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.

1, 5, 27, 191, 1732, 18690, 226300, 2964284, 41082774, 593967362, 8873943769, 136095567381, 2132329828638, 34008171994644, 550591656446061, 9029248417359913, 149726007326186129, 2507013639225903129, 42337830100883644650, 720436676774318943294, 12342627498327879008169

Offset

0,2

Comments

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

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

Links

Vaclav Kotesovec, Table of n, a(n) for n = 0..200

Formula

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

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.

Example

G.f.: A(x) = 1 + 5*x + 27*x^2 + 191*x^3 + 1732*x^4 + 18690*x^5 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1], 0,    0,     0,      0,       0,        0,         0, ...;
A^1: [1,  5],  27,   191,   1732,   18690,   226300,   2964284, ...;
A^2: [1, 10,   79],  652,   6103,   65014,   769509,   9862452, ...;
A^3: [1, 15,  156,  1508], 15138,  164232,  1933920,  24464106, ...;
A^4: [1, 20,  258,  2884,  31487], 355104,  4228676,  53345608, ...;
A^5: [1, 25,  385,  4905,  58425,  693015], 8452145, 107398205, ...;
A^6: [1, 30,  537,  7696,  99852, 1253100, 15791920],203842404, ...;
A^7: [1, 35,  714, 11382, 160293, 2133369, 27940444, 368826722], ...; ...
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):
(3*0)!/0!^3 = 1 = 1;
(3*1)!/1!^3 = 1 +  5 = 6;
(3*2)!/2!^3 = 1 + 10 +  79 = 90;
(3*3)!/3!^3 = 1 + 15 + 156 + 1508 = 1680;
(3*4)!/4!^3 = 1 + 20 + 258 + 2884 + 31487 = 34650;
(3*5)!/5!^3 = 1 + 25 + 385 + 4905 + 58425 +  693015 = 756756;
(3*6)!/6!^3 = 1 + 30 + 537 + 7696 + 99852 + 1253100 + 15791920 = 17153136; ...
RELATED SERIES.
From a main diagonal in the above array we can derive the sequence:
[1/1, 10/2, 156/3, 2884/4, 58425/5, 1253100/6, 27940444/7, ...] =
[1, 5, 52, 721, 11685, 208850, 3991492, 80086117, 1667185489, ...];
from which we can form the series G(x) = A(x*G(x)):
G(x) = 1 + 5*x + 52*x^2 + 721*x^3 + 11685*x^4 + 208850*x^5 + 3991492*x^6 +...
such that
(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 +...

Mathematica

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}]];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jan 15 2018, translated from 2nd PARI program *)

Prog

(PARI) /* By Definition: */
{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)}
for(n=0, 20, print1(a(n)*1!, ", "))
(PARI) /* Faster, using series reversion: */
{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));
for(i=1, n, G = 1 + intformal( (S3-1)*G/x - S3*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Crossrefs

Cf. A232606, A006480.

Sequence in context: A205774 A367048 A326094 * A240637 A023811 A126119

Adjacent sequences: A232680 A232681 A232682 * A232684 A232685 A232686

Keyword

nonn

Author

Paul D. Hanna, Nov 27 2013

Status

approved