A100363 -id:A100363 - cp's OEIS frontend

A100364 Numbers n such that the numbers of divisors of n,n+1,n+2 and n+3 are k,2k,4k,8k respectively for some k.

613, 757, 997, 1237, 2917, 6133, 9853, 15733, 19477, 22501, 24421, 25237, 26293, 26437, 28117, 31477, 33301, 35317, 44101, 45061, 46357, 51637, 53077, 56701, 59221, 61653, 61717, 64381, 65917, 66853, 68583, 70501, 70887, 72901, 75133

Offset: 1

Labos Elemer, Nov 19 2004

nonn

			n=613,614,615,616 have 2,4,8,16 divisors respectively.

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

Cf. A000005, A063446, A100363.

d0[x_] :=DivisorSigma[0, x];ta={{0}}; Do[g=n;s=d0[n];s1=d0[n+1];s2=d0[n+2];s3=d0[n+3] If[Equal[s1, 2*s]&&Equal[s2, 4*s]&&Equal[s3, 8*s], ta=Append[ta, n]; Print[{n, s, s1, s2, s3}]], {n, 1, 1000000}];{ta=Delete[ta, 1], g}
nd2Q[{a_,b_,c_,d_}]:={b,c,d}/a=={2,4,8}; Position[Partition[ DivisorSigma[ 0,Range[80000]],4,1],?(nd2Q[#]&)]//Flatten (* _Harvey P. Dale, Feb 25 2020 *)

A100365 Prime numbers q such that q, q+1, q+2, q+3, q+4 have 2, 4, 8, 16, 32 divisors respectively.

Original entry on oeis.org

1124581, 2101621, 2135701, 3829381, 5801701, 6097381, 6453541, 6535861, 6609781, 6799621, 6972661, 7055317, 7527061, 8281381, 8524981, 9412981, 9895141, 11455141, 11901781, 12043621, 12929941, 13749061, 14747701, 15504661

Offset: 1

Labos Elemer, Nov 19 2004

nonn

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

Cf. A000005, A063446, A100363, A100364.

Select[Prime[Range[1002000]],DivisorSigma[0,Range[0,4]+#]=={2,4,8,16,32}&] (* Harvey P. Dale, Oct 04 2019 *)

A339778 Numbers m such that numbers m, m + 1 and m + 2 have k, 2k and 3k divisors respectively.

Original entry on oeis.org

61, 73, 277, 421, 458, 493, 583, 1234, 1393, 1418, 1658, 1909, 1954, 2066, 2138, 2234, 2329, 2386, 2533, 2594, 2773, 2797, 2846, 3013, 3073, 3265, 3394, 3841, 4322, 4333, 4538, 4586, 4633, 4717, 4754, 4766, 5029, 5223, 5245, 5342, 5378, 5554, 5893, 5906, 6169

Offset: 1

Jaroslav Krizek, Dec 16 2020

nonn

Numbers m such that tau(m) = tau(m + 1) / 2 = tau(m + 2) / 3, where tau(k) = the number of divisors of k (A000005).
Corresponding values of tau(a(n)): 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 4, ...
Triplets of [tau(a(n)), tau(a(n) + 1), tau(a(n) + 2)] = [tau(a(n)), 2*tau(a(n)), 3*tau(a(n))]: [2, 4, 6], [2, 4, 6], [2, 4, 6], [2, 4, 6], [4, 8, 12], [4, 8, 12], [4, 8, 12], [4, 8, 12], [4, 8, 12], ...

			tau(61) = 2, tau(62) = 4, tau(63) = 6.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A000005, A100363.

Subsequence of A063446.

[m: m in [1..10^5] | #Divisors(m) eq #Divisors(m + 1) / 2 and #Divisors(m) eq #Divisors(m + 2) / 3]

Select[Range[6000], Equal @@ (DivisorSigma[0, # + {0, 1, 2}]/{1, 2, 3}) &] (* Amiram Eldar, Dec 16 2020 *)

isok(m) = my(nb = numdiv(m)); (numdiv(m+1) == 2*nb) && (numdiv(m+2) == 3*nb); \\ Michel Marcus, Dec 18 2020

A328137 Primes p such that p+1 is the product of two distinct primes and p+2 is the product of three distinct primes.

Original entry on oeis.org

193, 397, 613, 661, 757, 1093, 1237, 1453, 1657, 2137, 2341, 2593, 2917, 3217, 4177, 4621, 5233, 6121, 6133, 7057, 7537, 8101, 8317, 8353, 8677, 8893, 9013, 9721, 10957, 11677, 11701, 12421, 12433, 12541, 12853, 13933, 15277, 15733, 16033, 16381, 16417, 16633, 17257, 17293, 18013, 18253, 18481

Offset: 1

J. M. Bergot and Robert Israel, Oct 04 2019

nonn

All terms == 1 (mod 12).

Members k of A112998 such that k+2 is squarefree.

			a(3)=613 is in the sequence because 613 is prime, 614=2*307 is the product of two distinct primes, and 615=3*5*41 is the product of three distinct primes.

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

Contained in A005383, A100363 and A112998.

[p:p in PrimesUpTo(20000)| IsPrime((p+1) div 2) and IsSquarefree(p+2) and #PrimeDivisors(p+2) eq 3]; // Marius A. Burtea, Oct 04 2019

select(t -> isprime(t) and isprime((t+1)/2) and numtheory:-issqrfree(t+2) and numtheory:-bigomega(t+2)=3, [seq(i,i=1..10^5,12)]);

A100366 a(n) is the least prime number q such that q,q+1,q+2,q+3,...,q+n-1 have 2,4,8,...,2^n divisors respectively.

Original entry on oeis.org

2, 5, 193, 613, 1124581, 52071301, 213536830501

Offset: 1

Labos Elemer, Nov 19 2004

a(3), a(4), a(5) are the initial terms of A100363, A100364, A100365 resp.

Any run of 8 or more consecutive integers must include at least one number k of the form 8j+4; in the prime factorization of k, the prime factor 2 appears with multiplicity exactly 2, so the number of divisors of k is divisible by 3 (which is not a power of 2). Thus, there is no term a(8): the sequence is complete, ending with a(7). - Jon E. Schoenfield, Nov 12 2017

			a(4)=613: q=613 (a prime, hence two divisors), q+1 = 614 = 2*307 (4 divisors), q+2 = 615 = 3*5*41 (8 divisors), and q+3 = 616 = 2^3 * 7 * 11 (16 divisors).

Cf. A000005, A063446, A100363, A100364.

a(6)-a(7) from Donovan Johnson, Mar 23 2011

Keywords fini and full added and Example section edited by Jon E. Schoenfield, Nov 12 2017

A340229 Numbers m such that numbers m, m + 1, m + 2, m + 3 and m + 4 have k, 2k, 4k, 8k and 16k divisors respectively.

Original entry on oeis.org

1124581, 2101621, 2135701, 3829381, 5801701, 6097381, 6453541, 6535861, 6609781, 6799621, 6972661, 7055317, 7527061, 8281381, 8485502, 8524981, 8883326, 9412981, 9895141, 11455141, 11901781, 12043621, 12929941, 13749061, 14747701, 15150901, 15504661, 15533941

Offset: 1

Jaroslav Krizek, Jan 01 2021

nonn

Numbers m such that tau(m) = tau(m + 1)/2 = tau(m + 2)/4 = tau(m + 3)/8 = tau(m + 4)/16, where tau(k) = the number of divisors of k (A000005).
Quintuples of [tau(a(n)), tau(a(n) + 1), tau(a(n) + 2), tau(a(n) + 3), tau(a(n) + 4)] = [tau(a(n)), 2*tau(a(n)), 4*tau(a(n)), 8*tau(a(n)), 16*tau(a(n))]: [2, 4, 8, 16, 32], [2, 4, 8, 16, 32], [2, 4, 8, 16, 32], [2, 4, 8, 16, 32], [2, 4, 8, 16, 32], [2, 4, 8, 16, 32], ...
Corresponding values of numbers k: 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 4, 4, ...
Prime terms are in A100365; number 8485502 is the smallest composite term.
Subsequence of A063446, A100363 and A100364.

			tau(1124581) = 2, tau(1124582) = 4, tau(1124583) = 8, tau(1124584) = 16, tau (1124585) = 32.

Cf. A000005, A063446, A100363, A100364, A100365, A340230.

[m: m in [1..10^7] | #Divisors(m) eq #Divisors(m + 1) / 2 and #Divisors(m) eq #Divisors(m + 2) / 4 and #Divisors(m) eq #Divisors(m + 3) / 8 and #Divisors(m) eq #Divisors(m + 4) / 16]

isok(m) = vector(4, k, numdiv(m+k))/numdiv(m) == [2,4,8,16]; \\ Michel Marcus, Jan 02 2021

A340230 a(n) is the smallest number m such that numbers m, m + 1, m + 2, ..., m + n - 1 have k, 2k, 4k, 8k, ..., (2^(n-1))k divisors respectively.

Original entry on oeis.org

1, 1, 193, 613, 1124581, 52071301, 213536830501

Offset: 1

Jaroslav Krizek, Jan 01 2021

a(n) is the smallest number m such that tau(m) = tau(m + 1) / 2 = tau(m + 2) / 4 = tau(m + 3) / 8 = ... = tau(m + n - 1) / 2^(n - 1), where tau(k) = the number of divisors of k (A000005).

Conjecture: a(7) = 213536830501.

			a(4) = 613 because 613 is the smallest term of 4 consecutive numbers with this property: tau(613) = 2, tau(614) = 4, tau(615) = 8, tau(616) = 16.

Cf. A100366 (similar sequence for primes).

Cf. A000005, A063446, A100363, A100364, A100365, A340229.

isok(m, n) = my(nb=numdiv(m)); for (k=1, n-1, if (numdiv(m+k)/nb != 2^k, return(0))); return (1);
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Jan 05 2021

a(7), as conjectured by Jaroslav Krizek, from Martin Ehrenstein, Feb 06 2021

cp's OEIS Frontend

A100364 Numbers n such that the numbers of divisors of n,n+1,n+2 and n+3 are k,2k,4k,8k respectively for some k.

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A100365 Prime numbers q such that q, q+1, q+2, q+3, q+4 have 2, 4, 8, 16, 32 divisors respectively.

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A339778 Numbers m such that numbers m, m + 1 and m + 2 have k, 2k and 3k divisors respectively.

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A328137 Primes p such that p+1 is the product of two distinct primes and p+2 is the product of three distinct primes.

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A100366 a(n) is the least prime number q such that q,q+1,q+2,q+3,...,q+n-1 have 2,4,8,...,2^n divisors respectively.

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A340229 Numbers m such that numbers m, m + 1, m + 2, m + 3 and m + 4 have k, 2k, 4k, 8k and 16k divisors respectively.

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A340230 a(n) is the smallest number m such that numbers m, m + 1, m + 2, ..., m + n - 1 have k, 2*k, 4*k, 8*k, ..., (2^(n-1))*k divisors respectively.

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A340230 a(n) is the smallest number m such that numbers m, m + 1, m + 2, ..., m + n - 1 have k, 2k, 4k, 8k, ..., (2^(n-1))k divisors respectively.