A275233 - OEIS

%I #14 Jul 21 2016 00:22:54

%S 5,33,11,11,17,26,13,19,29,23,55,23,17,141,47,38,85,38,23,46,39,47,71,

%T 41,29,177,29,59,89,47,65,53,101,212,107,59,41,182,115,83,91,182,62,

%U 71,141,119,235,71,53,115,158,107,265,79,41,89,53,47,123,83,143,58,117,106,197

%N Least k such that k and k + n have same average of divisors.

%C Least k such that A000203(n+k) / A000005(n+k) = A000203(k) / A000005(k).

%C Records are a(1) = 5, a(2) = 33, a(11) = 55, a(14) = 141, a(26) = 177, a(34) = 212, a(47) = 235, a(53) = 265, a(93) = 285, a(98) = 366, a(102) = 442, a(107) = 535, a(158) = 902, a(166) = 2739, a(758) = 3424, a(863) = 4315, a(866) = 4470, a(1171) = 5855, a(1478) = 6790, a(1493) = 7465, a(1823) = 9115, a(2203) = 11015, a(3463) = 14795, a(3833) = 19165, a(6459) = 19379, a(6865) = 20257, a(7926) = 22918, a(9938) = 23161, .... - _Charles R Greathouse IV_, Jul 21 2016

%H Charles R Greathouse IV, <a href="/A275233/b275233.txt">Table of n, a(n) for n = 1..10000</a>

%e a(1) = 5 because sigma(5) / tau(5) = (1 + 5) / 2 = 3 and sigma(6) / tau(6) = (1 + 2 + 3 + 6) / 4 = 3.

%o (PARI) avdiv(n)=my(f=factor(n)); sigma(f)/numdiv(f)

%o a(n)=my(k=1); while(avdiv(k) != avdiv(k+n), k++); k \\ _Charles R Greathouse IV_, Jul 21 2016

%Y Cf. A000005, A000203, A003601.

%K nonn

%O 1,1

%A _Altug Alkan_, Jul 20 2016