%I #16 Oct 06 2022 10:31:56

%S 1,2,3,4,5,9,25,8,7,49,121,6,169,10,14,16,11,289,361,15,529,21,22,81,

%T 841,26,27,625,33,2401,14641,32,13,961,1369,34,1681,35,38,28561,1849,

%U 39,46,83521,51,130321,279841,12,2209,55,57,707281,58,923521,1874161,18

%N For n a power of 2, a(n) = n; otherwise, if 2^m is the greatest power of 2 not exceeding n and if k = n-2^m, then a(n) is the smallest number having d(a(k))+1 divisors which has not occurred earlier (d is the divisor counting function A000005).

%C Based on a similar recursion to that which produces the Doudna sequence, A005940. Conjectured to be permutation of the positive integers in which the primes appear in natural order.

%H Rémy Sigrist, <a href="/A357601/b357601.txt">Table of n, a(n) for n = 1..10000</a>

%H Rémy Sigrist, <a href="/A357601/a357601.gp.txt">PARI program</a>

%F a(2^n + 1) = prime(n + 1); n >= 0

%F A000005(a(n)) = A063787(n). - _Rémy Sigrist_, Oct 06 2022

%e a(9)=7 because k=1, and a(1)=1, which has 1 divisor, so we are looking for the smallest number not yet seen which has 2 divisors. This must be 7 because 2,3,5 have occurred already.

%t nn = 70; kk = 2^20; c[_] = False; to = Map[DivisorSigma[0, #] &, Range[kk]^2]; t = DivisorSigma[0, Range[kk]]; Do[Set[{m, k}, {1, n - 2^Floor[Log2[n]]}]; If[k == 0, Set[{a[n], c[n]}, {n, True}], d = 1 + DivisorSigma[0, a[k]]; If[OddQ[d], While[Nand[! c[m^2], to[[m]] == d], m++]; Set[{a[n], c[#]}, {#, True}] &[m^2], While[Nand[! c[m], t[[m]] == d], m++]; Set[{a[n], c[m]}, {m, True}]] ], {n, nn}]; Array[a, nn] (* _Michael De Vlieger_, Oct 05 2022 *)

%o (PARI) See Links section.

%Y Cf. A000005, A005940, A063787.

%K nonn

%O 1,2

%A _David James Sycamore_, Oct 05 2022