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A000285
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a(0) = 1, a(1) = 4, and a(n) = a(n-1) + a(n-2) for n>=2.
(Formerly M3246 N1309)
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25
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1, 4, 5, 9, 14, 23, 37, 60, 97, 157, 254, 411, 665, 1076, 1741, 2817, 4558, 7375, 11933, 19308, 31241, 50549, 81790, 132339, 214129, 346468, 560597, 907065, 1467662, 2374727, 3842389, 6217116, 10059505
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| a(n-1)=sum(P(4;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=3. These are the sums over the SW-NE diagonals in P(4;n,k), the (4,1) Pascal triangle A093561. Observation by Paul Barry (pbarry(AT)wit.ie), Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums in the Pascal (1,3) triangle A095660.
In general, for a Fibonacci sequence beginning with 1,b we have
a(n)=(2^(-1-n)((1-Sqrt[5])^n(1+Sqrt[5]-2b)+(1+Sqrt[5])^n(-1+Sqrt[5]+2b)))/Sqrt[5]. In this case we have b=4. - Herbert Kociemba(kociemba(AT)t-online.de), Dec 18 2011
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REFERENCES
| N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 53.
J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. Brousseau, Seeking the lost gold mine or exploring Fibonacci factorizations, Fib. Quart., 3 (1965), 129-130.
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LINKS
| T. D. Noe, Table of n, a(n) for n=0..500
Index entries for sequences related to linear recurrences with constant coefficients
Tanya Khovanova, Recursive Sequences
S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
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FORMULA
| G.f.: (1+3*x)/(1-x-x^2).
Row sums of A131775 starting (1, 4, 5, 9, 14, 23,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 14 2007
a(n)=2*Fibonacci(n)+Fibonacci(n+2) - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 05 2007
a(n)=((1+sqrt(5))^n-(1-sqrt(5))^n)/(2^n*sqrt(5))+ 3/2*((1+sqrt(5))^(n-1)-(1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=5. [From Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009]
a(n)=3*Fibonacci(n+2)-2*Fibonacci(n+1) [From Gary Detlefs (gdetlefs(AT)aol.com) Dec 21 2010]
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MAPLE
| A000285:=-(1+3*z)/(-1+z+z**2); [S. Plouffe in his 1992 dissertation.]
with(combinat):a:=n->2*fibonacci(n)+fibonacci(n+2): seq(a(n), n=0..34);
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MATHEMATICA
| a=1; lst={a}; s=6; Do[a=s-(a+1); AppendTo[lst, a]; s+=a, {n, 5!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Oct 27 2009]
LinearRecurrence[{1, 1}, {1, 4}, 40] (* or *) Table[(3*LucasL[n]- Fibonacci[n])/2, {n, 40}] (* From Harvey P. Dale, Jul 18 2011 *)
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PROG
| (Haskell)
a000285 n = a000285_list !! n
a000285_list = 1 : 4 : zipWith (+) a000285_list (tail a000285_list)
-- Reinhard Zumkeller, Apr 28 2011
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CROSSREFS
| Essentially the same as A104449.
a(n) = A101220(3, 0, n+1).
a(n) = A109754(3, n+1).
a(k) = A090888(2, k-1), for k > 0.
Cf. A131775.
Sequence in context: A096818 A038099 A120740 * A042031 A041493 A042765
Adjacent sequences: A000282 A000283 A000284 * A000286 A000287 A000288
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KEYWORD
| easy,nonn,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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