cp's OEIS Frontend

A244653 G.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = binomial(7*n,2*n).

%I A244653 #10 Jul 05 2014 02:19:54
%S A244653 1,20,280,3521,42945,521913,6379233,78640740,978172724,12270946072,
%T A244653 155139813381,1975245161155,25308115539770,326106155857041,
%U A244653 4223481710794292,54951230993010196,717942326681863941,9415448193554916520,123904268078599269723,1635676807214777434793
%N A244653 G.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = binomial(7*n,2*n).
%C A244653 Compare to Sum_{k=0..n} [x^k] 1/(1-x)^(6*n) = binomial(7*n,n).
%C A244653 Compare to Sum_{k=0..n} [x^k] 1/((1-x)*(1-2*x)^2)^n = binomial(4*n,2*n).
%H A244653 Vaclav Kotesovec, <a href="/A244653/b244653.txt">Table of n, a(n) for n = 0..170</a>
%F A244653 Recurrence: 3*n*(n+1)*(3*n+5)*(3*n+7)*(325*n^3 - 539*n^2 + 200*n - 28)*a(n) = 6*n*(26325*n^6 + 52866*n^5 - 3170*n^4 - 58740*n^3 - 48103*n^2 - 27502*n - 8876)*a(n-1) - (528125*n^7 + 627900*n^6 - 1376202*n^5 - 1623792*n^4 - 330807*n^3 + 114228*n^2 + 109732*n - 4704)*a(n-2) + 4*(177125*n^7 + 79020*n^6 - 878777*n^5 - 465945*n^4 + 871822*n^3 + 881769*n^2 + 181790*n - 53508)*a(n-3) - (n+2)*(404625*n^6 - 896280*n^5 - 811152*n^4 + 2278486*n^3 + 2725599*n^2 + 552482*n - 206976)*a(n-4) + 14*(n+1)*(n+2)*(4225*n^5 - 14157*n^4 + 29689*n^3 + 70969*n^2 + 21750*n - 7308)*a(n-5) + 49*(n-7)*n*(n+1)*(n+2)*(325*n^3 + 436*n^2 + 97*n - 42)*a(n-6). - _Vaclav Kotesovec_, Jul 04 2014
%F A244653 a(n) ~ c * (7+14/9*sqrt(21))^n / (sqrt(Pi)*n^(3/2)), where c = 43.267577625713256769244376361089321461925061695487162410160820989... . - _Vaclav Kotesovec_, Jul 04 2014
%e A244653 G.f.: A(x) = 1 + 20*x + 280*x^2 + 3521*x^3 + 42945*x^4 + 521913*x^5 +...
%e A244653 ILLUSTRATION OF INITIAL TERMS.
%e A244653 If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
%e A244653 A^0: [1],  0,    0,      0,        0,         0,           0, ...;
%e A244653 A^1: [1,  20], 280,   3521,    42945,    521913,     6379233, ...;
%e A244653 A^2: [1,  40,  960], 18242,   305130,   4733386,    70081627, ...;
%e A244653 A^3: [1,  60, 2040,  52163], 1122555,  21563619,   382898782, ...;
%e A244653 A^4: [1,  80, 3520, 113284,  2991220], 68901812,  1437454298, ...;
%e A244653 A^5: [1, 100, 5400, 209605,  6567125, 176797165], 4251203775, ...;
%e A244653 A^6: [1, 120, 7680, 349126, 12666270, 390658878, 10654434813],...; ...
%e A244653 then we can illustrate how the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals C(7*n,2*n):
%e A244653 C( 0, 0) = 1 = 1;
%e A244653 C( 7, 2) = 1 + 20 = 21;
%e A244653 C(14, 4) = 1 +  40 +  960 = 1001;
%e A244653 C(21, 6) = 1 +  60 + 2040 +  52163 = 54264;
%e A244653 C(28, 8) = 1 +  80 + 3520 + 113284 +  2991220 = 3108105;
%e A244653 C(35,10) = 1 + 100 + 5400 + 209605 +  6567125 + 176797165 = 183579396;
%e A244653 C(42,12) = 1 + 120 + 7680 + 349126 + 12666270 + 390658878 + 10654434813 = 11058116888; ...
%o A244653 (PARI) /* By Definition (slow): */
%o A244653 {a(n)=if(n==0, 1, ( binomial(7*n,2*n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j/1!)^n + x*O(x^k), k)))/n)}
%o A244653 for(n=0, 20, print1(a(n), ", "))
%o A244653 (PARI) /* Faster, using series reversion: */
%o A244653 {a(n)=local(B=sum(k=0, n+1, binomial(7*k,2*k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
%o A244653 for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
%o A244653 for(n=0, 30, print1(a(n), ", "))
%Y A244653 Cf. A232606, A232683, A232687, A244650, A244651, A244652, A244654.
%K A244653 nonn
%O A244653 0,2
%A A244653 _Paul D. Hanna_, Jul 03 2014