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

%I #12 Sep 03 2020 02:00:07

%S 1,1,13,907,153145,46602295,22140651001,15084920403375,

%T 13929456839705657,16740856184792482831,25396842996449548203625,

%U 47478179622583931337645823,107267415766722597731672066713,288206818852524037700531966913487

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

%C It is remarkable that this sequence consists entirely of integers.

%F Given o.g.f. A(x), define B(x) = A(x*B(x)), then B(x) is the o.g.f. of A337576 and satisfies [x^n] exp( n^2*(n+1)*x ) / B(x)^(n+1) = 0 for n>0.

%e O.g.f.: A(x) = 1 + x + 13*x^2 + 907*x^3 + 153145*x^4 + 46602295*x^5 + 22140651001*x^6 + 15084920403375*x^7 + 13929456839705657*x^8 + ...

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients of x^k/k! in exp( n*(n-1)^2 * x/A(x) ) begins:

%e n=0: [1, 0, 0, 0, 0, 0, 0, 0, ...];

%e n=1: [1, 0, 0, 0, 0, 0, 0, 0, ...];

%e n=2: [1, 2, 0, -160, -43520, -36711168, -67072065536, ...];

%e n=3: [1, 12, 120, 0, -293760, -234067968, -415963247616, ...];

%e n=4: [1, 36, 1224, 36288, 0, -792405504, -1355831322624, ...];

%e n=5: [1, 80, 6240, 467840, 31356160, 0, -3403785728000, ...];

%e n=6: [1, 150, 22200, 3229200, 456364800, 58514400000, 0, ...];

%e n=7: [1, 252, 63000, 15603840, 3817860480, 913835768832, 200316485182464, 0, ...]; ...

%e in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n*(n-1)^2*x/A(x) ) = 0 for n>0.

%e RELATED SERIES.

%e Define B(x) = A(x*B(x)), which begins

%e B(x) = 1 + x + 14*x^2 + 947*x^3 + 157190*x^4 + 47437866*x^5 + 22437363324*x^6 + 15246207565643*x^7 + ... + A337576(n)*x^n + ...

%e then the table of coefficients of x^k/k! in exp( n^2*(n+1)*x ) / B(x)^(n+1) begins:

%e n=0: [1, -1, -26, -5520, -3723384, -5652041280, -16083171669600, ...];

%e n=1: [1, 0, -54, -11200, -7486872, -11338403328, -32230618603040, ...];

%e n=2: [1, 9, 0, -18258, -11861352, -17522277048, -49272492906432, ...];

%e n=3: [1, 32, 916, 0, -17438424, -25288921344, -69043257103968, ...];

%e n=4: [1, 75, 5490, 363500, 0, -35101453320, -94993441197200, ...];

%e n=5: [1, 144, 20574, 2882400, 368064576, 0, -127110906431280, ...];

%e n=6: [1, 245, 59836, 14528010, 3470388768, 759773089152, 0, ...];

%e n=7: [1, 384, 147240, 56329472, 21453513648, 8058471570432, 2785824326725888, 0, ...]; ...

%e in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n^2*(n+1)*x ) / B(x)^(n+1) = 0 for n>0.

%e Also note that B(x) = (1/x)*Series_Reversion( x/A(x) ) and A(x) = B(x/A(x)).

%o (PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(m*(m-1)^2*x/Ser(A) ))[m+1]/(m*(m-1)^2) );A[n+1]}

%o for(n=0, 20, print1(a(n), ", "))

%Y Cf. A337576, A337457, A337577.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Sep 02 2020