A204061 G.f.: exp( Sum_{n>=1} A001333(n)^2 * x^n/n ) where A001333(n) = A002203(n)/2, one-half the companion Pell numbers.

1, 1, 5, 21, 101, 501, 2561, 13345, 70561, 377281, 2035285, 11059205, 60454005, 332138405, 1832677185, 10150115201, 56398558081, 314273655745, 1755700634981, 9830544087221, 55155558312901, 310027473436821, 1745567243959105, 9843160519978401, 55582528404717601

Offset: 0

Author

Paul D. Hanna, Jan 10 2012

Keywords

• nonn

Comments

a(n) == 1 (mod 5) iff n has no 2's in its base 5 expansion (A023729), otherwise a(n) == 0 (mod 5); this is a conjecture needing proof.

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 21*x^3 + 101*x^4 + 501*x^5 + 2561*x^6 +...
where log(A(x)) = x + 3^2*x^2/2 + 7^2*x^3/3 + 17^2*x^4/4 + 41^2*x^5/5 + 99^2*x^6/6 + 239^2*x^7/7 +...+ A001333(n)^2*x^n/n +...
The last digit of the terms in this sequence seems to be either a '1' or a '5':
by conjecture, a(n) == 0 (mod 5) whenever n has a 2 in its base 5 expansion;
if true, terms a(2*5^k) through a(3*5^k - 1) all end with digit '5' for k>=0.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..625

Crossrefs

Cf. A204062, A026933, A001333, A023729.

Programs

PARI

{A001333(n)=polcoeff((1-x)/(1-2*x-x^2+x*O(x^n)),n)}
{a(n)=polcoeff(exp(sum(k=1, n, A001333(k)^2*x^k/k)+x*O(x^n)), n)}

PARI

{a(n)=polcoeff(1/(sqrt(1+x+x*O(x^n))*(1-6*x+x^2+x*O(x^n))^(1/4)),n)}

Formula

G.f.: 1 / ( sqrt(1+x) * (1-6*x+x^2)^(1/4) ).

Self-convolution yields A026933: Sum_{k=0..n} a(n-k)*a(k) = Sum_{k=0..n} D(n-k,k)^2 where D(n,k) = A008288(n,k) are the Delannoy numbers.

a(n) ~ 2^(1/8) * GAMMA(3/4) * (3+2*sqrt(2))^(n+1/2) / (4 * Pi * n^(3/4)). - Vaclav Kotesovec, Oct 30 2014