A375462 - OEIS

A375462

a(1)=1; thereafter, if n is odd we require that a(n) be deficient and a(n) || a(n+1) abundant, otherwise a(n) abundant and a(n) || a(n+1) deficient. At each step, chose the smallest possible integer not yet in the sequence.

1, 12, 2, 20, 3, 18, 4, 40, 5, 60, 7, 36, 9, 24, 8, 70, 10, 56, 11, 48, 13, 30, 14, 72, 15, 42, 17, 80, 19, 84, 16, 96, 21, 54, 22, 88, 23, 100, 25, 90, 26, 104, 27, 66, 29, 112, 31, 108, 32, 120, 33, 78, 34, 132, 35, 140, 37, 156, 38, 144, 39, 102, 41, 160, 43

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OFFSET

1,2

LINKS

Table of n, a(n) for n=1..65.

EXAMPLE

a(1)=1 and n=1 is odd -> a(2)=12 because is the smallest abundant number not yet in the sequence such that a(1) || a(2) = 112 is abundant.

Now n=2 is even -> a(3)=2 because is the smallest deficient number not yet in the sequence such that a(2) || a(3) = 122 is deficient. And so on.

MAPLE

with(numtheory): P:=proc(q) local a, b, k, n; a:=[1];

for n from 2 to q do for k from 2 do if numboccur(k, a)=0

then b:=a[nops(a)]*10^length(k)+k;