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A326005 G.f.: Sum_{n>=0} (n+1)*(n+2)*(n+3)*(n+4)/4! * x^n * (1 + x^n)^n. 3

%I #8 Jul 02 2019 22:39:39

%S 1,5,20,35,100,126,330,330,775,820,1631,1365,3535,2380,5370,5136,9085,

%T 5985,16900,8855,21966,19580,29965,17550,60375,24381,58345,57205,

%U 90350,40920,152837,52360,164145,141120,175560,93801,404500,101270,280175,309050,503041,148995,714435,178365,748705,708946,633950,249900,1771645,295135,1120236,1155015,1760500,395010,2483110,905576,2622545,2036060,1744525,595665,6962328,677040,2343880

%N G.f.: Sum_{n>=0} (n+1)*(n+2)*(n+3)*(n+4)/4! * x^n * (1 + x^n)^n.

%C More generally, the following sums are equal:

%C (1) Sum_{n>=0} binomial(n+k-1, n) * r^n * (p + q^n)^n,

%C (2) Sum_{n>=0} binomial(n+k-1, n) * r^n * q^(n^2) / (1 - p*q^n*r)^(n+k),

%C for any fixed integer k; here, k = 5 and p = 1, q = x, r = x.

%F Generating functions.

%F (1) Sum_{n>=0} (n+1)*(n+2)*(n+3)*(n+4)/4! * x^n * (1 + x^n)^n.

%F (2) Sum_{n>=0} (n+1)*(n+2)*(n+3)*(n+4)/4! * x^(n*(n+1)) / (1 - x^(n+1))^(n+5).

%e G.f.: A(x) = 1 + 5*x + 20*x^2 + 35*x^3 + 100*x^4 + 126*x^5 + 330*x^6 + 330*x^7 + 775*x^8 + 820*x^9 + 1631*x^10 + 1365*x^11 + 3535*x^12 + 2380*x^13 + 5370*x^14 + 5136*x^15 + 9085*x^16 + 5985*x^17 + 16900*x^18 + 8855*x^19 + 21966*x^20 + ...

%e where we have the following series identity:

%e A(x) = 1 + 5*x*(1+x) + 15*x^2*(1+x^2)^2 + 35*x^3*(1+x^3)^3 + 70*x^4*(1+x^4)^4 + 126*x^5*(1+x^5)^5 + 210*x^6*(1+x^6)^6 + 330*x^7*(1+x^7)^7 + 495*x^8*(1+x^8)^8 + 715*x^9*(1+x^9)^9 +...

%e is equal to

%e A(x) = 1/(1-x)^5 + 5*x^2/(1-x^2)^6 + 15*x^6/(1-x^3)^7 + 35*x^12/(1-x^4)^8 + 70*x^20/(1-x^5)^9 + 126*x^30/(1-x^6)^10 + 210*x^42/(1-x^7)^11 + 330*x^56/(1-x^8)^12 +...

%o (PARI) {a(n) = my(A = sum(m=0,n, (m+1)*(m+2)*(m+3)*(m+4)/4! * x^m * (1 + x^m +x*O(x^n))^m)); polcoeff(A,n)}

%o for(n=0,120,print1(a(n),", "))

%o (PARI) {a(n) = my(A = sum(m=0,n, (m+1)*(m+2)*(m+3)*(m+4)/4! * x^m * x^(m^2) / (1 - x^(m+1) +x*O(x^n))^(m+5))); polcoeff(A,n)}

%o for(n=0,120,print1(a(n),", "))

%Y Cf. A217668 (k=1), A326002 (k=2), A326003 (k=3), A326004 (k=4).

%K nonn

%O 0,2

%A _Paul D. Hanna_, Jun 01 2019

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