[seqfan] Fibonacci-sums
Gottfried Helms
Annette.Warlich at t-online.de
Wed Aug 18 13:59:48 CEST 2010
Reading the wikipedia-article on fibonacci-numbers
I tried what would be the alternating sum of
all fibonacci-numbers. The sequence is diverging,
so one has to employ a method of divergent summation.
Happily the growthrate is not too strong, so simple
Euler-summation suffices.
Using fib(0)=0,fib(1)=1, fib(2)=1, fib(3)=2,... I got
inf
sum (-1)^k * fib(k) = -1 // Eulersummation
k=0
Well, it is interesting, that it gives such a simple value.
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It calls for extension. I used powers of the index as cofactors
(additionally inverting the sign)
inf
sum - (-1)^k * fib(k) * k^n = a01(n) // Eulersummation
k=0
and got the following sequence a(n), n beginning at 0:
a01(0...5) = 1,2,8,50,416,4322
The sequence is known in OEIS:
* A000557: From Fibonacci sums.
* 1, 2, 8, 50, 416, 4322, 53888, 783890, 13031936, 243733442
The commentary gives a finite formula
* FORMULA: E.g.f.: 1/(1-2*sinh(x)). - Vladeta Jovovic (vladeta(AT)eunet.rs), Jul 06 2002
and
* a(n) = Sum[k=0..n, k!*Stirling2(n, k)*Fibonacci(k+2)].
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If I start with the convention
fib(0)=1,fib(1)=1, fib(2)=2, fib(3)=3,... this gives
a02(0..5) = 1, 1, 5, 31, 257, 2671,
Also this sequence is known to OEIS:
* A000556: Expansion of exp(-x) / [ 1 - exp(x) + exp(-x) ].
* 1, 1, 5, 31, 257, 2671, 33305, 484471, 8054177, 150635551,
The commentary gives a computation by finite sums:
* FORMULA: Sum(k!*fibonacci(k + 1)*stirling2(n, k), k = 0 .. n).
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Using binomials as cofactors I get the fibonacci-numbers again:
inf
sum - (-1)^k * fib(k) * binomial(k,n) = a03(n) // Eulersummation
k=0
a03(0..5) = 1,2,3,5,8,13
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Hmm. It would be nice to see, whether we can define values for the
infinite non-alternating sums....
With cofactors 1/n^k , abs(n) > 1 I get
inf n
sum fib(k) /n^k = a04(n) = -----------
k=0 n^2 - n -1
whose denominator looks familiar in context with fibonaccimumbers.
It seems valid also for fractional/real values of n. Also this
agrees with formula a01, if I insert n=-1 . So possibly I can
assume, that analytical continuation will allow also to insert
n=1 and to get the result a04(1) = -1 .
But how could this be proved/dealt with?
Gottfried Helms
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Pari/GP:
{ fibseq(k,f0=0) = if(k==0
,n1=f0;n2=1;n3=n1+n2
,n1=n2;n2=n3;n3=n1+n2);
return(n1) }
a01(n) = - sumalt(k=0,(-1)^k* fibseq(k,0) * k^n )
a02(n) = sumalt(k=0,(-1)^k* fibseq(k,1) * k^n )
a03(n) = - sumalt(k=0,(-1)^k* fibseq(k,0) * binomial(k,n) )
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