A319836 O.g.f. A(x) satisfies: [x^n] 1/(1-x)^(n^3) / exp( n^2*A(x) ) = 0 for n >= 1.

1, 3, 82, 8809, 1968876, 737476237, 413835914885, 324291453995297, 338056326570381922, 452343107609773394401, 755809472347044012276336, 1542728249552891936516983561, 3778352523083207335101484460943, 10938158784764591996291933439895707, 36956866034584035700151967827780276701, 144148838902339708999595419616325771715713

Offset

1,2

Comments

It is remarkable that this sequence should consist entirely of integers.

Links

Paul D. Hanna, Table of n, a(n) for n = 1..200

Example

G.f.: A(x) = x + 3*x^2 + 82*x^3 + 8809*x^4 + 1968876*x^5 + 737476237*x^6 + 413835914885*x^7 + 324291453995297*x^8 + ...
The table of coefficients of x^k/k! in 1/(1-x)^(n^3)/exp(n^2*A(x)) begins
n=1: [1, 0, -5, -490, -211335, -236240596, -530964635645, ...];
n=2: [1, 4, 0, -2080, -877360, -961966464, -2146569842816, ...];
n=3: [1, 18, 297, 0, -2162835, -2310096834, -5017161130227, ...];
n=4: [1, 48, 2272, 98240, 0, -4547777536, -9705730181888, ...];
n=5: [1, 100, 9975, 980450, 88397225, 0, -16884222466625, ...];
n=6: [1, 180, 32400, 5814720, 1029708720, 168003828864, 0, ...]; ...
in which the coefficient of x^n in the n-th row forms a diagonal of zeros.
RELATED SERIES.
exp(A(x)) = 1 + x + 7*x^2/2! + 511*x^3/3! + 213529*x^4/4! + 237357241*x^5/5! + 532425292591*x^6/6! + 2089488422870647*x^7/7! + ...

Prog

(PARI) {a(n) = my(A=[1], m); for(i=1, n+1, A=concat(A, 0); m=#A; A[m] = Vec( 1/(1-x +x^2*O(x^m))^(m^3) * exp(-m^2*x*Ser(A)) )[m+1]/m^2 ); polcoeff( x*Ser(A), n)}
for(n=1, 20, print1(a(n), ", "))

Crossrefs

Cf. A317342, A319835, A319832.

Sequence in context: A037111 A067202 A203494 * A356503 A158995 A079652

Adjacent sequences: A319833 A319834 A319835 * A319837 A319838 A319839

Keyword

nonn,changed

Author

Paul D. Hanna, Sep 30 2018

Status

approved