int a=10000,b,c=2800,d,e,f[2801],g; int main(void){ for(;b-c;)f[b++]=a/5; for(;d=0,g=c*2;c-=14,printf("%.4d",e+d/a),e=d%a) for(b=c;d+=f[b]*a,f[b]=d%--g,d/=g--,--b;d*=b); }
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OK, so we'll start with popular publications on Fibs and pull the problems from there. Let's start with Wiki. Warming up:
Prove that F(n+1)*F(n-1) - F(n)^2 = (-1)^n
Examples:
Fibonacci numbers: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
8*21 - 13^2 = -1
13*34 - 21^2 = 1 etc.
P.S. OK, let's try it together.
H(n) = F(n+1)*F(n-1) - F(n)^2 =
= ( F(n)+F(n-1) )*( F(n)-F(n-2) ) - F(n)^2 =
= - F(n-2)*F(n) + ( F(n-1)*F(n) - F(n-1)*F(n-2) ) =
= - F(n-2)*F(n) + F(n-1)*( F(n) - F(n-2) ) =
= - F(n)*F(n-2) + F(n-1)^2 =
= - ( F(n)*F(n-2) - F(n-1)^2 ) =
= - H(n-1)
Hence H(n) = (-1)^(n-2)*H(n-(n-2)) =
= (-1)^n*H(2) =
= (-1)^n*( F(3)*F(1) - F(2)^2 ) =
= (-1)^n*( 2*1 - 1^2 ) =
= (-1)^n, as required to prove it.
A strange problem from the Mechmatov forum:
Find all pairs of numbers (m,n) such that 1+2+...+m = n!
A strange problem from the Mechmatov forum:
Find all pairs of numbers (m,n) such that 1+2+...+m = n!
By computer I could find it, but verbally :)
Not a problem, but just wondering if anyone has seen formulas for calculating the coefficients of a polynomial of degree 3 ? (without using matrices)
Not a problem, but just wondering if anyone has seen formulas for calculating the coefficients of a polynomial of degree 3 ? (without using matrices).
Always use a computer for that :) MaStak, are you so interested or for forex? If it's the latter, it's too weak :)
By the way, there is an interesting thing in Excel - it makes formulas on tabular data - equation of trendline, it shows approximation.
I'm just curious )) I tried to find it once, I got such three-storey expressions ))))
Excel? Yes, of course I do. Here's an example.
MaStak, the cubic regression formulas will help you here.
I don't think so, I don't even know the words.)
I found something similar to this somewhere, but it didn't seem to work,
Maybe the algorithm, maybe the implementation is to blame.
int a=10000,b,c=2800,d,e,f[2801],g;
int main(void){
for(;b-c;)f[b++]=a/5;
for(;d=0,g=c*2;c-=14,printf("%.4d",e+d/a),e=d%a)
for(b=c;d+=f[b]*a,f[b]=d%--g,d/=g--,--b;d*=b);
}
How does it work?