OFFSET
3,3
COMMENTS
A general congruence connected with the Banach matchboxes problem is the following: for k=1,2,...,(p-1)/2, Sum_{i=1..p-2k-1} 2^(i-1)*binomial(k-1+i,k) == 0 (mod p) (p is odd prime). If k=1 (case 1), then one can prove that the corresponding quotients are 2^(prime(n)-3) - A007663(n), n >= 2.
LINKS
Vladimir Shevelev, Banach matchboxes problem and a congruence for primes, arXiv:1110.5686 [math.HO], 2011.
MATHEMATICA
Array[Sum[2^(i - 1)*Binomial[i + 1, 2]/#, {i, # - 5}] &@ Prime@ # &, 19, 3] (* Michael De Vlieger, Dec 06 2018 *)
PROG
(PARI) a(n) = sum(i=1, prime(n)-5, 2^(i-1)*binomial(i+1, 2))/prime(n); \\ Michel Marcus, Dec 06 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Mar 03 2014
EXTENSIONS
More terms from Peter J. C. Moses, Mar 03 2014
STATUS
approved