OFFSET
1,2
COMMENTS
There is at least one infinite sequence that satisfies the parameters of the sequence, because it is always true that A075826(k) < k for any k. (For the statement to be false, the number of letters must be less than or equal to 0.)
All known values in this sequence are guaranteed to be in any infinite sequence which can be produced, since iterating A075826(k) where the starting value of k is any number greater than 11126 converges at 6890 or larger. This proves any infinite sequence must contain 6890 and all values returned by iterating A075826 down to 0.
LINKS
Michel Marcus, Table of n, a(n) for n = 1..11623
FORMULA
a(n-1) = A075826(a(n)).
EXAMPLE
A075826 yields the number of letters in US English spelling of each number subtracted from the number's value. For example, A075826(5) yields 1, because 5 - 4 (F-I-V-E) is 1. Our formula reveals that if 5 is in our sequence, it must come immediately after 1, which it does. However, 4, which must come immediately after 0, is not in our sequence, which we can prove because there exists no number k such that A075826(k) = 4. Each number less than the largest known value in this sequence produces a finite sequence unless it is in this sequence, because some number eventually is a dead end like 4.
PROG
(PARI) f(k) = k - A005589(k);
listd(nn) = {nn *= 2; my(vs = vector(nn)); my(list = List()); my(m=1, lbound = 0); listput(list, m); while (m < nn, if (vs[m] == 0, vs[m] = Vec(select(x->(x==m), vector(100, k, f(k+m-1)), 1)); if (#vs[m], vs[m] = apply(x->(x+m-1), vs[m]))); my(ok = 1, vc = vs[m]); if (! #vc, ok = 0, vc = select(x->(x>lbound), vc); if (! #vc, ok = 0); ); if (!ok, lbound = list[#list]; listpop(list); if (! #list, return()); m = list[#list]; , lbound = 0; m = vc[1]; listput(list, m); ); ); Vec(select(x->(x<=nn/2), list)); } \\ Michel Marcus, Aug 31 2022
CROSSREFS
KEYWORD
nonn,word
AUTHOR
Aidan Clarke, Aug 22 2022
STATUS
approved