|
|
A251758
|
|
Let n>=2 be a positive integer with divisors 1 = d_1 < d_2 < ... < d_k = n, and s = d_1*d_2 + d_2*d_3 + ... + d_(k-1)*d_k. The sequence lists the values a(n) = floor(n^2/s).
|
|
2
|
|
|
2, 3, 1, 5, 1, 7, 1, 2, 1, 11, 1, 13, 1, 2, 1, 17, 1, 19, 1, 2, 1, 23, 1, 4, 1, 2, 1, 29, 1, 31, 1, 2, 1, 4, 1, 37, 1, 2, 1, 41, 1, 43, 1, 2, 1, 47, 1, 6, 1, 2, 1, 53, 1, 4, 1, 2, 1, 59, 1, 61, 1, 2, 1, 4, 1, 67, 1, 2, 1, 71, 1, 73, 1, 2, 1, 6, 1, 79, 1, 2, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
2,1
|
|
COMMENTS
|
s is always less than n^2 and if n is a prime number then s divides n^2.
For n >= 2, the sequence has the following properties:
a(n) = n if n is prime.
a(n) = 1 if n is in A005843 and > 2;
a(n) <= 2 if n is in A016945 and > 3;
a(n) <= 4 if n is in A084967 and > 5;
a(n) <= 6 if n is in A084968 and > 7;
a(n) = 8: <= 35336848261, ...;
a(n) <= 10 if n is in A084969 and > 11;
a(n) <= 12 if n is in A084970 and > 13;
a(n) = 14: 6678671, ...;
This is different from A250480 (a(n) = n for all prime n, and a(n) = A020639(n) - 1 for all composite n), which thus satisfies the above conditions exactly, while with this sequence A020639(n)-1 gives only the guaranteed upper limit for a(n) at composite n. Note that the first different term does not occur until at n = 2431 = 11*13*17, for which a(n) = 9. (See the example below.)
Conjecture: Terms x, where a(x)=n, x=p#k/p#j, p#i is the i-th primorial, k>j is suitable large k and j is the number of primes less than n. As an example, n=9, x = p#7/p#4 = 2431. For n=10, x = p#6/p#4 = 143 although 121 = 11^2 is the least x where a(x)=10 (see formula section). For n=8, x = p#12/p#4, p#13/p#4, p#14/p#4, p#15/p#4, p#16/p#4, etc. But is p#12/p#4 the least such x? - Robert G. Wilson v, Dec 18 2014
First occurrence of n >= 1: 4, 2, 3, 25, 5, 49, 7, ??? <= 35336848261, 2431, 121, 11, 169, 13, 6678671, 7429, 289, 17, 361, 19, 31367009, 20677, 529, 23, ..., . - Robert G. Wilson v, Dec 18 2014
|
|
LINKS
|
|
|
FORMULA
|
a(n) = n iff n is prime. (End)
|
|
EXAMPLE
|
For n = 2431 = 11*13*17, we have (as the eight divisors of 2431 are [1, 11, 13, 17, 143, 187, 221, 2431]) a(n) = floor((2431*2431) / ((1*11)+(11*13)+(13*17)+(17*143)+(143*187)+(187*221)+(221*2431))) = floor(5909761/608125) = floor(9.718) = 9.
|
|
MAPLE
|
with(numtheory):nn:=100:
for n from 2 to nn do:
x:=divisors(n):n0:=nops(x):s:=sum('x[i]*x[i+1]', 'i'=1..n0-1):
z:=floor(n^2/s):printf(`%d, `, z):
od:
|
|
MATHEMATICA
|
f[n_] := Floor[ n^2/Plus @@ Times @@@ Partition[ Divisors@ n, 2, 1]]; Array[f, 81, 2] (* Robert G. Wilson v, Dec 18 2014 *)
|
|
CROSSREFS
|
Differs from A250480 for the first time at n = 2431, where a(2431) = 9, while A250480(2431) = 10.
Cf. A078730 (sum of products of two successive divisors of n).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|