cp's OEIS Frontend

A262512 Sequence gives the unique x for each term of A262511 which contains those numbers n that have exactly one solution to x - d(x) = n, where d(n) is the number of divisors of n (A000005).

%I A262512 #11 Oct 02 2015 12:14:08
%S A262512 6,5,8,7,11,14,18,20,17,24,22,23,27,32,34,35,40,43,46,50,47,51,57,58,
%T A262512 61,72,65,73,84,77,81,79,88,86,87,96,92,93,94,98,99,102,97,105,101,
%U A262512 103,120,107,114,116,119,123,125,130,135,137,143,154,160,151,155,158,164,163,175,173,177,184,179,187,198,200,191,194,193,204,210,197,203,216,206,209,212
%N A262512 Sequence gives the unique x for each term of A262511 which contains those numbers n that have exactly one solution to x - d(x) = n, where d(n) is the number of divisors of n (A000005).
%C A262512 Sequence is sorted by the magnitude of terms in A262511. Cf. also A262513.
%H A262512 Antti Karttunen, <a href="/A262512/b262512.txt">Table of n, a(n) for n = 1..10000</a>
%F A262512 a(n) = the least (and the only) such number k > A262511(n) that A049820(k) = A262511(n).
%F A262512 a(n) = A082284(A262511(n)).
%F A262512 a(n) = A262686(A262511(n)).
%o A262512 (Scheme) (define (A262512 n) (let ((s (A262511 n))) (let loop ((k s)) (if (= s (A049820 k)) k (loop (+ 1 k))))))
%Y A262512 Cf. A000005, A049820, A082284, A262511, A262686.
%Y A262512 Cf. A262513 (same sequence sorted into ascending order).
%K A262512 nonn
%O A262512 1,1
%A A262512 _Antti Karttunen_, Sep 25 2015