A069781 -id:A069781 - cp's OEIS frontend

A069782 Numbers k such that gcd(d(k^3), d(k)) = 2^w for some w.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107

Offset: 1

Labos Elemer, Apr 08 2002

The first missing integer is 432 (see in A069781).

			Below 100000 only 314 integers are missing, collected in A069781.

Andrew Howroyd, Table of n, a(n) for n = 1..10000
Tanya Khovanova, Non Recursions

Cf. A000005, A069780, A069781, A037992, A061701.

f[x_] := GCD[DivisorSigma[0, x^3], DivisorSigma[0, x]]; Do[s=f[n]; If[IntegerQ[Log[2, s]], Print[{n, s}]], {n, 1, 100000}]

is(n)=my(f=factor(n)[, 2], g=gcd(prod(i=1, #f, 3*f[i]+1), prod(i=1, #f, f[i]+1))); g>>valuation(g, 2)==1 \\ Charles R Greathouse IV, Oct 16 2015

A069780 a(n) = gcd(d(n^3), d(n)).

Original entry on oeis.org

1, 2, 2, 1, 2, 4, 2, 2, 1, 4, 2, 2, 2, 4, 4, 1, 2, 2, 2, 2, 4, 4, 2, 8, 1, 4, 2, 2, 2, 8, 2, 2, 4, 4, 4, 1, 2, 4, 4, 8, 2, 8, 2, 2, 2, 4, 2, 2, 1, 2, 4, 2, 2, 8, 4, 8, 4, 4, 2, 4, 2, 4, 2, 1, 4, 8, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 2, 2, 1, 4, 2, 4, 4, 4, 4, 8, 2, 4, 4, 2, 4, 4, 4, 4, 2, 2, 2, 1, 2, 8, 2, 8, 8

Offset: 1

Labos Elemer, Apr 08 2002

nonn

Terms are usually powers of 2. Smallest number m such that A069780(m)=2^n is A037992(n). The first n such that a(n) is not a power of 2 equals 432: a(432) = gcd(d(80621568), d(432)) = gcd(130,20) = 10.

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

Cf. A000005, A037992, A069781, A069782.

Table[GCD[DivisorSigma[0, n^3], DivisorSigma[0, n]], {n, 1, 500}]

a(n)=my(f=factor(n)[, 2]); gcd(prod(i=1, #f, 3*f[i]+1), prod(i=1, #f, f[i]+1)) \\ Charles R Greathouse IV, Oct 16 2015

a(n) = gcd(A000005(n^3), A000005(n)).

A069783 a(n) = gcd(d(n!^3), d(n!)), where d() is the number of divisors function.

Original entry on oeis.org

1, 2, 4, 8, 16, 2, 4, 32, 32, 10, 20, 8, 16, 32, 448, 448, 1792, 32, 64, 80, 3200, 1280, 2560, 320, 448, 1792, 25088, 101920, 203840, 128, 256, 4096, 81920, 112640, 2048, 8960, 17920, 1024, 2048, 5120, 10240, 5734400, 11468800, 1003520, 250880, 8960, 17920

Offset: 1

Labos Elemer, Apr 08 2002

Amiram Eldar, Table of n, a(n) for n = 1..1000

Cf. A000005, A000142, A069780, A069781, A069782.

Table[GCD[DivisorSigma[0, (n!)^3], DivisorSigma[0, n! ]], {n, 1, 100}]

a(n) = my(e = factor(n!)[, 2]); gcd(vecprod(apply(x -> x + 1, e)), vecprod(apply(x -> 3*x + 1, e))) \\ Amiram Eldar, Mar 07 2025

a(n) = A069780(n!).

A069784 Numbers m such that gcd(d((m!)^3), d(m!)) = 2^k, i.e., is a power of 2; d = A000005.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 14, 18, 19, 30, 31, 32, 35, 38, 39

Offset: 1

Labos Elemer, Apr 08 2002

From David A. Corneth, Jul 31 2017: (Start)
Theorem: There are no further terms.
Proof:
Let e_n(n, p) be the exponent of p in n!. The prime p has exponent e_(n, p) = n/p for sqrt(n) < p < n in n!. n/4 <= p < n/3, e_(n, p) = 3 so e_(n, p) * 3 = 9. and for n/5 <= p < n/4, e = 4. The gap g_n between prime(n) and prime(n+1) is about sqrt(n) * log(n). There is a gap of n/4 - n/5 = n/20 between n/5 and n/4. primepi(1000) = 168, so for n > 5*1000, the gap between n/5 and the next prime is about sqrt(168) * log(168) ~= 66. This is much less than n/20. No 40 <= m <= 15000 is in the sequence, which completes the proof. (End)

Cf. A000005, A000142, A069780, A069781, A069782, A069783.

Do[s=GCD[DivisorSigma[0, (n!)^3], DivisorSigma[0, n! ]]; If[IntegerQ[n/100], Print[{n}]]; If[IntegerQ[Log[2, s]], Print[n]], {n, 1, 10000}]

val(n, p) = my(r=0); while(n, r+=n\=p);r
is(n) = {my(p1 = p2 = 1); forprime(p=2, n, v = val(n, p); p1 *= (v + 1); p2 *= (3*v + 1)); g = gcd(p1, p2); g==2^(valuation(g, 2))} \\ David A. Corneth, Jul 31 2017

Keywords fini and full added by David A. Corneth, Jul 31 2017

A069785 a(n) = A061680(n!).

Original entry on oeis.org

1, 1, 1, 1, 1, 15, 15, 3, 5, 135, 135, 99, 99, 9, 63, 21, 21, 459, 459, 135, 19, 15, 15, 15, 21, 189, 189, 585, 585, 18225, 18225, 675, 15, 135, 891, 8505, 25515, 81, 81, 7695, 7695, 1575, 1575, 4725, 6615, 40635, 40635, 945, 1215, 3645, 3645, 151875, 151875

Offset: 1

Labos Elemer, Apr 09 2002

			Observe cases when consecutive terms are equal: n={1,2,3,4,6,10,...,78,80,82,88,96}.

Amiram Eldar, Table of n, a(n) for n = 1..1000

Cf. A000005, A000142, A061680, A061701, A048691, A069780, A069781, A069782, A069783, A069784.

a[n_] := Module[{e = FactorInteger[n!][[;;, 2]]}, GCD[Times @@ (2*e+1), Times @@ (e+1)]]; Array[a, 100] (* Amiram Eldar, Dec 02 2023 *)

a(n) = {my(e = factor(n!)[,2]); gcd(vecprod(apply(x -> 2*x+1, e)), vecprod(apply(x -> x+1, e)));} \\ Amiram Eldar, Dec 02 2023

a(n) = A061680(A000142(n)). - Amiram Eldar, Dec 02 2023

cp's OEIS Frontend

A069782 Numbers k such that gcd(d(k^3), d(k)) = 2^w for some w.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A069780 a(n) = gcd(d(n^3), d(n)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A069783 a(n) = gcd(d(n!^3), d(n!)), where d() is the number of divisors function.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A069784 Numbers m such that gcd(d((m!)^3), d(m!)) = 2^k, i.e., is a power of 2; d = A000005.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

Extensions

A069785 a(n) = A061680(n!).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula