A078730 Sum of products of two successive divisors of n.

0, 2, 3, 10, 5, 26, 7, 42, 30, 62, 11, 116, 13, 114, 93, 170, 17, 242, 19, 280, 171, 266, 23, 476, 130, 366, 273, 528, 29, 713, 31, 682, 399, 614, 285, 1070, 37, 762, 549, 1150, 41, 1342, 43, 1264, 873, 1106, 47, 1916, 350, 1562, 921, 1752, 53, 2186, 665, 2166, 1143

Offset: 1

Author

Vladeta Jovovic, Dec 20 2002

Keywords

• nonn

Comments

a(n) = dot_product (d_1,d_2,...,d_(tau(n)-1))*(d_2,d_3,...d_tau(n)), where d_1

a(n) = n if and only if n is prime. - Robert Israel, May 05 2014

The 4th problem of the 43rd International Mathematical Olympiad asked to prove that a(n) < n^2 and determine when a(n) is a divisor of n^2? (answer is: iff n is prime). - Bernard Schott, Nov 28 2022

Links

• Robert Israel, Table of n, a(n) for n = 1..10000.
• The IMO Compendium, Problem 4, 43rd IMO 2002, Glasgow, UK.
• Index to sequences related to Olympiads.

Crossrefs

Cf. A078713(n) = 2*A001157(n)-2*a(n)-n^2-1.

Cf. A027750.

Programs

Haskell

a078730 n = sum $ zipWith (*) ps $ tail ps where ps = a027750_row n
-- Reinhard Zumkeller, Dec 20 2014

Maple

A078730:= proc(n)
local d,i;
d:= numtheory[divisors](n);
add(d[i]*d[i+1], i=1..nops(d)-1)
end proc;
seq(A078730(n),n=1..100); # Robert Israel, May 05 2014

Mathematica

f[n_] := Module[{d, l, s, i}, d = Divisors[n]; l = Length[d]; s = 0; For[i = 1, i < l, i++, s = s + d[[i + 1]]*d[[i]]]; s]; Table[ f[n], {n, 1, 100}]
f[n_] := Plus @@ Times @@@ Partition[ Divisors@ n, 2, 1]; Array[f, 57] (* Robert G. Wilson v, Dec 18 2014 *)

PARI

a(n) = my(d = divisors(n)); sum(k=1, #d-1, d[k]*d[k+1]); \\ Michel Marcus, Feb 15 2015

Formula

a(p^e) = p * (p^(2e) - 1) / (p^2 -1). - Bernard Schott, Nov 28 2022

n <= a(n) < n^2 for n > 1. a(n)/n^2 can be arbitrarily close to n (proof: let n be divisible by the numbers up to k, for large enough k). a(n) > n^(3/2) for n composite. - Charles R Greathouse IV, Nov 29 2022

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