A393170 - OEIS

%I #20 Mar 15 2026 10:41:57

%S 1,1,1,4,31,348,5042,88688,1824547,42820188,1126268086,32763077560,

%T 1043332897134,36074556857688,1345273816275396,53805147477433032,

%U 2297089670073393203,104254628953344306652,5012154610038856594222,254446503765279301626680,13601683265701053731130458,763693471378260675464295240

%N G.f. A(x) satisfies [x^n] 1/(A(x) - n*x) = 0 for n >= 1.

%C Conjecture: for n > 0, a(n) is odd iff n is a power of 2.

%C Conjecture: a(n) == 2 (mod 4) iff n/2 is a sum of two distinct powers of 2 (A018900).

%C a(n) ~ c*n^n where c appears to exist; c = 0.1469010938679... at n = 3000. What is an approximate value of this constant (if it exists and is finite)?

%C c = (exp(1) - 1)^2 / exp(3) = 0.1469959430660808807868671958665538372622476... - _Vaclav Kotesovec_, Mar 15 2026

%H Paul D. Hanna, <a href="/A393170/b393170.txt">Table of n, a(n) for n = 0..1024</a>

%F G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

%F (1) [x^n] 1/(A(x) - n*x) = 0 for n >= 1.

%F (2) [x^n] 1/A(x) = -Sum_{k=1..n} n^k * [x^(n-k)] 1/A(x)^(k+1) for n >= 1.

%e G.f.: A(x) = 1 + x + x^2 + 4*x^3 + 31*x^4 + 348*x^5 + 5042*x^6 + 88688*x^7 + 1824547*x^8 + 42820188*x^9 + 1126268086*x^10 + ...

%e where [x^n] 1/(A(x) - n*x) = 0 for n >= 1.

%e ILLUSTRATION OF DEFINITION.

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

%e   n = 0: [1,-1,  0,  -3,  -24,  -290, -4368,  -78799, ...];

%e   n = 1: [1, 0, -1,  -4,  -30,  -340, -4965,  -87756, ...];

%e   n = 2: [1, 1,  0,  -5,  -40,  -414, -5744,  -98745, ...];

%e   n = 3: [1, 2,  3,   0,  -42,  -506, -6801, -112744, ...];

%e   n = 4: [1, 3,  8,  17,    0,  -490, -7872, -130251, ...];

%e   n = 5: [1, 4, 15,  52,  146,     0, -7253, -145284, ...];

%e   n = 6: [1, 5, 24, 111,  480,  1690,     0, -129301, ...];

%e   n = 7: [1, 6, 35, 200, 1110,  5786, 24591,       0, ...];

%e   n = 8: [1, 7, 48, 325, 2168, 14094, 86224,  430041, 0, ...]; ...

%e in which the main diagonal, the coefficient of x^n in row n, is all zeros after the initial term.

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

%o A[#A] = polcoef(1/(Ser(A) - m*x), m) ); A[n+1]}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A393173.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Mar 05 2026