|
|
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).
|
|
10
|
|
|
0, 1, 93, 571, 16143, 79333, 1755225, 160251339, 705725473, 57691858003, 1057609507815, 4500326662525, 80662044522801, 5995948088798691, 437230824840308493, 1820340203482736875, 130228506669621162901, 2230237339841166071433, 9214275012380069727751
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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
|
|
|
|