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

0,3

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.

Links

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

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

Example

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.

Prog

(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)}

Crossrefs

Cf. A204062, A026933, A001333, A023729.

Sequence in context: A050897 A338673 A199215 * A362556 A046633 A280623

Adjacent sequences: A204058 A204059 A204060 * A204062 A204063 A204064

Keyword

nonn

Author

Paul D. Hanna, Jan 10 2012

Status

approved