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A319833
O.g.f. A(x) satisfies: [x^n] (1+x)^(n^2) / exp( n*A(x) ) = 0 for n >= 1.
3
1, 0, 4, 47, 911, 23333, 746033, 28562859, 1272633121, 64632002944, 3683087863632, 232628276945861, 16125634803535583, 1216918911172600488, 99302746177547793376, 8712136543494539283783, 817730822993776296035305, 81762932151410802804444230, 8676175483883206768183587574, 973819892539539309636148202259
OFFSET
1,3
COMMENTS
It is remarkable that this sequence should consist entirely of integers.
EXAMPLE
G.f.: A(x) = x + 4*x^3 + 47*x^4 + 911*x^5 + 23333*x^6 + 746033*x^7 + 28562859*x^8 + 1272633121*x^9 + 64632002944*x^10 + ...
The table of coefficients of x^k/k! in (1+x)^(n^3) / exp( n*A(x) ) begins
n=1: [1, 0, -1, -22, -1131, -109076, -16778045, -3756839514, ...];
n=2: [1, 2, 0, -56, -2632, -241152, -36192704, -7985674720, ...];
n=3: [1, 6, 27, 0, -5139, -449838, -63628713, -13580480196, ...];
n=4: [1, 12, 128, 1088, 0, -776864, -109331648, -22054994112, ...];
n=5: [1, 20, 375, 6430, 90485, 0, -176125925, -36313601350, ...];
n=6: [1, 30, 864, 23688, 603864, 12838464, 0, -55928077344, ...];
n=7: [1, 42, 1715, 67844, 2580333, 92216278, 2774895151, 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 + x^2/2! + 25*x^3/3! + 1225*x^4/4! + 115201*x^5/5! + 17478841*x^6/6! + 3880928521*x^7/7! + 1182410321905*x^8/8! + ...
PROG
(PARI) {a(n) = my(A=[1], m); for(i=1, n+1, A=concat(A, 0); m=#A; A[m] = Vec( (1+x +x^2*O(x^m))^(m^3) * exp(-m*x*Ser(A)) )[m+1]/m ); polcoeff( x*Ser(A), n)}
for(n=1, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 28 2018
STATUS
approved