|
|
A188163
|
|
Smallest m such that A004001(m) = n.
|
|
15
|
|
|
1, 3, 5, 6, 9, 10, 11, 13, 17, 18, 19, 20, 22, 23, 25, 28, 33, 34, 35, 36, 37, 39, 40, 41, 43, 44, 46, 49, 50, 52, 55, 59, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 77, 78, 79, 81, 82, 84, 87, 88, 89, 91, 92, 94, 97, 98, 100, 103, 107, 108, 110, 113, 117, 122
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
It is not hard to show that a(n) exists for all n, and in particular a(n) < 2^n. - Charles R Greathouse IV, Jan 13 2013
Positions of records in A004001. After 1 the positions where A004001 increases (by necessity by one).
An answer to the question of R. J. Mathar above: This sequence is equal to A088359 with prepended 1. This follows because at each of its unique values (terms of A088359), A004001 must grow, but it can grow nowhere else. See Kubo and Vakil paper and especially the illustrations of Q and R-trees on pages 229-230 (pages 5 & 6 in PDF) and also in sequence A265332.
Obviously A004001 can obtain unique values only at points which form a subset (A266399) of this sequence.
(End)
|
|
LINKS
|
|
|
FORMULA
|
Other identities. For all n >= 1:
|
|
MAPLE
|
for a from 1 do
return a;
end if;
end do:
|
|
MATHEMATICA
|
h[1] = 1; h[2] = 1; h[n_] := h[n] = h[h[n-1]] + h[n - h[n-1]];
a[n_] := For[m = 1, True, m++, If[h[m] == n, Return[m]]];
|
|
PROG
|
(Haskell)
import Data.List (elemIndex)
import Data.Maybe (fromJust)
a188163 n = succ $ fromJust $ elemIndex n a004001_list
(Scheme)
|
|
CROSSREFS
|
Cf. A087686 (complement, apart from the initial 1).
Cf. A004001 (also the least monotonic left inverse of this sequence).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|