OFFSET
0,5
COMMENTS
Consider the Fibonacci sequence f(j+1)=f(j)+f(j-1) starting with f(0)=prime(n) and some f(1)=k>=0. Count how many nonprimes follow after the initial prime f(0), before the next prime term. Obviously, if f(1) is a positive multiple of f(0), then all following terms are positive multiples of f(0) and therefore composite; in this case we set T(n,k)=0.
This is somehow complementary to the search of primes in arithmetic progression, where one looks for the length of chains of prime terms.
EXAMPLE
The square array is:
n ( p) k=0 1 2 3 4 5 6 7
1 ( 2) : 1,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,...
2 ( 3) : 1,2,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,...
3 ( 5) : 1,2,0,0,2,0,1,0,1,2,0,0,1,0,1,0,2,0,1,0,0,...
4 ( 7) : 1,3,0,0,1,0,1,0,2,3,1,0,1,0,0,2,1,0,2,0,2,...
5 (11) : 1,2,0,0,2,0,1,0,1,2,2,0,1,0,4,2,2,0,1,0,1,... etc.
The entry of the array is the least index >= 0 after which a prime is reached, in the Fibonacci sequence f(0)=p=prime(n), f(1)=k, f(j+1)=f(j)+f(j-1).
For n=4 (f(0)=prime(4)=7), k=0 yields f=(7,0,7,7,14,....): T(n,k)=1 nonprime (0), f(1+1)=7 is prime.
k=1 yields f=(7,1,8,9,17,....): T(n,k)=3 nonprimes (1,8,9), f(3+1)=17 is prime.
k=2 yields f=(7,2,9,11,....): T(n,k)=0 nonprimes, since f(0+1)=2 is already prime.
PROG
(PARI) T(n, k)=if(k%n=prime(n), !isprime(k) & for(c=1, 1e9, ispseudoprime(k=n+0+n=k)&return(c)), !k)
for(n=1, 19, for(k=0, n-1, print1(T(k+1, n-k-1)", "))) /* to list antidiagonals starting at (1, k=n-1) and ending at (n, k=0) */
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
M. F. Hasler, Dec 13 2010
STATUS
approved