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A092687
First column and main diagonal of triangle A092686, in which the convolution of each row with {1,2} produces a triangle that, when flattened, equals the flattened form of A092686.
7
1, 2, 6, 16, 46, 132, 384, 1120, 3278, 9612, 28236, 83072, 244752, 722048, 2132704, 6306304, 18666190, 55300732, 163968612, 486528288, 1444571068, 4291629384, 12756459936, 37934818112, 112855778768, 335867740704, 999895548736
OFFSET
0,2
COMMENTS
Conjecture: Limit n->infinity a(n)^(1/n) = 3. - Vaclav Kotesovec, Jun 29 2015
LINKS
FORMULA
G.f. satisfies: A(x) = A( x^2/(1-2x) )/(1-2x). Recurrence: a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*2^(n-2k)*a(k). - Paul D. Hanna, Jul 10 2006
MATHEMATICA
m = 27; A[_] = 1; Do[A[x_] = A[x^2/(1-2x)]/(1-2x) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* Jean-François Alcover, Nov 03 2019 *)
PROG
(PARI) T(n, k)=if(n<0||k>n, 0, if(n==0&k==0, 1, if(n==1&k<=1, 2, if(k==n, T(n, 0), 2*T(n-1, k)+T(n-1, k+1)))))
a(n)=T(n, 0)
for(n=0, 30, print1(a(n), ", "))
(PARI) a(n)=local(A=1+x); for(i=0, n\2, A=subst(A, x, x^2/(1-2*x+x*O(x^n)))/(1-2*x)); polcoeff(A, n) \\ Paul D. Hanna, Jul 10 2006
(PARI) /* Using Recurrence: */
a(n)=if(n==0, 1, sum(k=0, n\2, binomial(n-k, k)*2^(n-2*k)*a(k)))
for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Jul 10 2006
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 04 2004
STATUS
approved