tribonacci numbers
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From: Chip Eastham on 25 Mar 2010 11:08 On Mar 25, 9:27 am, emath <emath...(a)gmail.com> wrote: > On 25 Mart, 14:57, Chip Eastham <hardm...(a)gmail.com> wrote: > > > > > On Mar 25, 8:24 am, emath <emath...(a)gmail.com> wrote: > > > > On 25 Mart, 14:06, Dan Cass <dc...(a)sjfc.edu> wrote: > > > > > > Hi, > > > > > Is there any formulae for sum of squares of fibonacci > > > > > numbers from 0 > > > > > to n. > > > > > Well if f(1)=1 and f(2)=1 and later f(n)=f(n-1)+f(n-2), > > > > as it is usually defined, then one may extend > > > > to f(0)=0 and preserve the recursion. > > > > So summing the squares "from 0th to nth" > > > > is the same as summing from the 1st to nth. > > > > Then f(1)^2 + f(2)^2 + ... + f(n)^2 > > > > comes out the same as f(n)*f(n+1). > > > > Thank you Dan. But I need a formulae for the sum of squares of > > > tribonacci numbers from. > > > Tribonacci sequence is defined as follows: > > > T_0=0, > > > T_1=1, > > > T_2=1, > > > T_3=2, > > > T_n=T_{n-1}+T_{n-2}+T_{n-3} for n\geq 4. > > > In principle such problems may be solved by > > explicitly expressing so-called tribonacci > > numbers: T_k = a*q^k + b*r^k + c*s^k where > > q,r,s are the roots of the characteristic > > polynomial X^3 - X^2 - X - 1 and constants > > a,b,c are chosen to satisfy the initial > > conditions T_1 = T_2 = 1 and T_0 = 0. > > Then we also have an explict expression > > for (T_k)^2 and it follows that the sum > > of the first n squares is a combination > > of six (finite) geometric series. > > > regards, chip > > Thank you. But what do you mean by > combination of six geometric series. Hi, emath: I mean that since: (T_k)^2 = (a*q^k + b*r^k + c*s^k)^2 = (a^2)*(q^2)^k + (b^2)*(r^2)^k + (c^2)*(s^2)^k + 2ab*(q*r)^k + 2ac*(q*s)^k + 2bc*(r*s)^k summing T_k from k = 1 to n (say) would amount to summing the six corresponding geometric series. Note that T_0 is zero and can be omitted from the sum without loss of generality. regards, chip
From: Henry on 25 Mar 2010 11:29 On 25 Mar, 13:27, emath <emath...(a)gmail.com> wrote: > On 25 Mart, 14:57, Chip Eastham <hardm...(a)gmail.com> wrote: > > On Mar 25, 8:24 am, emath <emath...(a)gmail.com> wrote: > > > Thank you Dan. But I need a formulae for the sum of squares of > > > tribonacci numbers from. > > > Tribonacci sequence is defined as follows: > > > T_0=0, > > > T_1=1, > > > T_2=1, > > > T_3=2, > > > T_n=T_{n-1}+T_{n-2}+T_{n-3} for n\geq 4. > > > In principle such problems may be solved by > > explicitly expressing so-called tribonacci > > numbers: T_k = a*q^k + b*r^k + c*s^k where > > q,r,s are the roots of the characteristic > > polynomial X^3 - X^2 - X - 1 and constants > > a,b,c are chosen to satisfy the initial > > conditions T_1 = T_2 = 1 and T_0 = 0. > > Then we also have an explict expression > > for (T_k)^2 and it follows that the sum > > of the first n squares is a combination > > of six (finite) geometric series. > > > Thank you. But what do you mean by combination of six geometric series. He means the sum of six terms of the form u + u*v + u*v^2 + ... + u*v^(n-1), i.e. of the form u*(1-v^n)/(1-v), which can also be the sum of six terms of the form w*v^n plus a constant. Two of the roots of the polynomial X^3 - X^2 - X - 1 are complex, which would make an exact formula slightly more complicated. The real root is about 1.8392867552... and the square of that is about 3.3829757679... as Valeri said earlier so an approximate formula for the sum of squares seems to be about 0.1604897563*3.3829757679^n
From: Robert Israel on 25 Mar 2010 17:07 emath <emathgzl(a)gmail.com> writes: > On 25 Mart, 13:56, emath <emath...(a)gmail.com> wrote: > > Hi, > > Is there any formulae for sum of squares of fibonacci numbers from 0 > > to n. > > Sorry, my problem is for tribonacci numbers. See <http://www.research.att.com/~njas/sequences/A107239>. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Robert Israel on 26 Mar 2010 03:30
Robert Israel <israel(a)math.MyUniversitysInitials.ca> writes: > emath <emathgzl(a)gmail.com> writes: > > > On 25 Mart, 13:56, emath <emath...(a)gmail.com> wrote: > > > Hi, > > > Is there any formulae for sum of squares of fibonacci numbers from 0 > > > to n. > > > > Sorry, my problem is for tribonacci numbers. > > See <http://www.research.att.com/~njas/sequences/A107239>. Explicitly, the formula is 1/4 - 1/11 sum_{r} (3+7r+5r^2)/(3r^2-2r-1) r^(-n) - 1/44 sum_{s} (-1-2s-9s^2)/(3s^2+2s+3) s^(-n) where the first sum is over the three roots r of the polynomial z^3 - z^2 - z - 1 and the second sum is over the three roots s of the polynomial z^3 + z^2 + 3 z - 1. Or in Maple's notation, 1/4 -1/11*sum((3+7*_R+5*_R^2)/(3*_R^2-2*_R-1)*_R^(-n), _R = RootOf(_Z^3-_Z^2-_Z-1)) -1/44*sum((-1-2*_R-9*_R^2)/(3*_R^2+2*_R+3)*_R^(-n), _R = RootOf(_Z^3+_Z^2+3*_Z-1)) -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada |