A359390 Sequence lists the numbers k such that bottom entry is an integer in the ratio d(i+1)/d(i) triangle of the elements in the divisors of n, where d(1) < d(2) < ... < d(q) denote the divisors of k.

1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, 29, 31, 32, 36, 37, 41, 43, 47, 49, 53, 59, 61, 64, 67, 71, 73, 79, 81, 83, 89, 97, 100, 101, 103, 107, 109, 113, 121, 125, 127, 128, 131, 137, 139, 144, 149, 151, 157, 163, 167, 169, 173, 179, 181, 191, 193

Offset: 1

Author

Michel Lagneau, Jan 03 2023

Keywords

• nonn

Comments

The corresponding integer bottom entry is 1 if k is nonprime or k if k is prime. [It is very likely that this is true, but no proof has yet been given. - Jianing Song, Jan 22 2023]

We observe that a(n) = A323306(n) for n = 1..50. But a(51) = 144 does not belong to that sequence.

Note that the bottom rational is Product_{i=1..q} d(i) ^ (binomial(q-1,i-1) * (-1)^(q-i)). - Kevin Ryde, Jan 03 2023

Given n, let 1 = M(1,1) < M(1,2) < ... < M(1,d) = n be the divisors of n, and M(i,j) = M(i-1,j+1)/M(i-1,j) for 2 <= i <= d, 1 <= j <= d+1-i. Since M(1,d+1-j) = n/M(1,j) for 1 <= j <= d, we have M(i,d+2-i-j) = M(i,j) for even i, 1 <= j <= d+1-i, and M(i,d+2-i-j) = 1/M(i,j) for odd i > 1, 1 <= j <= d+1-i. If n is a square, then d is odd, so M(d,1) = 1/M(d,1) => M(d,1) = 1. This shows that all square numbers are terms. Note that all powers of primes (A000961) are trivially terms. It seems that the squares and the powers of primes are the only terms. - Jianing Song, Jan 03 2023

Examples

			36 is a term because the triangle of the elements d(i+1)/d(i) has bottom entry 1:
  [1, 2, 3, 4, 6, 9, 12, 18, 36]
  [2, 3/2, 4/3, 3/2, 3/2, 4/3, 3/2, 2]
  [3/4, 8/9, 9/8, 1, 8/9, 9/8, 4/3]
  [32/27, 81/64, 8/9, 8/9, 81/64, 32/27]
  [2187/2048, 512/729, 1, 729/512, 2048/2187]
  [1048576/1594323, 729/512, 729/512, 1048576/1594323]
  [1162261467/536870912, 1, 536870912/1162261467]
  [536870912/1162261467, 536870912/1162261467]
  [1].
6 is not a term because the triangle of the elements d(i+1)/d(i) has bottom entry 16/9.
  [1, 2, 3, 6]
  [2, 3/2, 2]
  [3/4, 4/3]
  [16/9]
		

Crossrefs

Cf. A323306. Contains A000290 and A000961 as subsequences (and conjectured to be the union of these two sequences).

Programs

Mathematica

Lst={}; Table[d=Divisors[n]; While[Length[d]>1,d=Ratios[d]]; If[d[[1]]==Floor[d[[1]]],AppendTo[Lst,n]],{n,300}]; Lst

PARI

isA359390(n) = my(L = factor(n), w = #L~, v=divisors(n), q=#v); for(i_d=1, q-1, for(i_p=1, w, L[i_p,2] += binomial(q-1,i_d-1) * (-1)^(q-i_d) * valuation(v[i_d], L[i_p,1]))); for(i_p=1, w, if(L[i_p,2]<0, return(0))); return(1) \\ Jianing Song, Jan 22 2023, based on the formula provided by Kevin Ryde