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A014335
Exponential convolution of Fibonacci numbers with themselves (divided by 2).
12
0, 0, 1, 3, 11, 35, 115, 371, 1203, 3891, 12595, 40755, 131891, 426803, 1381171, 4469555, 14463795, 46805811, 151466803, 490156851, 1586180915, 5132989235, 16610702131, 53753361203, 173949530931, 562912506675, 1821623137075, 5894896300851, 19076285150003
OFFSET
0,4
COMMENTS
It can be noticed that A014335/A011782 is an "autosequence", that is a sequence which is identical to its inverse binomial transform, except for alternating signs. - Jean-François Alcover, Jun 15 2016
FORMULA
a(n) = A014334(n)/2.
G.f.: x^2/((1-x)*(1-2*x-4*x^2)). - Vladeta Jovovic, Mar 05 2003
E.g.f.: exp(x)*(cosh(sqrt(5)*x)-1)/5. - Vladeta Jovovic, Sep 01 2004
From Benoit Cloitre, Sep 25 2004: (Start)
a(n+1) = Sum_{i=0..n} A000045(i)*2^(i-1).
a(n) = (1/5)*(2^(n-1)*A000032(n) - 1). (End)
a(n) = 2*a(n-1) + 4*a(n-2) + 1, a(0)=0; a(1)=0. - Zerinvary Lajos, Dec 14 2008
G.f.: G(0)*x^2/(2*(1-x)^2), where G(k)= 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n) = (A203579(n) - 2)/5. - Vladimir Reshetnikov, Oct 06 2016
MAPLE
a[0]:=0:a[1]:=0:for n from 2 to 50 do a[n]:=2*a[n-1]+4*a[n-2]+1 od: seq(a[n], n=0..29); # Zerinvary Lajos, Dec 14 2008
# second Maple program:
a:= n-> (<<0|1|0>, <0|0|1>, <-4|2|3>>^n)[1, 3]:
seq(a(n), n=0..30); # Alois P. Heinz, Oct 04 2016
MATHEMATICA
LinearRecurrence[{3, 2, -4}, {0, 0, 1}, 41] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
Table[(2^n LucasL[n] - 2)/10, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 06 2016 *)
PROG
(Magma) [(2^n*Lucas(n)-2)/10: n in [0..40]]; // G. C. Greubel, Jan 06 2023
(SageMath) [(2^n*lucas_number2(n, 1, -1) -2)/10 for n in range(41)] # G. C. Greubel, Jan 06 2023
CROSSREFS
Cf. (partial sums of) A063727.
Column k=2 of A346415.
Sequence in context: A107683 A259400 A320087 * A147474 A247417 A222286
KEYWORD
nonn,easy
STATUS
approved