Problems & Puzzles: Puzzles

 Puzzle 834. A274071. Here we ask to expand the sequence A274041, that was born from the entry 1443 of the always interesting Claudio Meller's site. The A274041 sequence has been defined this way: "a(n) is the least possible sum of a sequence consisting of exactly prime(i) multiples of prime(i) with i = 1 to n." Currently the published sequence has only four terms, a(n)= 6, 20, 87, 304,... Example: for n=3 a(3) = 87, (3,9,5,10,15,20,25): this sequence has only two multiples of 2, only three multiples of 3 and only five multiples of 5; moreover this is the sequence for n=3 with the smallest sum of the terms of the sequence. Q1: Can you check if a(5)=1398 as suggested by Mmonchi and myself in the Meller's site? Q2: Please send the following next a(n) terms and the integers composing the sums reported for each of your a(n) terms.

Contributions came from Jan van Delden and Carlos Rivera

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Jan wrote:

Q1:

The first few terms are
a[1]=6 [2, 4]
a[2]=20 [2, 3, 6, 9]
a[3]=87 [3, 5, 9, 10, 15, 20, 25]
a[4]=304 [5, 7, 14, 15, 21, 25, 35, 49, 63, 70]
a[5]=1398 [5, 7, 11, 22, 25, 33, 35, 49, 55, 77, 91, 99, 119, 121, 143, 154, 165, 187]

These solutions have the following property:

The number of prime factors of each term are <=3.
If in a[i] in one term a prime p[j] is used with j>i it is used only once.
All terms in a[i] are <= p[i]*p[j_max].
The small primes and the largest primes are “linked” [Like 2*11 and 3*11 in a[5]].
Small primes and their powers seem to disappear.

Q2:

a[6]=3718 [7, 11, 13, 22, 26, 33, 35, 39, 49, 55, 65, 77, 91, 117, 119, 121, 133, 143, 169, 187, 209, 221, 247, 253, 275, 299, 325, 377]
a[7]=9223 [7, 11, 13, 17, 39, 49, 51, 55, 65, 77, 85, 91, 119, 121, 133, 143, 153, 169, 187, 209, 221, 247, 253, 289, 299, 319, 323, 341, 374, 377, 391, 403, 425, 442, 481, 493, 527, 595, 629]

I’m not completely sure whether these numbers are minima.

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Carlos wrote:

As a matter of fact Mmonchi found and reported on April 30, 2016 in the Meller pages, that a(6)=3653, smaller than the a[6]=3718 reported by Jan above.

3653 = {7, 11, 13, 26, 35, 39, 49, 55, 65, 77, 91, 117, 121, 143, 169, 187, 209, 221, 247, 253, 273, 275, 286, 299, 385}

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