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, 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
Offset: 1
Keywords
Examples
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.
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
- Michael De Vlieger, Fan style binary tree showing a(n), n = 1..2^13, with a color function representing primes in red, proper prime powers in gold, squarefree composites in green, and numbers neither squarefree nor prime powers in blue and magenta, where the latter also represents powerful numbers that are not prime powers.
- Michael De Vlieger, Log log scatterplot of a(n), n = 1..10000, using the same color function as immediately above.
- Rémy Sigrist, PARI program
Programs
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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 *)
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PARI
\\ See Links section.
Formula
a(2^n + 1) = prime(n + 1); n >= 0
Comments