A275418 Numbers n such that n - 1 has exactly as many odd divisors as n + 1.

3, 4, 6, 11, 12, 13, 18, 21, 23, 25, 27, 30, 34, 39, 42, 45, 47, 56, 57, 60, 72, 75, 81, 86, 87, 92, 93, 94, 95, 99, 102, 105, 108, 109, 117, 123, 124, 131, 135, 138, 139, 142, 144, 147, 150, 155, 159, 160, 165, 169, 177, 180, 184, 186, 192, 193, 198, 202, 204, 207, 213, 214, 216

Offset: 1

Author

Juri-Stepan Gerasimov, Jul 27 2016

Keywords

• nonn
• easy

Comments

Numbers n > 1 such that d(2n - 2) + d(n + 1) = d(2n + 2) + d(n - 1) where d = A000005.

Conjectures:

(1) There are only finitely many terms n such that A001227(n - 1) = A001227(n + 1) is odd: 3, 99, 577, 3363, ... (see A276188).

(2) There are only finitely many terms n such that A001227(n - 1) = A001227(n) = A001227(n + 1) = 2: 6, 11, 12, 13, 23, 47, 192, 193, 383, 786432, ... (see also A181490-A181493, A276136).

(3) There are only finitely many prime terms p such that A001227(p - 1) = A001227(p + 1) is prime: 11, 13, 23, 47, 193, 383, 577, ... (see also A275598).

I don't find any more for conjecture #3 up to 10^10. - Charles R Greathouse IV, Aug 22 2016

Examples

			3 is in this sequence because 2 and 4 both have only one odd divisor, 1.
4 is in this sequence because 3 and 5 both have exactly two odd divisors each (1 and 3 for the former, 1 and 5 for the latter).
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Supersequence of A014574.

Cf. A000005, A001227, A067888, A181490-A181493, A273401, A275598, A276136, A276188.

Programs

Magma

[n: n in [2..216] | NumberOfDivisors(2*(n-1))+ NumberOfDivisors(n+1) eq NumberOfDivisors(2*(n+1))+ NumberOfDivisors(n-1)];

Maple

N:= 1000: # to get all terms < N
nod:= proc(n) numtheory:-tau(n/2^padic:-ordp(n,2)) end proc:
X:= map(nod,[$1..N]):
select(t -> X[t+1]=X[t-1], [$2..N-1]); # Robert Israel, Aug 04 2016

Mathematica

f[n_] := Count[Divisors@ n, k_ /; OddQ@ k]; Select[Range[2, 240], f[# - 1] == f[# + 1] &] (* Michael De Vlieger, Jul 28 2016 *)
Flatten[Position[Partition[Table[Count[Divisors[n],?OddQ],{n,300}],3,1],?(#[[1]]==#[[3]]&),{1},Heads->False]]+1 (* Harvey P. Dale, Nov 02 2016 *)

PARI

a001227(n) = sumdiv(n, d, d%2);
is(n) = a001227(n-1)==a001227(n+1) \\ Felix Fröhlich, Jul 27 2016

PARI

is(n)=numdiv((n-1)>>valuation(n-1,2)) == numdiv((n+1)>>valuation(n+1,2)) \\ Charles R Greathouse IV, Jul 29 2016

Extensions

Name edited by Alonso del Arte, Aug 23 2016