|
%I #44 Oct 02 2015 12:19:22
%S 0,119143,119147,119163,119225,119227,119921,119923,120081,120095,
%T 120097,120101,120281,120293,120349,120399,120707,120747,120891,
%U 120895,120903,120917,120919,121443,121551,121823,122079,122261,122263,122273,122277,122813,122961,123205,123213,123223,123237,123257,123765,24660543,24660549,24662311,24662329,24663759,24664997,24665023,24665351
%N Numbers n such that there is no other number k for which A155043(k) = A155043(n).
%C Starting offset is zero, because a(0) = 0 is a special case in this sequence.
%C Numbers where A155043 takes a unique value. (Those values are given by A262508.)
%C Numbers n such that there does not exist any other number h from which one could reach zero in exactly the same number of steps as from n by repeatedly applying the map where k is replaced by k - A000005(k) = A049820(k). Thus in the tree where zero is the root and parent-child relation is given by A049820(child) = parent, all the numbers > n+t (where t is a small value depending on n) have n as their common ancestor. As it is guaranteed that there is at least one infinite path in such a tree, any n in this sequence can be neither a leaf nor any other vertex in a finite side-tree, as then at least one node in the infinite part would have the same distance to the root, thus it must be that n itself is in the infinite part and thus has an infinite number of descendant vertices. Also, for the same reason, the tree cannot branch to two infinite parts from any ancestor of n (which are nodes nearer to the root of the tree, zero).
%C From the above it follows that if this sequence is infinite, then A259934 is guaranteed to be the only infinite sequence which starts with a(0)=0 and satisfies the condition A049820(a(k)) = a(k-1) for all k>=1, where A049820(n) = n-d(n) and d(n) is the number of divisors of n (A000005). This is a sufficient condition for the uniqueness of A259934, although not necessary. See e.g. A179016 which is the unique infinite solution to a similar problem, even though A086876 contains no ones after its two initial terms.
%C Is it possible for any even terms to occur after zero? If not, then apart from zero, this would be a subsequence of A262517.
%H Antti Karttunen, <a href="/A262509/b262509.txt">Table of n, a(n) for n = 0..68</a>
%F a(n) = A261089(A262508(n)) = A262503(A262508(n)) = A259934(A262508(n)).
%o (PARI)
%o \\ Compute A262508 and A262509 at the same time:
%o allocatemem((2^31)+(2^30));
%o \\ The limits used are quite ad hoc. Beware of the horizon-effect if you change these.
%o \\ As a post-check, test that A262509(n) = A259934(A262508(n)) for all the terms produced by this program.
%o uplim1 = 43243200 + 672; \\ = A002182(54) + A002183(54).
%o uplim2 = 36756720; \\ = A002182(53).
%o uplim3 = 10810800; \\
%o v155043 = vector(uplim1);
%o v262503 = vector(uplim3);
%o v262507 = vector(uplim3);
%o v155043[1] = 1; v155043[2] = 1;
%o for(i=3, uplim1, v155043[i] = 1 + v155043[i-numdiv(i)]);
%o A155043 = n -> if(!n,n,v155043[n]);
%o for(i=1, uplim1, v262503[v155043[i]] = i; v262507[v155043[i]]++; if(!(i%1048576),print1(i,", ")));
%o A262503 = n -> if(!n,n,v262503[n]);
%o A262507 = n -> if(!n,1,v262507[n]);
%o k=0; for(n=0, uplim3, if((1==A262507(n)) && (A262503(n) <= uplim2), write("b262508.txt", k, " ", n); write("b262509.txt", k, " ", A262503(n)); k++));
%o (Scheme) (define (A262509 n) (A261089 (A262508 n)))
%Y Cf. A000005, A049820, A070319, A155043, A262508.
%Y Subsequence of A259934, A261089, A262503 and A262513.
%Y Cf. A262510 (gives the parent nodes for the terms a(1) onward), A262514, A262516, A262517.
%Y Cf. also A086876, A179016.
%K nonn
%O 0,2
%A _Antti Karttunen_, Sep 25 2015
|
