|
Search: id:A101257
|
|
|
| A101257 |
|
Remainder when the least divisor of n greater than the square root of n (A033677(n)) is divided by the greatest divisor of n less than the square root of n (A033676(n)). |
|
+0
1
|
|
| 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 2, 0, 0, 0, 0, 1, 1, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 0, 2, 1, 2, 0, 0, 1, 1, 3, 0, 1, 0, 3, 4, 1, 0, 2, 0, 0, 2, 1, 0, 3, 1, 1, 1, 1, 0, 4, 0, 1, 2, 0, 3, 5, 0, 1, 2, 3, 0, 1, 0, 1, 0, 3, 4, 1, 0, 2, 0, 1, 0, 5, 2, 1, 2, 3, 0, 1, 6, 3, 1, 1, 4, 4, 0, 0, 2, 0, 0, 5, 0, 5, 1
(list; graph; listen)
|
|
|
OFFSET
|
1,15
|
|
|
COMMENT
|
Given n points, sort them into the most-square rectangular point lattice possible. Now sort the points into square point lattices of dimension equal to the lesser dimension of the earlier rectangle. a(n) is the number of points left over. a(n) is trivially 0 for prime numbers n (the most-square and only rectangular point lattice on a prime number of points is a linear point lattice). a(n) != 0 iff n is a member of A080363
|
|
LINKS
|
Eric Weisstein et al., "Point Lattice."
Eric Weisstein's World of Mathematics, "Divisor."
|
|
EXAMPLE
|
a(6)=1 because the least divisor of 6 greater than sqrt(6) is 3, the greater divisor of 6 less than sqrt(6) is 2 and 3 mod 2 = 1
|
|
MATHEMATICA
|
num[n_] := If[OddQ[DivisorSigma[0, n]], Sqrt[n], Divisors[n][[DivisorSigma[0, n]/2 + 1]]] den[n_] := If[OddQ[DivisorSigma[0, n]], Sqrt[n], Divisors[n][[DivisorSigma[0, n]/2]]] Table[Mod[num[n], den[n]], {n, 1, 128}]
|
|
CROSSREFS
|
Cf. A033676, A033677, A080363.
Sequence in context: A141846 A035202 A128616 this_sequence A144629 A025907 A024157
Adjacent sequences: A101254 A101255 A101256 this_sequence A101258 A101259 A101260
|
|
KEYWORD
|
base,nonn
|
|
AUTHOR
|
Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 17 2004
|
|
|
Search completed in 0.002 seconds
|
