A303563 G.f. A(x) satisfies: 0 = [x^(n-1)] 1 / A(x)^(n^2-n+1) for n>2.

1, 1, 4, 21, 132, 840, 6798, 26187, 982794, -23411010, 1229958800, -63402693620, 3727765284702, -241049598495378, 17055417754898346, -1310070866036785677, 108572840286328367574, -9656468474317765916970, 917437511140569561151848, -92733586081750860360411954, 9936829948115042380890921976, -1125196473407637775842431681496, 134258639558360961220987962351588

Offset: 0

Author

Paul D. Hanna, Apr 27 2018

Keywords

• sign

Comments

a(n) is odd iff n = 2^k - 1 for k >= 0 (conjecture).

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 21*x^3 + 132*x^4 + 840*x^5 + 6798*x^6 + 26187*x^7 + 982794*x^8 - 23411010*x^9 + 1229958800*x^10 + ...
such that the coefficient of x^(n-1) in 1/A(x)^(n^2-n+1) equals zero for n>2.
RELATED SERIES.
1/A(x) = 1 - x - 3*x^2 - 14*x^3 - 85*x^4 - 504*x^5 - 4424*x^6 - 6796*x^7 - 878157*x^8 + ... + (-1)^n*A230218(n)*x^n + ...
A'(x)/A(x) = 1 + 7*x + 52*x^2 + 427*x^3 + 3286*x^4 + 32938*x^5 + 108718*x^6 + 7379059*x^7 - 221516750*x^8 + 12494464362*x^9 - 710385713212*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in 1/A(x)^(n^2-n+1) begins:
n=1: [1, -1, -3, -14, -85, -504, -4424, -6796, -878157, ...];
n=2: [1, -3, -6, -25, -153, -819, -8664, 18360, ...];
n=3: [1, -7, 0, -7, -98, -210, -10122, 141525, ...];
n=4: [1, -13, 39, 0, -78, 819, -15483, 380952, ...];
n=5: [1, -21, 147, -364, 0, 2457, -35805, 821916, ...];
n=6: [1, -31, 372, -2139, 5580, 0, -91698, 1792947, ...];
n=7: [1, -43, 774, -7525, 42097, -125517, 0, 4097298, ...];
n=8: [1, -57, 1425, -20482, 185877, -1089270, 3791298, 0, ...]; ...
in which the main diagonal equals zeros after the initial terms, illustrating that 0 = [x^(n-1)] 1/A(x)^(n^2-n+1) for n>2.
RELATED TABLES.
The table of coefficients of x^k in (x*A(x)^n)' / A(x)^(n^2+1) begins:
n=1: [1, 0, 3, 28, 255, 2016, 22120, 40776, ...];
n=2: [1, -1, 2, 25, 255, 1911, 25992, -67320, ...];
n=3: [1, -4, 0, 2, 70, 240, 15906, -283050, ...];
n=4: [1, -9, 15, 0, 18, -441, 13101, -439560, ...];
n=5: [1, -16, 77, -104, 0, -468, 15345, -547944, ...];
n=6: [1, -25, 228, -897, 1260, 0, 14790, -636207, ...];
n=7: [1, -36, 522, -3850, 14685, -23352, 0, -571716, ...];
n=8: [1, -49, 1025, -11858, 81525, -324870, 598626, 0,  ...]; ...
in which the main diagonal equals zeros after the initial terms, illustrating that 0 = [x^(n-1)] (x*A(x)^n)' / A(x)^(n^2+1) for n>2.
The table of coefficients of x^k in (x*A(x)^n)' / A(x)^(n^2) begins:
n=1: [1, 1, 7, 52, 427, 3286, 32938, 108718, ...];
n=2: [1, 0, 5, 44, 399, 3016, 35670, -3960, ...];
n=3: [1, -3, 0, 7, 120, 630, 19906, -265455, ...];
n=4: [1, -8, 10, 0, 21, -456, 13950, -450240, ...];
n=5: [1, -15, 65, -70, 0, -539, 16215, -566100, ...];
n=6: [1, -24, 207, -748, 882, 0, 16887, -665604, ...];
n=7: [1, -35, 490, -3451, 12299, -17017, 0, -645000, ...];
n=8: [1, -48, 980, -11008, 72870, -274880, 451776, 0, ...]; ...
in which the main diagonal equals zeros after the initial term, illustrating that 0 = [x^(n-1)] (x*A(x)^n)' / A(x)^(n^2) for n>1.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A303562, A230218, A292877.

Programs

PARI

{a(n) = my(A=[1, 1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( (x*Ser(A)^m)'/Ser(A)^(m^2+1) )[m] ); A[n+1]}
for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) satisfies:

(1) 0 = [x^(n-1)] 1 / A(x)^(n^2-n+1) for n>2.

(2) 0 = [x^(n-1)] (x*A(x))' / A(x)^(n^2-n+2) for n>2.

(3) 0 = [x^(n-1)] (x*A(x)^n)' / A(x)^(n^2+1) for n>2.

(4) 0 = [x^(n-1)] (x*A(x)^(n-1))' / A(x)^(n^2) for n>2.

(5) 0 = [x^(n-1)] (x*A(x)^n)' / A(x)^(n^2) for n>1.