OFFSET
1,1
COMMENTS
a(1) = 2 since b(2) = 2 does not divide s(1) = 1.
For n > 2, 1 < gcd(b(a(n)), s(a(n)-1)) < b(a(n)).
For n > 2, both b(a(n)) and s(a(n)-1) are necessarily composite, since prime p either divides or is coprime to n. Furthermore, both b(a(n)) and s(a(n))-1) have at least 2 distinct prime factors.
Indices of records in A124652 except {1, 2, 3, 5} are in this sequence.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Log log scatterplot of A124652(n), n = 1..10^5, showing A124652(a(n)) in red.
EXAMPLE
a(2) = 4 since b(4) = 4 does not divide s(3) = 6.
a(3) = 6 since b(6) = 9 does not divide s(5) = 15.
a(4) = 8 since b(8) = 8 does not divide s(7) = 30.
a(5) = 9 since b(9) = 16 does not divide s(8) = 38, etc.
Table of b(k) and s(k-1), where k = a(n), n = 2..12. Asterisked k denote terms such that rad(b(k)) | rad(s(k-1)); k = 73 and k = 4316 are the only other known indices where the terms have this quality.
k b(k) s(k-1)
----------------------------------------------------------
4 4 = 2^2 6 = 2 * 3
6 9 = 3^2 15 = 3 * 5
8 8 = 2^3 30 = 2 * 3 * 5
9 16 = 2^4 38 = 2 * 19
10* 12 = 2^2 * 3 54 = 2 * 3^3
14* 28 = 2^2 * 7 98 = 2 * 7^2
19 32 = 2^5 216 = 2^3 * 3^3
21 81 = 3^4 279 = 3^2 * 31
23 20 = 2^2 * 5 370 = 2 * 5 * 37
25 169 = 13^2 403 = 13 * 31
32 49 = 7^2 728 = 2^3 * 7 * 13
...
73* 100 = 2^2 * 5^2 4800 = 2^6 * 3 * 5^2
...
4316* 4720 = 2^4 * 5 * 59 30806850 = 2 * 3 * 5^2 * 59^3
MATHEMATICA
nn = 120; c[_] := False;
rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
f[x_] := Select[Range[x], Divisible[x, rad[#]] &];
Array[Set[{a[#], c[#]}, {#, True}] &, 2]; s = a[1] + a[2];
{2}~Join~Reap[Do[
r = f[s]; k = SelectFirst[r, ! c[#] &];
If[! Divisible[s, k], Sow[i]];
c[k] = True;
s += k, {i, 3, nn}] ][[-1, 1]]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael De Vlieger, May 05 2024
STATUS
approved