A238692 -id:A238692 - cp's OEIS frontend

A238693 Quotients connected with the Banach matchboxes problem: Sum_{i=1..prime(n)-5} 2^(i-1)*binomial(i+1,2)/prime(n) (case 2).

0, 1, 93, 571, 16143, 79333, 1755225, 160251339, 705725473, 57691858003, 1057609507815, 4500326662525, 80662044522801, 5995948088798691, 437230824840308493, 1820340203482736875, 130228506669621162901, 2230237339841166071433, 9214275012380069727751

Offset: 3

Vladimir Shevelev, Mar 03 2014

nonn

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.

Vladimir Shevelev, Banach matchboxes problem and a congruence for primes, arXiv:1110.5686 [math.HO], 2011.

Cf. A007663, A007619, A238692.

Array[Sum[2^(i - 1)*Binomial[i + 1, 2]/#, {i, # - 5}] &@ Prime@ # &, 19, 3] (* Michael De Vlieger, Dec 06 2018 *)

a(n) = sum(i=1, prime(n)-5, 2^(i-1)*binomial(i+1,2))/prime(n); \\ Michel Marcus, Dec 06 2018

More terms from Peter J. C. Moses, Mar 03 2014

cp's OEIS Frontend

A238693 Quotients connected with the Banach matchboxes problem: Sum_{i=1..prime(n)-5} 2^(i-1)*binomial(i+1,2)/prime(n) (case 2).

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