|
|
A355713
|
|
Numbers k such that k and k+1 have the same sum of 5-smooth divisors.
|
|
1
|
|
|
175, 2224, 2575, 4975, 7024, 9424, 9775, 11824, 12175, 14224, 14575, 16975, 19024, 21424, 21775, 23824, 24175, 26224, 26575, 28975, 31024, 33424, 33775, 35824, 36175, 38224, 38575, 40975, 43024, 45424, 45775, 47824, 48175, 50224, 50575, 52975, 55024, 57424, 57775
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Equivalently, numbers k such that the largest 5-smooth divisors of k and k+1, A355582(k) and A355582(k+1), have the same sum of divisors (A000203).
For all the terms k, both k and k+1 are not squarefree: each of the two largest 5-smooth divisors, of k and k+1, cannot be squarefree, since the squarefree 5-smooth numbers are the divisors of 30 = 2*3*5 (A018255) whose values of sigma (A000203), {1, 3, 4, 6, 12, 18, 24, 72}, are not shared with sigma of any other 5-smooth number.
Apparently, all the terms are of only two types: numbers k such that A355582(k) = 16 and A355582(k+1) = 25, or numbers k such that A355582(k) = 25 and A355582(k+1) = 16. Both types are infinite sequences: The first type is the sequence of numbers of the form 2224 + 2400*m, where m is not congruent to 1 (mod 5), and the second type is the sequence of numbers of the form 175 + 2400*m, where m is not congruent to 3 (mod 5). If there are no other terms, then this sequence is a linear recurrence with a signature (1,0,0,0,0,0,0,1,-1). The question of the existence of other types is equivalent to the question of the existence of two coprime 5-smooth numbers other than 16 and 25 whose sums of divisors are equal.
Are there runs of 3 consecutive numbers with the same sum of 5-smooth divisors? There are no such runs below 5*10^10.
|
|
LINKS
|
|
|
EXAMPLE
|
|
|
MATHEMATICA
|
f[p_, e_] := If[p > 5, 1, (p^(e + 1) - 1)/(p - 1)]; s[1] = 1; s[n_] := Times @@ (f @@@ FactorInteger[n]); Select[Range[10^5], s[#] == s[# + 1] &]
|
|
PROG
|
(PARI) s(n) = (2^(valuation(n, 2) + 1) - 1) * (3^(valuation(n, 3) + 1) - 1) * (5^(valuation(n, 5) + 1) - 1) / 8;
s1 = s(1); for(k = 2, 6e4, s2 = s(k); if(s1 == s2, print1(k-1, ", ")); s1 = s2);
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|