cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A359493 Numbers k such that the bottom entry in the ratio d(i)/d(i+1) triangle of the elements in the divisors of n, where d(1) < d(2) < ... < d(q) denote the divisors of k, is equal to 1.

Original entry on oeis.org

1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 225, 243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1331, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936
Offset: 1

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Author

Michel Lagneau, Jan 03 2023

Keywords

Comments

Subsequence of A001597, but A001597(20) = 216, A001597(41) = 1000, A001597(53) = 1728 are not here.
Observation according to a comment by Bernard Schott: 216, 1000, 1728, 2744, 5832, ... are the first terms of A124581 (abundant cubes), hence the following conjecture: terms in A124581 do not belong to this sequence.

Examples

			100 is a term because the d(i)/d(i+1) triangle has bottom entry 1:
  [1, 2, 4, 5, 10, 20, 25, 50, 100]
  [1/2, 1/2,  4/5, 1/2, 1/2, 4/5, 1/2, 1/2]
  [1, 5/8, 8/5, 1, 5/8, 8/5, 1]
  [8/5, 25/64, 8/5, 8/5, 25/64, 8/5]
  [512/125, 125/512, 1, 512/125, 125/512]
  [262144/15625, 125/512, 125/512, 262144/15625]
  [134217728/1953125, 1, 1953125/134217728]
  [134217728/1953125, 134217728/1953125]
  [1]
6 is not a term because the d(i)/d(i+1) triangle has bottom entry 9/16.
  [1, 2, 3, 6]
  [1/2, 2/3, 1/2]
  [3/4, 4/3]
  [9/16]
		

Crossrefs

Programs

  • Mathematica
    Lst={}; Table[d=Divisors[n]; While[Length[d]>1,d=Ratios[Reverse[d]]]; If[d[[1]]==Floor[d[[1]]],AppendTo[Lst,n]],{n,2000}]; Lst
  • PARI
    ratios(v) = { my(u=vector(#v-1)); for(i=1,#u,u[i] = v[i]/v[1+i]); (u); };
    isA359493(n) = { my(ds=divisors(n)); while(#ds>1, ds = ratios(ds)); (1==ds[1]); }; \\ Antti Karttunen, Jan 04 2023
    
  • PARI
    is(n) = { if(!(ispower(n) || n==1), return(0)); my(f = factor(n), d = divisors(f), m = Map(), i, j, nv, e, fd); for(i = 1, #d, e = (-1)^i * binomial(#d-1, i-1); fd = factor(d[i]); for(j = 1, #fd~, if(mapisdefined(m, fd[j, 1]), nv = mapget(m, fd[j, 1]); mapput(m, fd[j, 1], nv + e * fd[j, 2]) , mapput(m, fd[j, 1], e * fd[j, 2]) ) ) ); for(i = 1, #f~, if(mapget(m, f[i, 1]) != 0, return(0) ) ); return(1) } \\ David A. Corneth, Jan 07 2023
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