A268066 -id:A268066 - cp's OEIS frontend

A268037 Numbers k such that the number of divisors of k+2 divides k and the number of divisors of k divides k+2.

4, 30, 48, 110, 208, 270, 320, 368, 510, 590, 688, 750, 1070, 1216, 1328, 1566, 1808, 2030, 2190, 2510, 2670, 2768, 3008, 3088, 3728, 4110, 4208, 4430, 4528, 4688, 4698, 4910, 5008, 5696, 5870, 5886, 5968, 6128, 6592, 6846, 7088, 7310, 7790, 8384, 9008, 9230, 9390, 9488, 9534, 9710

Offset: 1

Waldemar Puszkarz, Apr 27 2016

nonn

One can call such pairs {n, n+2} mutually (or amicably) step 2 refactorable numbers.
All terms are even.
Proof. Let us suppose n is odd. Then so is n+2 and the number of their divisors has to be odd, which implies they are squares (see the proof in A268066). Since the separation between closest squares is 2n+1, it is always greater than 2, except for n=0, when it is 1.
Contains 48 + 160*k if 3 + 10*k and 5 + 16*k are prime. Dickson's conjecture implies that there are infinitely many of these. - Robert Israel, May 09 2016

			4 is a term because its number of divisors (3) divides 6=4+2 and the number of divisors of 6 (4) divides 4.

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

Cf. A000005 (number of divisors), A033950 (refactorable numbers), A272353, A269781 (related sequences).

select(n -> n mod numtheory:-tau(n+2)=0 and (n+2) mod numtheory:-tau(n) = 0,
2*[$1..10000]); # Robert Israel, May 09 2016

lst={}; Do[ If[ Divisible[n, DivisorSigma[0, n+2]]&&Divisible[n+2, DivisorSigma[0, n]], AppendTo[lst, n]], {n, 12000}]; lst
Select[Range[12000], Divisible[#, DivisorSigma[0, # + 2]] && Divisible[# + 2, DivisorSigma[0, #]] &]

for(n=1, 12000, (n%numdiv(n+2)==0)&&((n+2)%numdiv(n)==0)&&print1(n ", "))

from sympy import divisors
def ok(n): return n%len(divisors(n+2)) == 0 and (n+2)%len(divisors(n)) == 0
print(list(filter(ok, range(1, 9711)))) # Michael S. Branicky, Apr 30 2021

A269818 Numbers coprime to the number of their even divisors.

Original entry on oeis.org

1, 2, 8, 32, 50, 98, 128, 162, 200, 242, 338, 392, 512, 578, 722, 968, 1058, 1352, 1458, 1568, 1682, 1922, 2048, 2312, 2450, 2592, 2738, 2888, 3200, 3362, 3698, 3872, 4232, 4418, 4802, 5408, 5618, 6050, 6728, 6962, 7442, 7688, 8192, 8450, 8978, 9248, 9800, 10082, 10368, 10658, 10952, 11552, 11858, 12482, 12800

Offset: 1

Waldemar Puszkarz, Mar 05 2016

nonn

This sequence is characterized by the following property (theorem).
Theorem. If n is coprime to the number of its even divisors, then n is 1 or of the form 2m^2, m>0.
Proof. If n is odd, its number of even divisors is 0 and since gcd(n,0)=|n| (for any n), n must be 1 to be coprime to 0. If n is even, then it is of the form 2^k*p^a*q*^b*...*r^c, where p, q, r are odd primes and k, a, b, c are positive integers, and its sum of even divisors is k*(1+a)*(1+b)*...*(1+c). The latter number can be coprime to an even number only if it is odd, implying that k must be odd and a, b, ..., c must be even; thus n is twice a square.

			For n = 3, a(3) = 8 is a member for the number of even divisors of 8, (2,4,8), is 3, which is coprime with 8.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A183063 (number of even divisors), A046642 (numbers coprime to the number of their divisors), A269870 (counterpart for the number of odd divisors), A268066 (related sequence).

Select[Range@13000, CoprimeQ[#, Length@Select[Divisors[#], EvenQ]]&]

for(n=1, 13000, gcd(n, if(n%2, 0, numdiv(n/2)))==1&&print1(n, ", "))

A341779 Numbers k such that k and k+1 are both anti-tau numbers (A046642).

Original entry on oeis.org

3, 4, 15, 16, 64, 100, 195, 196, 255, 256, 483, 484, 676, 783, 784, 1023, 1024, 1155, 1156, 1295, 1296, 1443, 1444, 1599, 1600, 1936, 2116, 2703, 2704, 3363, 3364, 3844, 4096, 4623, 4624, 4899, 4900, 5183, 5184, 5476, 5776, 6399, 6400, 6723, 6724, 7395, 7396

Offset: 1

Amiram Eldar, Feb 19 2021

nonn

Since the even anti-tau numbers (A268066) are square numbers, all the terms are either of the form 4*k^2 - 1 or 4*k^2.

			3 is a term since 3 and 4 are both anti-tau numbers: gcd(3, tau(3)) = gcd(3, 2) = 1 and gcd(4, tau(4)) = gcd(4, 3) = 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Subsequence of A046642 and A081350.

Cf. A000010, A009191, A114617, A268066, A341780.

antiTauQ[n_] := CoprimeQ[n, DivisorSigma[0, n]]; s = {}; Do[k = 4*n^2; If[antiTauQ[k], If[antiTauQ[k - 1], AppendTo[s, k - 1]]; If[antiTauQ[k + 1], AppendTo[s, k]]], {n, 1, 50}]; s

A341780 Starts of runs of 3 consecutive anti-tau numbers (A046642).

Original entry on oeis.org

3, 15, 195, 255, 483, 783, 1023, 1155, 1295, 1443, 1599, 2703, 3363, 4623, 4899, 5183, 6399, 6723, 7395, 7743, 8463, 8835, 10815, 11235, 11663, 12099, 12543, 15375, 16383, 16899, 17955, 18495, 20163, 24963, 25599, 26895, 27555, 31683, 33855, 35343, 36099, 37635

Offset: 1

Amiram Eldar, Feb 19 2021

nonn

Since the even anti-tau numbers (A268066) are square numbers, all the terms are of the form 4*k^2 - 1, and there cannot be a run of more than 3 consecutive anti-tau numbers.

			3 is a term since 3, 4 and 5 are all anti-tau numbers: gcd(3, tau(3)) = gcd(3, 2) = 1, gcd(4, tau(4)) = gcd(4, 3) = 1 and gcd(5, tau(5)) = gcd(5, 2) = 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Subsequence of A000466, A046642 and A341779.

Cf. A000010, A009191, A268066.

antiTauQ[n_] := CoprimeQ[n, DivisorSigma[0, n]]; Select[4*Range[100]^2 - 1, AllTrue[# + {0, 1, 2}, antiTauQ] &]

cp's OEIS Frontend

A268037 Numbers k such that the number of divisors of k+2 divides k and the number of divisors of k divides k+2.

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A269818 Numbers coprime to the number of their even divisors.

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A341779 Numbers k such that k and k+1 are both anti-tau numbers (A046642).

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Mathematica

A341780 Starts of runs of 3 consecutive anti-tau numbers (A046642).

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Mathematica