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A081186
4th binomial transform of (1,0,1,0,1,...), A059841.
14
1, 4, 17, 76, 353, 1684, 8177, 40156, 198593, 986404, 4912337, 24502636, 122336033, 611148724, 3054149297, 15265963516, 76315468673, 381534296644, 1907542343057, 9537324294796, 47685459212513, 238423809278164, 1192108586037617, 5960511549128476
OFFSET
0,2
COMMENTS
Binomial transform of A007582.
a(n) is a companion to A005059(n): a(n) + A005059(n) = 5^n; e.g. a(4) = A005059(4) = 353 + 272 = 625 = 5^4. - Gary W. Adamson, Jun 30 2006
Number of words of length n from an alphabet of 5 letters in which a chosen letter appears an even number of times. - James Mahoney, Feb 03 2012 [See a comment in A007582, also for crossrefs. for the 1- to 11-letter word cases. - Wolfdieter Lang, Jul 17 2017]
The sequence of fractions x(n) = a(n+1)/a(n) satisfies a simple recurrence x(n+1) = 108 - (815 - 1500 / x(n-1)) / x(n) known as Muller's recurrence. It is used for the demonstration of an unexpected failure of floating-point computations. - Andrey Zabolotskiy, Sep 17 2019
FORMULA
a(n) = 8*a(n-1) - 15*a(n-2) with n>1, a(0)=1, a(1)=4.
G.f.: (1-4*x)/((1-3*x)*(1-5*x)).
a(n) = (3^n + 5^n)/2.
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k)*4^(n-2*k).
E.g.f.: exp(4*x) * cosh(x). - Paul Barry, Oct 06 2004
EXAMPLE
Say the alphabet is {a,b,c,d,e} and we want to know how many words of length one and two contain c an even number of times. a(1) = 4, which we can see by the four words {(a),(b),(d),(e)} and a(2) = 17, which we can see by the seventeen words {(a,a), (a,b), (a,d), (a,e), (b,a), (b,b), (b,d), (b,e), (c,c), (d,a), (d,b), (d,d), (d,e), (e,a), (e,b), (e,d), (e,e)}. - James Mahoney, Feb 03 2012
MAPLE
seq( (3^n + 5^n)/2, n=0..30); # G. C. Greubel, Dec 26 2019
MATHEMATICA
CoefficientList[Series[(1-4x)/((1-3x)(1-5x)), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 07 2013 *)
LinearRecurrence[{8, -15}, {1, 4}, 30] (* Harvey P. Dale, Apr 13 2019 *)
PROG
(Magma) [3^n/2+5^n/2: n in [0..25]]; // Vincenzo Librandi, Aug 07 2013
(PARI) vector(31, n, (3^(n-1) + 5^(n-1))/2 ) \\ G. C. Greubel, Dec 26 2019
(Sage) [(3^n + 5^n)/2 for n in (0..25)] # G. C. Greubel, Dec 26 2019
(GAP) List([0..25], n-> (3^n + 5^n)/2); # G. C. Greubel, Dec 26 2019
CROSSREFS
Sequence in context: A081910 A379823 A026773 * A239204 A005572 A202879
KEYWORD
nonn,easy
AUTHOR
Paul Barry, Mar 11 2003
STATUS
approved