OFFSET
0,1
COMMENTS
Conjecture: a(n) is odd iff n is of the form 2^k - 1 (k >= 0).
Conjecture: a(3*n+1) == A000108(n) (mod 3) for n >= 0 (see A111019 and A111020), with a(3*n) == a(3*n+2) == 0 (mod 3) for n >= 0.
The 10-adic integer sum A(x=10) = 9 + 19*10 + 48*10^2 + 135*10^3 + 406*10^4 + 1278*10^5 + ... equals ...999999999, the 10-adic integer equivalent of -1. This is because the 10-adic integer C = Sum_{n>=0} A000108(n)*10^n (A395834) satisfies C = 1 + 10*C^2 and so C * (10*C - 1) = ...999999999. The motivation for the g.f. 10*C(x)^2 - C(x) of this sequence was to illustrate a sequence with the 10-adic integer value Sum_{n>=0} a(n)*10^n == -1.
FORMULA
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x) = 10*C(x)^2 - C(x), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
(2) A(x) = (10 - 21*x - (10-x)*sqrt(1 - 4*x)) / (2*x^2).
(3) a(n) = 10*binomial(2*n+2,n)/(n+1) - binomial(2*n+1,n)/(2*n+1).
(4) Sum_{k=0..n} a(k)*10^k = 10^(n+1)*A000108(n+1) - 1 for n >= 0.
(5) Sum_{n>=0} a(n)*10^n = ...9999999 == -1 as a 10-adic integer sum.
EXAMPLE
G.f.: A(x) = 9 + 19*x + 48*x^2 + 135*x^3 + 406*x^4 + 1278*x^5 + 4158*x^6 + 13871*x^7 + 47190*x^8 + 163098*x^9 + ...
such that A(x) = 10*C(x)^2 - C(x) where C(x) begins
C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + 1430*x^8 + ... + A000108(n)*x^n + ...
PROG
(PARI) {a(n) = 10*binomial(2*n+2, n)/(n+1) - binomial(2*n+1, n)/(2*n+1)}
for(n=0, 30, print1(a(n), ", "))
(PARI) {a(n) = my(A = (10 - 21*x - (10-x)*sqrt(1-4*x +x^3*O(x^n))) / (2*x^2)); polcoef(GF=A, n)}
{upto(n) = a(n); Vec(GF)}
upto(30)
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, May 17 2026
STATUS
approved