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%I #54 Jan 28 2023 12:17:00
%S 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,
%T 53,59,61,64,67,71,73,79,81,83,89,97,100,101,103,107,109,113,121,125,
%U 127,128,131,137,139,144,149,151,157,163,167,169,173,179,181,191,193
%N 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.
%C 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]
%C We observe that a(n) = A323306(n) for n = 1..50. But a(51) = 144 does not belong to that sequence.
%C 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
%C 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
%e 36 is a term because the triangle of the elements d(i+1)/d(i) has bottom entry 1:
%e [1, 2, 3, 4, 6, 9, 12, 18, 36]
%e [2, 3/2, 4/3, 3/2, 3/2, 4/3, 3/2, 2]
%e [3/4, 8/9, 9/8, 1, 8/9, 9/8, 4/3]
%e [32/27, 81/64, 8/9, 8/9, 81/64, 32/27]
%e [2187/2048, 512/729, 1, 729/512, 2048/2187]
%e [1048576/1594323, 729/512, 729/512, 1048576/1594323]
%e [1162261467/536870912, 1, 536870912/1162261467]
%e [536870912/1162261467, 536870912/1162261467]
%e [1].
%e 6 is not a term because the triangle of the elements d(i+1)/d(i) has bottom entry 16/9.
%e [1, 2, 3, 6]
%e [2, 3/2, 2]
%e [3/4, 4/3]
%e [16/9]
%t Lst={}; Table[d=Divisors[n]; While[Length[d]>1,d=Ratios[d]]; If[d[[1]]==Floor[d[[1]]],AppendTo[Lst,n]],{n,300}]; Lst
%o (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_
%Y Cf. A323306. Contains A000290 and A000961 as subsequences (and conjectured to be the union of these two sequences).
%K nonn
%O 1,2
%A _Michel Lagneau_, Jan 03 2023