A078730 -id:A078730 - cp's OEIS frontend

A251758 Let n>=2 be a positive integer with divisors 1 = d_1 < d_2 < ... < d_k = n, and s = d_1d_2 + d_2d_3 + ... + d_(k-1)*d_k. The sequence lists the values a(n) = floor(n^2/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, 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

Offset: 2

Michel Lagneau, Dec 08 2014

nonn

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
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

			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.

Michel Lagneau, Table of n, a(n) for n = 2..10000
International Mathematical Olympiad, Problems, IMO-2002, Problem 4.

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).

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:

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

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)

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

A136193 Irregular array read by rows: row n contains the products of each pair of consecutive positive divisors of n.

Original entry on oeis.org

2, 3, 2, 8, 5, 2, 6, 18, 7, 2, 8, 32, 3, 27, 2, 10, 50, 11, 2, 6, 12, 24, 72, 13, 2, 14, 98, 3, 15, 75, 2, 8, 32, 128, 17, 2, 6, 18, 54, 162, 19, 2, 8, 20, 50, 200, 3, 21, 147, 2, 22, 242, 23, 2, 6, 12, 24, 48, 96, 288, 5, 125, 2, 26, 338, 3, 27, 243

Offset: 2

Leroy Quet, Dec 20 2007; corrected Jan 20 2008

The first listed row is row 2. Row n contains d(n)-1 (= A032741(n)) terms, where d(n) is the number of positive divisors of n.

			The positive divisors of 20 are 1,2,4,5,10,20. 1*2=2. 2*4=8. 4*5=20. 5*10=50. 10*20=200. So row 20 is (2,8,20,50,200).
The first few rows of the triangle are:
2;
3;
2, 8;
5;
2, 6, 18;
7;
2, 8, 32;
...

Cf. A032741, A078730, A136195, A136181.

with(numtheory): a:=proc(n) local div: div:=divisors(n): seq(div[j]*div[j+1], j=1..tau(n)-1) end proc: for n from 2 to 25 do a(n) end do; # yields sequence as a two-dimensional array - Emeric Deutsch, Jan 08 2008

Flatten[Table[Times@@@Partition[Divisors[n],2,1],{n,30}]]  (* Harvey P. Dale, Apr 23 2011 *)

tabf(nn) = {for (n = 2, nn, d = divisors(n); for (i = 1, #d - 1, print1(d[i]*d[i+1], ", ");););} \\ Michel Marcus, Feb 10 2014

More terms from Emeric Deutsch, Jan 08 2008

More terms from Michel Marcus, Feb 10 2014

cp's OEIS Frontend

A251758 Let n>=2 be a positive integer with divisors 1 = d_1 < d_2 < ... < d_k = n, and s = d_1d_2 + d_2d_3 + ... + d_(k-1)*d_k. The sequence lists the values a(n) = floor(n^2/s).

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A136193 Irregular array read by rows: row n contains the products of each pair of consecutive positive divisors of n.

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cp's OEIS Frontend

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).

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Maple

Mathematica

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Extensions

A136193 Irregular array read by rows: row n contains the products of each pair of consecutive positive divisors of n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Extensions

A251758 Let n>=2 be a positive integer with divisors 1 = d_1 < d_2 < ... < d_k = n, and s = d_1d_2 + d_2d_3 + ... + d_(k-1)*d_k. The sequence lists the values a(n) = floor(n^2/s).