|
|
A180919
|
|
a(n) = n^2 + 731*n + 1.
|
|
1
|
|
|
1, 733, 1467, 2203, 2941, 3681, 4423, 5167, 5913, 6661, 7411, 8163, 8917, 9673, 10431, 11191, 11953, 12717, 13483, 14251, 15021, 15793, 16567, 17343, 18121, 18901, 19683, 20467, 21253, 22041, 22831, 23623, 24417, 25213, 26011, 26811
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
Consider all sequences of numbers of the form m^2+h*m+1 (with h natural number and m = 0,1,2,3,4,5,...) which contain exactly 7 squares; the present sequence has the smallest value of h. Note that for 6 squares the smallest h is 23 and for 8 squares the smallest h is 37.
For n < 365^2, the squares of the form n^2+731*n+1 are 1, 239121, 2653641, 24413481, 220255281, 1982831841, 17846020921; for n > 365^2-1 we have (n+365)^2 < n^2+731*n+1 < (n+366)^2 and therefore n^2+731*n+1 cannot be a square.
|
|
LINKS
|
|
|
FORMULA
|
G.f.: (1+730*x-729*x^2)/(1-x)^3.
a(2*n-1) - a(n) - a(n-1) = A142463(n-1) for n>0.
a(0)=1, a(1)=733, a(2)=1467; for n>2, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Harvey P. Dale, Oct 14 2012
|
|
MATHEMATICA
|
Table[n^2 + 731 n + 1, {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {1, 733, 1467}, 40] (* Harvey P. Dale, Oct 14 2012 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|