%I
%S 19,24,27,29,38,40,54,56,57,58,59,76,79,87,88,89,95,104,108,109,114,
%T 116,118,120,133,135,136,139,145,149,152,158,168,171,174,177,178,179,
%U 184,189,190,199,203,209,216,218,228,229,232,236,237,239,247,248
%N Numbers whose sum of divisors is divisible by 10.
%C Of the first 10^2, 10^6, and 10^10 positive integers, 17%, 40.8%, and 48.4%, respectively, are in the sequence (See Robert G. Wilson's table below). So the conjecture here is that if this trend continues then as the numbers approach infinity, the percentage of those numbers whose sum of divisors is divisible by 10 approaches 100%.
%C From _David A. Corneth_, Jan 01 2018: (Start)
%C If a(n) and m > 0 are coprime then a(n) * m is in the sequence.
%C A030433 is a subsequence.
%C Let p(n, d) be a prime ending in base10 digit d and m != n gives p(m, d) != p(n, d). For p(n, 2) and p(n, 5) we must have n = 1 (one unique value). Then we could describe families of terms as factorizations in terms of these primes. 24 would give the family p(1, 2)^3 * p(1, 3).
%C For d coprime to 10 we could replace p(n, d) with p(n, 10  d) and give another family of terms. Constructing terms could then be done by selecting primes for p(n, 1), p(n, 3), p(n, 7) and p(n, 9) from A030430, A030431, A030432 and A030433 respectively. (End)
%C From _Robert G. Wilson v_, Jan 03 2017: (Start)
%C The following array has for n powers of 10 and k is the number of integers <= 10^n which are == k (mod 10).
%C Array begins:
%C ========================================================================================
%C n\k 0 1 2 3 4 5 6 7 8 9
%C 
%C _1  0 1 1 2 1 1 1 1 2 0
%C _2  17 6 16 4 21 2 12 4 17 1
%C _3  274 17 178 16 177 7 145 7 173 6
%C _4  3352 47 1690 44 1727 23 1498 27 1563 29
%C _5  37709 130 16012 144 16069 77 14649 87 15022 101
%C _6  408270 406 151936 424 152079 249 141527 293 144481 335
%C _7  4327266 1255 1451931 1341 1452017 787 1368110 923 1395278 1092
%C _8  45278675 3876 13963299 4206 13963783 2494 13268616 2998 13508556 3497
%C _9  469680154 11975 134976927 13185 134999718 7903 129084019 9684 131205200 11235
%C 10 4842279472 37545 1310133910 41349 1310381140 25017 1259051325 31136 1277983443 35663
%C ... (End)
%C From _Robert Israel_, Jan 07 2018: (Start)
%C A number with prime factorization Product_j (p_j)^(e_j) is in the sequence if and only if
%C 1) e_j is odd for some odd p_j, and
%C 2) For some j, either p_j == 1 (mod 5) and e_j == 4 (mod 5), or p_j == 2 or 3 (mod 5) and e_j == 3 (mod 4), or p_j == 4 (mod 5) and e_j is odd.
%C By the strong form of Dirichlet's theorem, the product of 11/p for primes == 4 (mod 5) is 0, which implies that the asymptotic density of the sequence is 1.
%C (End)
%D I. Niven, H. S. Zuckermann and H. L. Montgomery, An Introduction to the Theory of Numbers, John Wiley and Sons, 1991, pages 420.
%H Zoltan Galantai, <a href="/A297575/b297575.txt">Table of n, a(n) for n = 1..1591</a>
%H Zoltan Galantai, <a href="/A297575/a297575.txt">List of numbers up to 3 million whose sum of divisors is divisible by 10</a>
%e 19 is in the sequence, since sigma(19) = 20, which is divisible by 10.
%e 20 is not in the sequence as sigma(20) = 42, which isn't divisible by 10.
%p select(n > numtheory:sigma(n) mod 10 = 0, [$1..1000]); # _Robert Israel_, Jan 02 2018
%t Select[Range@250, Mod[DivisorSigma[1, #], 10] == 0 &] (* _Robert G. Wilson v_, Jan 03 2018 *)
%o (PARI) is(n) = sigma(n) % 10 == 0 \\ _David A. Corneth_, Jan 01 2018
%Y Cf. A000203, A008592, A030430, A030431, A030432, A030433, A291422, A292217.
%K nonn
%O 1,1
%A _Zoltan Galantai_, Jan 01 2018
