A273401 Numbers n such that n and n + 1 have exactly the same number of odd divisors.

1, 5, 6, 10, 11, 12, 13, 19, 22, 23, 28, 37, 40, 43, 46, 47, 49, 52, 54, 58, 61, 65, 67, 69, 73, 77, 79, 82, 84, 88, 96, 103, 106, 110, 112, 114, 119, 129, 132, 136, 140, 148, 151, 154, 155, 157, 163, 166, 172, 178, 182, 185, 186, 191, 192, 193, 203, 204, 211, 215, 216, 219, 220, 221

Offset: 1

Author

Juri-Stepan Gerasimov, May 26 2016

Keywords

• nonn

Comments

If A001227(n) = A001227(n*2^m) for m >= 0 then:

1) A001227(n) is equal to number of ways to write 2n - 1 as (4*x + 2)*y + 4*x + 1 where x and y are nonnegative integers;

2) A001227(n) is equal to number of distinct values of k if k/(2n-1) + 1 divides (k/(2n - 1))^(k/(2n - 1)) + k, (k/(2n - 1))^k + k/(2n - 1) and k^(k/(2n - 1)) + k/(2n - 1).

Examples

			5 and 6 have both two odd divisors: (1 and 5) and (1 and 3) respectively; so 5 is a term in the sequence.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001227, A206581 (primes in a(n)).

Programs

Maple

A001227:= n -> numtheory:-tau(n)/(1+padic:-ordp(n,2)):
R:= map(A001227,[$1..1000]):
ListTools:-SearchAll(0,A001227[2..-1]-A001227[1..-2]); # Robert Israel, May 27 2016

Mathematica

Select[Range@ 221, First@ Differences@ Map[Count[Divisors@ #, ?OddQ] &, {#, # + 1}] == 0 &] (* _Michael De Vlieger, Jun 26 2016 *)
SequencePosition[Table[Count[Divisors[n],?OddQ],{n,250}],{x,x_}] [[All, 1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Apr 06 2019 *)

PARI

lista(nn) = for (n=1, nn, if (sumdiv(n, d, d%2) == sumdiv(n+1, d, d%2), print1(n, ", "))); \\ Michel Marcus, May 27 2016

PARI

is(n)=numdiv(n>>valuation(n,2))==numdiv((n+1)>>valuation(n+1,2)) \\ Charles R Greathouse IV, Jul 15 2016