tribonacci numbers
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From: Chip Eastham on 25 Mar 2010 08:57 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
From: emath on 25 Mar 2010 09:27 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- Alýntýyý gizle - > > - Alýntýyý göster - Thank you. But what do you mean by combination of six geometric series.
From: Valeri Astanoff on 25 Mar 2010 10:38 On 25 mar, 12:57, emath <emath...(a)gmail.com> wrote: > 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. Good day, Have a look at Sloane's integer sequences, and at Plouffe's inverter. Let f(n) the sum of squares of tribonacci numbers. Then (asymptotically) f(n) = tau * f(n-1) with tau real zero of 1 + x + 3*x^2 - x^3 tau = (1/3)(3+(6(9-sqrt(33)))^(1/3)+(6(9+sqrt(33)))^(1/3)) tau = 3.3829757679... A (very!) appproximate formula is then f(n) = 31514*tau^(n-1) [ beginning at n=10 for a better accuracy ] hth -- Valeri Astanoff
From: Valeri Astanoff on 25 Mar 2010 10:40 On 25 mar, 15:38, Valeri Astanoff <astan...(a)gmail.com> wrote: > On 25 mar, 12:57, emath <emath...(a)gmail.com> wrote: > > > 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. > > Good day, > > Have a look at Sloane's integer sequences, > and at Plouffe's inverter. > Let f(n) the sum of squares of tribonacci numbers. > Then (asymptotically) f(n) = tau * f(n-1) > with tau real zero of 1 + x + 3*x^2 - x^3 > tau = (1/3)(3+(6(9-sqrt(33)))^(1/3)+(6(9+sqrt(33)))^(1/3)) > tau = 3.3829757679... > > A (very!) appproximate formula is then f(n) = 31514*tau^(n-1) > [ beginning at n=10 for a better accuracy ] > > hth > > -- > Valeri Astanoff Sorry! Please read : 31514*tau^(n - 10) v.a.
From: Gottfried Helms on 25 Mar 2010 11:01
Am 25.03.2010 12:56 schrieb emath: > Hi, > Is there any formulae for sum of squares of fibonacci numbers from 0 > to n. I have a funny idea, however it does not sum the squares but only the original values. Maybe you can find a translation to the squares-problem. Anyway - here it goes... --------------------------------------------------------- I look at the problem using matrices. Call the tribonacci- generation-matrix T: T = 1 1 0 1 0 1 1 0 0 Then [1,0,0] * T = [1,1,0] [1,1,0] * T = [2,1,1] [2,1,1] * T = [4,2,1] ... [13,7,4] * T = [24,13,7] ... and so on, giving the sequence of tribonacci-numbers [0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, ...] --------------------- This is the same as using the powers of T. If we index the tribonacci-numbers by t(0), t(1),t(2),... then [t(2),t(1),t(0)] * T^n = [t(2+n),t(1+n),t(n)] If we want the sum of tribonacci-numbers, we can do [1,1,0] * (T^0 + T^1 + T^2 + ... + T^n) = [s(2,n+2),s(1,n+1),s(0,n)] where s(a,b) = sum of tribonacci-numbers with indexes a to b If we want a contiguous segment, then we could express the parenthese containing the powers of T as difference of two geometric series: [1,1,0] * (T^0 + T^1 + T^2 + T^3 + T^4 + ... - (T^4 + T^5 + ...) = [s(2,inf) - s(6,inf), #, # ] Now the geometric series of T seems to make sense in this case, we can define a matrix TZeta analoguously to that of the closed form of a geometric series of a scalar: TZeta = T^0 + T^1 + T^2 + T^3 + T^4 + ... = ( I - T )^-1 to rewrite the above: [1,1,0] * ( T^0 * TZeta - T^4 * TZeta) = [s(2,inf) - s(6,inf), # , # ] or even simpler: [1,1,0] * (T^0 - T^4) * TZeta = [s(2,inf) - s(6,inf), *, * ] This gives the remarkably simple matrix TZeta = -1/2 -1/2 -1/2 -1 0 0 -1/2 -1/2 1/2 --------------------------------------------- The above geometric series have a different head; we can define the geometric series beginning at the a'th power as T^a*TZeta. Recall the sequence: [0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, ...] We get the following results for some examples [1,1,0] * (T^0 - T^4) * TZeta = [14, 8, 4 ] where 14=1+2+4+7 8=1+1+2+4 4=0+1+1+2 [1,1,0] * (T^2 - T^4) * TZeta = [11, 6, 3 ] where 11= 4+7 6= 2+4 3= 1+2 [1,1,0] * (T^100 - T^200) * TZeta = [115321754909823101063964621922292627920038147909653186 , 62699171068839331460199441502898752019054568568280184 , 34088850415028854295167352915976781995775535601697088 ] where we found the sums of the tribonacci-numbers t(102)+...+t(202) and so on. (note, that we could use the eigen-decomposition of T to calculate that high powers (*)) Using this method, we have the infinite "sums" of all tribonacci-numbers sum {k=2,inf} t(k) = -3/2 sum {k=1,inf} t(k) = -1/2 sum {k=0,inf} t(k) = -1/2 if we simply apply [1,1,0]*TZeta = [-3/2, -1/2, -1/2] ------------ Well, as I said in the beginning: don't know how to apply this to the sum-of-squared-tribonacci-problem. Perhaps we can find another not too complicated transfer-matrix which produces the squared-tribonaccis in the analoguous way. Gottfried Helms (*) The eigen-decomposition allows then the continuation to fractional indexes as well as far as the fractional powers for the eigenvalues are meaningful. |