

A275233


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


1



5, 33, 11, 11, 17, 26, 13, 19, 29, 23, 55, 23, 17, 141, 47, 38, 85, 38, 23, 46, 39, 47, 71, 41, 29, 177, 29, 59, 89, 47, 65, 53, 101, 212, 107, 59, 41, 182, 115, 83, 91, 182, 62, 71, 141, 119, 235, 71, 53, 115, 158, 107, 265, 79, 41, 89, 53, 47, 123, 83, 143, 58, 117, 106, 197
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OFFSET

1,1


COMMENTS

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


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000


EXAMPLE

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


PROG

(PARI) avdiv(n)=my(f=factor(n)); sigma(f)/numdiv(f)
a(n)=my(k=1); while(avdiv(k) != avdiv(k+n), k++); k \\ Charles R Greathouse IV, Jul 21 2016


CROSSREFS

Cf. A000005, A000203, A003601.
Sequence in context: A125709 A203112 A174471 * A043068 A194405 A299518
Adjacent sequences: A275230 A275231 A275232 * A275234 A275235 A275236


KEYWORD

nonn


AUTHOR

Altug Alkan, Jul 20 2016


STATUS

approved



