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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 = 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). 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 (list; graph; refs; listen; history; text; internal format)
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
Rémy Sigrist, PARI program
FORMULA
a(2^n + 1) = prime(n + 1); n >= 0
A000005(a(n)) = A063787(n). - Rémy Sigrist, Oct 06 2022
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
Sequence in context: A162374 A323289 A355374 * A065885 A274331 A177064
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified March 5 09:44 EST 2024. Contains 370545 sequences. (Running on oeis4.)