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Integers k such that Sum_{i=1..t-1} d(i)/d(i+1) is prime, where d(1), ..., d(t) are the divisors of k in ascending order.
1

%I #21 Feb 07 2022 02:39:40

%S 16,64,729,1024,1536,6250,9375,16384,19683,39366,1179648,4194304,

%T 6770688,9765625,14348907,29229255,39062500,67108864,125000000,

%U 128472708,335544320,1337982976,10460353203

%N Integers k such that Sum_{i=1..t-1} d(i)/d(i+1) is prime, where d(1), ..., d(t) are the divisors of k in ascending order.

%C The corresponding primes are 2, 3, 2, 5, 13, 5, 5, 7, 3, 11, 41, 11, 89, 2, 5, 37, 19, 13, 53, 37, ...

%C a(n) is a power of 2 for n = 1, 2, 4, 8, 12, 18, ... with the corresponding primes 2, 3, 5, 7, 11, 13, ...

%C a(n) is a perfect square for n = 1, 2, 3, 4, 8, 12, 14, 17, 18, ... with the corresponding primes 2, 3, 2, 5, 7, 11, 2, 19, 13, ...

%e 64 is in the sequence because the divisors of 64 are {1, 2, 4, 8, 16, 32, 64} and 1/2 + 2/4 + 4/8 + 8/16 + 16/32 + 32/64 = 3 is prime.

%t Do[s=0;Do[s=s+Divisors[n][[i]]/Divisors[n][[i+1]],{i,1,Length[Divisors[n]]-1}];If[PrimeQ[s]&&!PrimeQ[n],Print[n]],{n,10^6}]

%t Select[Range[40000],PrimeQ[Total[#[[1]]/#[[2]]&/@Partition[ Divisors[ #],2,1]]]&] (* The program generates the first 10 terms of the sequence. To generate more, increase the Range constant. *) (* _Harvey P. Dale_, Feb 06 2022 *)

%o (Python)

%o from sympy import isprime, divisors

%o from fractions import Fraction

%o def ok(n):

%o divs = divisors(n)

%o f = sum(Fraction(dn, dd) for dn, dd in zip(divs[:-1], divs[1:]))

%o return f.denominator == 1 and isprime(f.numerator)

%o print([k for k in range(1, 40000) if ok(k)]) # _Michael S. Branicky_, Feb 06 2022

%Y Subsequence of A227993.

%K nonn,more

%O 1,1

%A _Michel Lagneau_, Feb 25 2015

%E a(20) inserted and a(22)-a(23) from _Michael S. Branicky_, Feb 06 2022 using A227993

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Last modified September 23 18:05 EDT 2024. Contains 376178 sequences. (Running on oeis4.)