A350410 - OEIS

%I #5 Dec 30 2021 01:57:25

%S -1,1,3,12,116,-534,-19083,4653416,8687989872,-128647855586426,

%T -17178731504646708274,22932060477876906089601744,

%U 336748983792218322012229442319058,-59340140012235302782558826850918965528904,-135935979554634873389993183019614435570206186506348,4359616519227623971178976912185258573259921937219097427608884

%N G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies: 0 = [x^n] Sum_{m=0..n} A(x)^(m^2) / m!, for n > 1.

%C This sequence is conjectured to consist entirely of integers.

%e G.f.: A(x) = -1 + x + 3*x^2 + 12*x^3 + 116*x^4 - 534*x^5 - 19083*x^6 + 4653416*x^7 + 8687989872*x^8 - 128647855586426*x^9 + ...

%e The table of coefficients of x^k in A(x)^(n^2), k >= 0, begins:

%e n=1: [-1, 1, 3, 12, 116, -534, -19083, 4653416, ...];

%e n=2: [1, -4, -6, -16, -301, 3720, 72702, -18849144, ...];

%e n=3: [-1, 9, -9, -24, 486, -11844, -124941, 43265088, ...];

%e n=4: [1, -16, 72, -32, -1116, 27216, 75136, -78475712, ...];

%e n=5: [-1, 25, -225, 800, 1050, -58320, 280175, 123329600, ...];

%e n=6: [1, -36, 522, -3792, 11259, 90792, -1375170, -171395640, ...];

%e n=7: [-1, 49, -1029, 11956, -79184, 141414, 3229737, 203778120, ...];

%e n=8: [1, -64, 1824, -30336, 319504, -1977792, 1553280, -204791552, ...]; ...

%e in which, by definition, the following sums along the columns equal zero:

%e 0 = (3)/1! + (-6)/2! ;

%e 0 = (12)/1! + (-16)/2! + (-24)/3! ;

%e 0 = (116)/1! + (-301)/2! + (486)/3! + (-1116)/4! ;

%e 0 = (-534)/1! + (3720)/2! + (-11844)/3! + (27216)/4! + (-58320)/5! ;

%e 0 = (-19083)/1! + (72702)/2! + (-124941)/3! + (75136)/4! + (280175)/5! + (-1375170)/6! ;

%e 0 = (4653416)/1! + (-18849144)/2! + (43265088)/3! + (-78475712)/4! + (123329600)/5! + (-171395640)/6! + (203778120)/7! ; ...

%e One may continue the above pattern to determine all the terms of this sequence.

%o (PARI) {a(n) = my(A=[-1,1]); for(i=1,n, A=concat(A,0);

%o A[#A] = (-1)^(#A-1)*(#A-2)! * polcoeff( sum(m=0,#A-1, Ser(A)^(m^2) / m!) ,#A-1) );A[n+1]}

%o for(n=0,15,print1(a(n),", "))

%Y Cf. A350411, A350412.

%K sign

%O 0,3

%A _Paul D. Hanna_, Dec 29 2021