|
%I #22 Apr 14 2015 10:26:21
%S 1512,7560,16632,19656,25704,28728,34776,37800,43848,44928,46872,
%T 55944,61992,65016,71064,80136,83160,89208,92232,98280,101304,107352,
%U 110376,119448,125496,128520,134568,143640,146664,152712,155736,161784,164808,170856,173880,182952,189000,192024,198072,207144,210168,216216
%N Numbers n for which there exists k > n such that A000203(k) = A000203(n) and A007947(k) = A007947(n), where A000203 gives the sum of divisors, and A007947 gives the squarefree kernel of n.
%C None of the terms are squarefree, because if there were such x, then we would have rad(x) = x, and for any value k > x such that rad(k) = x we would have k = y*x, for some strictly positive integer y, and in that case sigma(k) > sigma(x). Thus all terms are members of sequence A013929.
%C None of the terms in range a(1) .. a(6589) occur in A255335. Are the sequences disjoint forever?
%H Antti Karttunen, <a href="/A255334/b255334.txt">Table of n, a(n) for n = 1..6589</a>
%F a(n) = A255424(n) * A255425(n).
%o (PARI)
%o A007947(n) = factorback(factorint(n)[, 1]); \\ _Andrew Lelechenko_, May 09 2014
%o isA255334(n) = { my(r=A007947(n), s=sigma(n), k=n+r); while(k<s, if((sigma(k) == s)&&(A007947(k) == r), return(1), k = k+r)); return(0); };
%o i=0; for(n=1, 2^25, if(isA255334(n), i++; write("b255334.txt", i, " ", n)))
%o (Scheme, with _Antti Karttunen_'s IntSeq-library. Quite naive and slow implementation.)
%o (define A255334 (MATCHING-POS 1 1 isA255334?))
%o (define (isA255334? n) (let ((sig_n (A000203 n)) (rad_n (A007947 n))) (let loop ((try (+ n rad_n))) (cond ((>= try sig_n) #f) ((and (= sig_n (A000203 try)) (= rad_n (A007947 try))) #t) (else (loop (+ try rad_n)))))))
%Y Subsequence of A013929.
%Y Cf. A000203, A007947.
%Y Cf. also A255423 (gives the corresponding k), A255335 (same sequence sorted into ascending order, with duplicates removed), A255412 [gives sigma(a(n))], A255424 [gives rad(a(n))], A255425, A254035, A254791.
%K nonn
%O 1,1
%A _Antti Karttunen_, Mar 23 2015
|
