

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_(k1)*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
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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 ith 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
n^2/s is only an integer iff n is prime.  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

Michel Lagneau, Table of n, a(n) for n = 2..10000
International Mathematical Olympiad, IMO2002, Problem 4.


FORMULA

a(n) <= A250480(n), and especially, for all composite n, a(n) < A020639(n). [Cf. the Comments section above.]  Antti Karttunen, Dec 09 2014
From Robert G. Wilson v, Dec 18 2014: (Start)
a(n) = floor(n^2/A078730(n));
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..n01):
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

Cf. A000040 (prime numbers), A005843 (even numbers), A016945 (6n+3), A084967 (GCD( 5k, 6) =1), A084968 (GCD( 7k, 30) =1), A084969 (GCD( 11k, 30) =1), A084970 (Numbers whose smallest prime factor is 13).
Cf. also A020639 (the smallest prime divisor), A055396 (its index) and arrays A083140 and A083221 (Sieve of Eratosthenes).
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).
Sequence in context: A220137 A053815 A194446 * A250480 A326584 A166333
Adjacent sequences: A251755 A251756 A251757 * A251759 A251760 A251761


KEYWORD

nonn


AUTHOR

Michel Lagneau, Dec 08 2014


EXTENSIONS

Comments section edited by Antti Karttunen, Dec 09 2014
Instances of n for which a(n) = 8 and 14 found by Robert G. Wilson v, Dec 18 2014


STATUS

approved



