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%I #20 Jan 15 2018 08:53:27
%S 1,1,3,7,20,66,244,980,4182,18674,86353,410541,1996214,9888844,
%T 49760925,253767097,1309154825,6822023553,35865392690,190038440422,
%U 1014015337209,5444707218851,29401289997403,159584901816255,870267544114291,4766246752344215,26206635040151511
%N G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals Sum_{k=0..n} C(n,k)^3 = A000172(n) (Franel numbers), for n>=0.
%C Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = Sum_{k=0..n} C(n,k)^2 = (2*n)!/n!^2.
%C a(n+1)/a(n) tends to 6.0295... - _Vaclav Kotesovec_, Jan 22 2014
%H Paul D. Hanna and Vaclav Kotesovec, <a href="/A232687/b232687.txt">Table of n, a(n) for n = 0..460</a> (first 200 terms from Paul D. Hanna)
%F Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = Sum_{k=0..n} C(n,k)^3 = A000172(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^(2*n)/(1-2*x)^(3*n+1) = Sum_{n>=0} A000172(n)*x^n.
%e G.f.: A(x) = 1 + x + 3*x^2 + 7*x^3 + 20*x^4 + 66*x^5 + 244*x^6 + 980*x^7 +...
%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, 0, ...;
%e A^1: [1, 1], 3, 7, 20, 66, 244, 980, 4182, ...;
%e A^2: [1, 2, 7], 20, 63, 214, 789, 3124, 13112, ...;
%e A^3: [1, 3, 12, 40], 138, 492, 1848, 7326, 30531, ...;
%e A^4: [1, 4, 18, 68, 255], 960, 3716, 14920, 62295, ...;
%e A^5: [1, 5, 25, 105, 425, 1691], 6785, 27805, 117165, ...;
%e A^6: [1, 6, 33, 152, 660, 2772, 11560], 48588, 207774, ...;
%e A^7: [1, 7, 42, 210, 973, 4305, 18676, 80746],351792, ...;
%e A^8: [1, 8, 52, 280, 1378, 6408, 28916, 128808, 573311], ...; ...
%e then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals Sum_{k=0..n} C(n,k)^3 = A000172(n):
%e A000172(0) = 1 = 1;
%e A000172(1) = 1 + 1 = 2;
%e A000172(2) = 1 + 2 + 7 = 10;
%e A000172(3) = 1 + 3 + 12 + 40 = 56;
%e A000172(4) = 1 + 4 + 18 + 68 + 255 = 346;
%e A000172(5) = 1 + 5 + 25 + 105 + 425 + 1691 = 2252;
%e A000172(6) = 1 + 6 + 33 + 152 + 660 + 2772 + 11560 = 15184; ...
%t Franel[n_] := Sum[Binomial[n, k]^3, {k, 0, n}];
%t a[0] = 1; a[n_] := Module[{B, G}, B = Sum[Franel[k]*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[(B-1)* (G/x)-B*G^2, x]]; SeriesCoefficient[x/InverseSeries[x*G, x], {x, 0, n}]];
%t Table[a[n], {n, 0, 26}] (* _Jean-François Alcover_, Jan 15 2018, translated from 2nd PARI program *)
%o (PARI) /* By Definition (slow): */
%o {Franel(n)=sum(k=0,n,binomial(n,k)^3)}
%o {a(n)=if(n==0, 1, (Franel(n) - 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), ", "))
%o (PARI) /* Faster, using series reversion: */
%o {Franel(n)=sum(k=0,n,binomial(n,k)^3)}
%o {a(n)=local(B=sum(k=0, n+1, Franel(k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
%o for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A232606, A232683, A232690, A000172.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 05 2013
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