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%I #18 Nov 03 2019 19:46:30
%S 0,1,93,571,16143,79333,1755225,160251339,705725473,57691858003,
%T 1057609507815,4500326662525,80662044522801,5995948088798691,
%U 437230824840308493,1820340203482736875,130228506669621162901,2230237339841166071433,9214275012380069727751
%N Quotients connected with the Banach matchboxes problem: Sum_{i=1..prime(n)-5} 2^(i-1)*binomial(i+1,2)/prime(n) (case 2).
%C 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.
%H Vladimir Shevelev, <a href="http://arxiv.org/abs/1110.5686">Banach matchboxes problem and a congruence for primes</a>, arXiv:1110.5686 [math.HO], 2011.
%t Array[Sum[2^(i - 1)*Binomial[i + 1, 2]/#, {i, # - 5}] &@ Prime@ # &, 19, 3] (* _Michael De Vlieger_, Dec 06 2018 *)
%o (PARI) a(n) = sum(i=1, prime(n)-5, 2^(i-1)*binomial(i+1,2))/prime(n); \\ _Michel Marcus_, Dec 06 2018
%Y Cf. A007663, A007619, A238692.
%K nonn
%O 3,3
%A _Vladimir Shevelev_, Mar 03 2014
%E More terms from _Peter J. C. Moses_, Mar 03 2014
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