A251758 - OEIS

%I #77 Sep 09 2017 19:27:44

%S 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,

%T 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,

%U 1,4,1,67,1,2,1,71,1,73,1,2,1,6,1,79,1,2,1

%N 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).

%C s is always less than n^2 and if n is a prime number then s divides n^2.

%C For n >= 2, the sequence has the following properties:

%C a(n) = n if n is prime.

%C a(n) = 1 if n is in A005843 and > 2;

%C a(n) <= 2 if n is in A016945 and > 3;

%C a(n) <= 4 if n is in A084967 and > 5;

%C a(n) <= 6 if n is in A084968 and > 7;

%C a(n) = 8: <= 35336848261, ...;

%C a(n) <= 10 if n is in A084969 and > 11;

%C a(n) <= 12 if n is in A084970 and > 13;

%C a(n) = 14: 6678671, ...;

%C 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.)

%C 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

%C n^2/s is only an integer iff n is prime. - _Robert G. Wilson v_, Dec 18 2014

%C 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

%H Michel Lagneau, <a href="/A251758/b251758.txt">Table of n, a(n) for n = 2..10000</a>

%H International Mathematical Olympiad, <a href="http://www.imo-official.org/problems.aspx">IMO-2002, Problem 4.

%F a(n) <= A250480(n), and especially, for all composite n, a(n) < A020639(n). [Cf. the Comments section above.] - _Antti Karttunen_, Dec 09 2014

%F From _Robert G. Wilson v_, Dec 18 2014: (Start)

%F a(n) = floor(n^2/A078730(n));

%F a(n) = n iff n is prime. (End)

%e 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.

%p with(numtheory):nn:=100:

%p for n from 2 to nn do:

%p    x:=divisors(n):n0:=nops(x):s:=sum('x[i]*x[i+1]','i'=1..n0-1):

%p    z:=floor(n^2/s):printf(`%d, `,z):

%p od:

%t f[n_] := Floor[ n^2/Plus @@ Times @@@ Partition[ Divisors@ n, 2, 1]]; Array[f, 81, 2] (* _Robert G. Wilson v_, Dec 18 2014 *)

%Y 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).

%Y Cf. also A020639 (the smallest prime divisor), A055396 (its index) and arrays A083140 and A083221 (Sieve of Eratosthenes).

%Y Differs from A250480 for the first time at n = 2431, where a(2431) = 9, while A250480(2431) = 10.

%Y Cf. A078730 (sum of products of two successive divisors of n).

%K nonn

%O 2,1

%A _Michel Lagneau_, Dec 08 2014

%E Comments section edited by _Antti Karttunen_, Dec 09 2014

%E Instances of n for which a(n) = 8 and 14 found by _Robert G. Wilson v_, Dec 18 2014