A204061 - OEIS

%I #16 Oct 30 2014 10:05:51

%S 1,1,5,21,101,501,2561,13345,70561,377281,2035285,11059205,60454005,

%T 332138405,1832677185,10150115201,56398558081,314273655745,

%U 1755700634981,9830544087221,55155558312901,310027473436821,1745567243959105,9843160519978401,55582528404717601

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

%C 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.

%H Paul D. Hanna, <a href="/A204061/b204061.txt">Table of n, a(n) for n = 0..625</a>

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

%F 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.

%F 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

%e G.f.: A(x) = 1 + x + 5*x^2 + 21*x^3 + 101*x^4 + 501*x^5 + 2561*x^6 +...

%e 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 +...

%e The last digit of the terms in this sequence seems to be either a '1' or a '5':

%e by conjecture, a(n) == 0 (mod 5) whenever n has a 2 in its base 5 expansion;

%e if true, terms a(2*5^k) through a(3*5^k - 1) all end with digit '5' for k>=0.

%o (PARI) {A001333(n)=polcoeff((1-x)/(1-2*x-x^2+x*O(x^n)),n)}

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

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

%Y Cf. A204062, A026933, A001333, A023729.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jan 10 2012