%I #13 Nov 25 2022 11:05:14
%S 175,2224,2575,4975,7024,9424,9775,11824,12175,14224,14575,16975,
%T 19024,21424,21775,23824,24175,26224,26575,28975,31024,33424,33775,
%U 35824,36175,38224,38575,40975,43024,45424,45775,47824,48175,50224,50575,52975,55024,57424,57775
%N Numbers k such that k and k+1 have the same sum of 5-smooth divisors.
%C Numbers k such that A355584(k) = A355584(k+1).
%C 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).
%C 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.
%C 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.
%C Are there runs of 3 consecutive numbers with the same sum of 5-smooth divisors? There are no such runs below 5*10^10.
%H Amiram Eldar, <a href="/A355713/b355713.txt">Table of n, a(n) for n = 1..10000</a>
%e 175 is a term since A355584(175) = A355584(176) = 31.
%t 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] &]
%o (PARI) s(n) = (2^(valuation(n, 2) + 1) - 1) * (3^(valuation(n, 3) + 1) - 1) * (5^(valuation(n, 5) + 1) - 1) / 8;
%o s1 = s(1); for(k = 2, 6e4, s2 = s(k); if(s1 == s2, print1(k-1,", ")); s1 = s2);
%Y Cf. A000203, A002961, A018255, A355584.
%Y Subsequence of A013929 and A068781.
%Y Similar sequences: A002961, A064115, A064125, A293183, A306985, A333949.
%K nonn
%O 1,1
%A _Amiram Eldar_, Jul 15 2022