 + 18690*x^5 +...

%e ILLUSTRATION OF INITIAL TERMS.

%e If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:

%e A^0: [1], 0, 0, 0, 0, 0, 0, 0, ...;

%e A^1: [1, 5], 27, 191, 1732, 18690, 226300, 2964284, ...;

%e A^2: [1, 10, 79], 652, 6103, 65014, 769509, 9862452, ...;

%e A^3: [1, 15, 156, 1508], 15138, 164232, 1933920, 24464106, ...;

%e A^4: [1, 20, 258, 2884, 31487], 355104, 4228676, 53345608, ...;

%e A^5: [1, 25, 385, 4905, 58425, 693015], 8452145, 107398205, ...;

%e A^6: [1, 30, 537, 7696, 99852, 1253100, 15791920],203842404, ...;

%e A^7: [1, 35, 714, 11382, 160293, 2133369, 27940444, 368826722], ...; ...

%e 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 (3*0)!/0!^3 = 1 = 1;

%e (3*1)!/1!^3 = 1 + 5 = 6;

%e (3*2)!/2!^3 = 1 + 10 + 79 = 90;

%e (3*3)!/3!^3 = 1 + 15 + 156 + 1508 = 1680;

%e (3*4)!/4!^3 = 1 + 20 + 258 + 2884 + 31487 = 34650;

%e (3*5)!/5!^3 = 1 + 25 + 385 + 4905 + 58425 + 693015 = 756756;

%e (3*6)!/6!^3 = 1 + 30 + 537 + 7696 + 99852 + 1253100 + 15791920 = 17153136; ...

%e RELATED SERIES.

%e From a main diagonal in the above array we can derive the sequence:

%e [1/1, 10/2, 156/3, 2884/4, 58425/5, 1253100/6, 27940444/7, ...] =

%e [1, 5, 52, 721, 11685, 208850, 3991492, 80086117, 1667185489, ...];

%e from which we can form the series G(x) = A(x*G(x)):

%e G(x) = 1 + 5*x + 52*x^2 + 721*x^3 + 11685*x^4 + 208850*x^5 + 3991492*x^6 +...

%e such that

%e (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 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 Table[a[n], {n, 0, 20}] (* _Jean-François Alcover_, Jan 15 2018, translated from 2nd PARI program *)

%o (PARI) /* By Definition: */

%o {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 for(n=0, 20, print1(a(n)*1!, ", "))

%o (PARI) /* Faster, using series reversion: */

%o {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 for(i=1, n, G = 1 + intformal( (S3-1)*G/x - S3*G^2)); polcoeff(x/serreverse(x*G), n)}

%o for(n=0, 30, print1(a(n), ", "))

%Y Cf. A232606, A006480.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Nov 27 2013

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