Sum of Fibonacci numbers

Question

I found this task here .

Given the ith (1<=i<=35) Fibonacci number F(i) calculate the sum of the ith till i+9th number F(i)+F(i+1)+...+F(i+9) and the last digit of the i+246th one F(i+246)

I have been trying to solve this using python and some tricks(Binnet's formula and a tricky recurrence):

 f=lambda n:((1+5**.5)**n-(1-5**.5)**n)/(2**n*5**.5)
 exec"n=input();print int(55*f(n)+88*f(n+1)+f(n+6)%10);"*input()

but I didn't yet managed to squeeze thought the give source code limit which is 111 and mine is 115,any hints how to improve my solution?

I am a rather newbie to python so any sort of help resulting in a successful solution will be much appreciated.

Thanks,

Answer 1

Did you try to use this sum formula?

http://en.wikipedia.org/wiki/Fibonacci_number#Second_identity ("Second Identity")?

Answer 2

f = lambda n,t=5**.5:((1+t)**n-(1-t)**n)/(2**n*t) etc. spends 8 characters ,t=5**.5 to gain 12: three lots of 5**.5 -> t . That's a saving of 4 characters, which seems to be what you require.

[EDITED to correct a typo; I had 2*n instead of 2**n in the denominator.]

You can save a few more characters with a different twist on Binet's formula: f=lambda n:round((1+5**.5)**n/5**.5/2**n) .

Answer 3

Here is the 110 solution, I had to rewrite the formula though and used @Gareth's suggestion:

p=5**.5
f=lambda n:((1+p)**n-(1-p)**n)/(2**n*p)
exec "n=input();print int(f(n+11)-f(n+1)+f(n+6)%10);"*input()

Saving another symbol, 109 now (manipulating with n and getting rid of +11 ):

p=5**.5
f=lambda n:((1+p)**n-(1-p)**n)/(2**n*p)
exec "n=input()+6;print int(f(n+5)-f(n-5)+f(n)%10);"*input()

Edit: New way to calculate particular number, saves another 4 symbols and allows to avoid int() :

def f(n):exec"a=b=1;"+"a,b=b,a+b;"*(n-1);return a
exec "n=input()+6;print f(n+5)-f(n-5)+f(n)%10;"*input()

Answer 4

Sorry I did not read your question properly before posting. I am glad you at least found some use in it.

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I don't know Python, but in Mathematica, as generic as possible:

f[1] = 1;
f[2] = 1;
f[x_] := f[x] = f[x - 1] + f[x - 2]

t = 0;

n = 35;

For[i = 0, i <= 9, i++, t += f[n + i]]

t += f[n + 246] ~Mod~ 10

Or, in terse Mathematica, still without using Fibonacci function:

f[1|2]=1;a:f@x_:=a=f[x-1]+f[x-2];Sum[f[#+x],{x,0,9}]+f[#+246]~Mod~10&

Answer 5

p=5**.5
f=lambda n:((1+p)**n-(1-p)**n)/(2**n*p)
exec"n=input();print 55*f(n)+88*f(n+1)+f(n+6)%10;"*input()

106 chars as long you don't care about int() function and accept a float

Answer 6

This one prints the Fibonacci series up to n.

def fib(n):    
    a, b = 0, 1
    while a < n:
        print(a, end=' ')
        a, b = b, a+b
        print()