A276136 -id:A276136 - cp's OEIS frontend

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

Juri-Stepan Gerasimov, Jul 27 2016

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

			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).

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

Supersequence of A014574.

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

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

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

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 *)

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

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

Name edited by Alonso del Arte, Aug 23 2016

A369329 Numbers whose neighbors have a prime number as their greatest odd divisor.

Original entry on oeis.org

4, 6, 11, 12, 13, 18, 21, 23, 25, 27, 30, 39, 42, 45, 47, 57, 60, 72, 75, 81, 87, 93, 95, 102, 105, 108, 117, 123, 135, 138, 147, 150, 159, 165, 177, 180, 192, 193, 198, 207, 213, 225, 228, 240, 270, 273, 282, 297, 303, 312, 315, 327, 333, 345, 348, 357, 383, 385, 387

Offset: 1

Juri-Stepan Gerasimov, Jan 20 2024

nonn

Does this sequence contain a finite or infinite number of squares, powers of 3, even numbers, ...?

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Supersequence of A014574 and A276136.

Cf. A038550, A273401, A275418.

[k: k in [2..400] | #Divisors(2*k-2)-#Divisors(k-1) eq 2 and #Divisors(2*k+2)-#Divisors(k+1) eq 2];

filter:= proc(n) local m;
    andmap(m -> isprime(m/2^padic:-ordp(m,2)), [n-1,n+1])
end proc:
select(filter, [$1..1000]); # Robert Israel, Jan 24 2024

q[n_] := PrimeQ[n / 2^IntegerExponent[n, 2]]; Select[Range[400], And @@ q /@ {# - 1, # + 1} &] (* Amiram Eldar, Jan 20 2024 *)
Mean/@SequencePosition[Table[If[PrimeQ[Select[Divisors[n],OddQ][[-1]]],1,0],{n,400}],{1,,1}] (* _Harvey P. Dale, Jul 30 2025 *)

A276188 Numbers k > 1 such that the number of odd divisors of k-1 is odd and is equal to the number of odd divisors of k+1.

Original entry on oeis.org

3, 99, 577, 3363

Offset: 1

Juri-Stepan Gerasimov, Aug 23 2016

Conjecture: this sequence is finite.

Any further terms are greater than 10^10. - Charles R Greathouse IV, Aug 22 2016

			99 is in this sequence because there are 3 odd divisors 1, 7 and 49 of 98 and there are 3 odd divisors 1, 5 and 25 of 100, and 3 is odd.

Cf. A275418, A275598, A276136.

[n: n in [2..100000] | NumberOfDivisors(2*(n-1))- NumberOfDivisors(n-1) eq NumberOfDivisors(2*(n+1))-NumberOfDivisors(n+1) and ((NumberOfDivisors(2*(n+1))- NumberOfDivisors(n+1)) mod 2) eq 1 ];

odo[n_]:=Module[{c=Select[Divisors[n],OddQ]},If[OddQ[Length[c]],Length[c],0]]; Flatten[ Position[ Partition[Array[odo,3500],3,1],?(AllTrue[{#[[1]],#[[3]]},OddQ]&&#[[1]]==#[[3]]&),1,Heads->False]]+1 (* _Harvey P. Dale, Apr 07 2023 *)

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

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A369329 Numbers whose neighbors have a prime number as their greatest odd divisor.

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A276188 Numbers k > 1 such that the number of odd divisors of k-1 is odd and is equal to the number of odd divisors of k+1.

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