|
|
A125270
|
|
Coefficient of x^2 in polynomial whose zeros are 5 consecutive primes starting with the n-th prime.
|
|
0
|
|
|
1358, 3954, 10478, 22210, 43490, 78014, 129530, 206650, 324350, 466270, 621466, 853742, 1132130, 1436690, 1870850, 2388050, 2886370, 3440410, 4133410, 4904906, 5926654, 7195670, 8425430, 9792950, 11040910, 12098990, 13917898, 16097810
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Sums of all distinct products of 3 out of 5 consecutive primes, starting with the n-th prime; value of 3rd elementary symmetric function on the 5 consecutive primes.
|
|
LINKS
|
|
|
FORMULA
|
Let p = Prime(n), q = Prime(n+1), r = Prime(n+2), s = Prime(n+3) and t = Prime(n+4). Then a(n) = p q (r+s+t) + (p + q) r (s + t) + (p + q + r) s t.
|
|
MATHEMATICA
|
a = {}; Do[AppendTo[a, (Prime[x] Prime[x + 1] Prime[x + 2] + Prime[x] Prime[x + 1] Prime[x + 3] + Prime[x] Prime[x + 1] Prime[x + 4] + Prime[x] Prime[x + 2] Prime[x + 3] + Prime[x] Prime[x + 2] Prime[x + 4] + Prime[x] Prime[x + 3] Prime[x + 4] + Prime[x + 1] Prime[x + 2] Prime[x + 3] + Prime[x + 1] Prime[x + 2] Prime[x + 4] + Prime[x + 1] Prime[x + 3] Prime[x + 4] + Prime[x + 2] Prime[x + 3] Prime[x + 4])], {x, 1, 100}]; a
fcp[{p_, q_, r_, s_, t_}]:=p*q(r+s+t)+(p+q)r(s+t)+(p+q+r)s*t; fcp/@Partition[ Prime[ Range[40]], 5, 1] (* Harvey P. Dale, Sep 05 2014 *)
|
|
CROSSREFS
|
Cf. A001043, A034961, A034963, A034964, A127333, A127334, A127335, A127336, A127337, A127338, A127339, A127340, A127341, A127342, A127343, A127345, A127346, A127347, A127348, A127349, A127351, A037171, A034962, A034965, A082246, A082251, A070934, A006094, A046301, A046302, A046303, A046324, A046325, A046326, A046327, A127489.
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|