

A357601


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 = n2^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).


1



1, 2, 3, 4, 5, 9, 25, 8, 7, 49, 121, 6, 169, 10, 14, 16, 11, 289, 361, 15, 529, 21, 22, 81, 841, 26, 27, 625, 33, 2401, 14641, 32, 13, 961, 1369, 34, 1681, 35, 38, 28561, 1849, 39, 46, 83521, 51, 130321, 279841, 12, 2209, 55, 57, 707281, 58, 923521, 1874161, 18
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OFFSET

1,2


COMMENTS

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.


LINKS



FORMULA

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


EXAMPLE

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.


MATHEMATICA

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 *)


PROG

(PARI) See Links section.


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



