cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 41-47 of 47 results.

A287800 Numbers n such that phi(n) * tau(n) divides n^2, but neither tau(n) nor phi(n) divides n.

Original entry on oeis.org

900, 2400, 3840, 6480, 7200, 11520, 13056, 39168, 42240, 79200, 83232, 96000, 126720, 145200, 153600, 157440, 174240, 195840, 207360, 288000, 300000, 317520, 326592, 387840, 435600, 460800, 472320, 480000, 900000, 971520, 1056000, 1161600, 1163520, 1228800, 1440000
Offset: 1

Views

Author

Bernard Schott, Jun 01 2017

Keywords

Comments

The GCD of the first 43 terms is 12. The GCD of the first 166 terms is 4. The GCD of a(2) through a(166) is 16. - David A. Corneth, Jun 01 2017

Examples

			For n = 900, tau(900) = 27, phi(900) = 240 and 900^2/(27 * 240) = 125, but 900/27 = 33.33333 and 900/240 = 3.75.

Links

Crossrefs

Cf. A000005, A000010, A007694, A033950, A235353.

Programs

  • Magma
    [k:k in [1..1500000]| k^2 mod (EulerPhi(k) *NumberOfDivisors(k)) eq 0 and (k mod  EulerPhi(k) ne 0) and (k mod NumberOfDivisors(k) ne 0)]; // Marius A. Burtea, Dec 30 2019
  • Maple
    for n from 1 to 100000 do p(n):=n^2/(tau(n)*phi(n));
if p(n)=floor(p(n)) and n/tau(n)<>floor(n/tau(n)) and n/phi(n)<>floor(n/phi(n)) then print(n,p(n),phi(n),tau(n)) else fi; od:
  • Mathematica
    Select[Range[10^6], Function[n, And[Divisible[n^2, #1 #2], NoneTrue[{#1, #2}, Divisible[n, #] &]] & @@ {DivisorSigma[0, n], EulerPhi[n]}]] (* Michael De Vlieger, Jun 01 2017 *)
  • PARI
    is(n) = n^2 % (numdiv(n)*eulerphi(n)) == 0 && n % numdiv(n) != 0 && n % eulerphi(n) % n!=0 \\ David A. Corneth, Jun 01 2017

A303691 a(n) is the number of 3-smooth numbers k such that prime(n)-k is also a prime number, where prime(n) stands for the n-th prime.

Original entry on oeis.org

0, 1, 2, 2, 4, 3, 3, 5, 5, 6, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 6, 8, 9, 6, 5, 8, 7, 9, 8, 9, 6, 8, 8, 9, 7, 9, 8, 8, 10, 8, 8, 11, 8, 6, 10, 12, 10, 9, 9, 11, 9, 8, 8, 8, 8, 11, 9, 9, 8, 9, 8, 12, 7, 8, 7, 10, 8, 7, 9, 9, 10, 9, 10, 8, 9, 10, 11, 9, 11, 7, 8, 13
Offset: 1

Views

Author

Lei Zhou, Jun 25 2018

Keywords

Comments

Conjecture: a(n)>0 for all n>1.

Examples

			List of 3-smooth numbers from A003586: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, ...
For n=2, the 2nd prime is 3, 3-1=2 is prime. This is the only case. So a(2)=1;
For n=3, the 3rd prime is 5, 5-2=3 and 5-3=2 are prime.  So a(3)=2;
...
For n=10, the 10th prime is 29, 29-6=23, 29-12=17, 29-16=13, 29-18=11, 29-24=5, and 29-27=2, 6 valid numbers found, so a(10)=6.

Crossrefs

Cf. A000040, A003586, A007694.

Programs

  • Mathematica
    g = {1}; Table[p = Prime[n]; While[l = Length[g]; g[[l]] < p, pos = l + 1; While[pos--; c2 = g[[pos]]*2; c2 > g[[l]]]; c2 = g[[pos + 1]]*2; pos = l + 1; While[pos--; c3 = g[[pos]]*3; c3 > g[[l]]]; c3 = g[[pos + 1]]*3; c = Min[c2, c3]; AppendTo[g, c]]; ct = 0; i = 0; While[i++; cn = g[[i]]; cn < p, If[PrimeQ[p - cn], ct++]]; ct, {n, 1, 82}]
  • PARI
    is_a003586(n) = n<5||vecmax(factor(n, 5)[, 1])<5;
a(n) = my(p=prime(n)); sum(k=1, p, is_a003586(k) && isprime(p-k)); \\ Michel Marcus, Jul 03 2018

A342867 a(n) is the least number k such that the continued fraction for phi(k)/k contains exactly n elements.

Original entry on oeis.org

1, 2, 3, 15, 35, 33, 65, 215, 221, 551, 455, 2001, 3417, 3621, 11523, 16705, 16617, 69845, 107545, 157285, 324569, 358883, 1404949, 1569295, 3783970, 3106285, 7536065, 12216295, 10589487, 24038979, 57759065, 51961945, 177005465, 131462695, 741703701, 1467144445
Offset: 1

Views

Author

Amiram Eldar, Mar 27 2021

Keywords

Comments

a(n) is the least number k such that A342866(k) = n.
All the terms above 3 are composite numbers.

Crossrefs

Cf. A000010, A007694, A076512, A109395, A342866.
Cf. A071865 (similar, with sigma(k)/k).

Programs

  • Mathematica
    f[n_] := Length @ ContinuedFraction[EulerPhi[n]/n]; seq[max_] := Module[{s = Table[0, {max}], c = 0, n  = 1, i}, While[c < max, i = f[n]; If[i <= max && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[20]
  • PARI
    a(n) = my(k=1); while (#contfrac(eulerphi(k)/k) != n, k++); k; \\ Michel Marcus, Mar 30 2021

Formula

a(2) = 2 since 2 is the least number k such that A342866(k) = 2.

A364157 Numbers whose rounded-down (floor) mean of prime factors (with multiplicity) is 2.

Original entry on oeis.org

2, 4, 6, 8, 12, 16, 18, 24, 32, 36, 40, 48, 54, 64, 72, 80, 96, 108, 120, 128, 144, 160, 162, 192, 216, 224, 240, 256, 288, 320, 324, 360, 384, 432, 448, 480, 486, 512, 576, 640, 648, 672, 720, 768, 800, 864, 896, 960, 972, 1024, 1080, 1152, 1280, 1296, 1344
Offset: 1

Views

Author

Gus Wiseman, Jul 18 2023

Keywords

Examples

			The terms together with their prime factors begin:
   2 = 2
   4 = 2*2
   6 = 2*3
   8 = 2*2*2
  12 = 2*2*3
  16 = 2*2*2*2
  18 = 2*3*3
  24 = 2*2*2*3
  32 = 2*2*2*2*2
  36 = 2*2*3*3
  40 = 2*2*2*5
  48 = 2*2*2*2*3
  54 = 2*3*3*3
  64 = 2*2*2*2*2*2
  72 = 2*2*2*3*3
  80 = 2*2*2*2*5
  96 = 2*2*2*2*2*3

Crossrefs

Without multiplicity we appear to have A007694.
Prime factors are listed by A027746, indices A112798.
Positions of 2's in A126594, positions of first appearances A364037.
For prime indices and ceiling we have A363950, counted by A026905.
For prime indices we have A363954 (or A363949), counted by A363745.
A078175 lists numbers with integer mean of prime factors.
A123528/A123529 gives mean of prime factors, indices A326567/A326568.
A316413 ranks partitions with integer mean, counted by A067538.
A363895 gives floor of mean of distinct prime factors.
A363943 gives floor of mean of prime indices, ceiling A363944.
Cf. A001222, A056239, A327473, A327476, A360013, A360015, A363488, A363945.

Programs

  • Mathematica
    prifacs[n_]:=If[n==1,{},Flatten[ConstantArray@@@FactorInteger[n]]];
Select[Range[100],Floor[Mean[prifacs[#]]]==2&]

A368217 a(n) is the first number == 1 (mod n) that is the product of n primes, counted by multiplicity.

Original entry on oeis.org

2, 9, 28, 81, 176, 15625, 288, 6561, 1792, 137781, 17920, 244140625, 30720, 7971615, 311296, 43046721, 1492992, 3814697265625, 2752512, 3486784401, 38797312, 242137805625, 28311552, 59604644775390625, 184549376, 51684605176023, 2583691264, 63546645708225, 9512681472, 41858774825571336448888891
Offset: 1

Views

Author

Robert Israel, Dec 17 2023

Keywords

Comments

a(n) is the first number k == 1 (mod n) such that A001222(k) = n.
A053669(n)^n <= a(n) <= A034694(n).
If n is in A007694 then a(n) = A053669(n)^n.

Examples

			a(4) = 81 because 81 == 1 (mod 4) and 81 = 3^4 is the product of 4 primes, counted by multiplicity, and no smaller number works.

Links

Crossrefs

Cf. A001222, A007694, A034694, A053669.

Programs

  • Maple
    f:= proc(n) uses priqueue; local p, x, Aprimes, v;
    initialize(Aprimes);
    p:= 2;
    while n mod p = 0 do p:= nextprime(p) od:
    insert([-p^n,p,0],Aprimes);
    do
      v:= extract(Aprimes);
      x:= -v[1];
      if x mod n = 1 then return x fi;
      if v[3] < n then
        insert([v[1],v[2],v[3]+1],Aprimes);
        p:= nextprime(v[2]);
        while n mod p = 0 do p:= nextprime(p) od;
        x:= x * (p/v[2])^(n-v[3]);
        insert([-x,p,v[3]],Aprimes);
      fi;
    od;
end proc:
f(1):= 2:
map(f, [$1..30]);

A385748 Numbers k such that A384247(k) divides k.

Original entry on oeis.org

1, 2, 6, 8, 12, 24, 32, 54, 96, 108, 128, 192, 216, 240, 384, 486, 512, 864, 972, 1536, 1728, 1944, 2048, 2160, 3072, 3456, 4374, 6000, 6144, 7776, 8192, 8748, 13824, 15552, 17496, 19440, 24576, 27648, 31104, 32768, 39366, 49152, 54000, 55296, 61440, 65280, 69984
Offset: 1

Views

Author

Amiram Eldar, Jul 08 2025

Keywords

Comments

(2^(2^k)-1) * 2^(2^k) is a term for k = 0..5.
Apparently, the only prime factors of any term are 2 and the Fermat primes (A019434), i.e., A092506.
Apparently, except for n = 1, a(n) / A384247(a(n)) is either 2 or 3.

Examples

			  n | a(n) | a(n) / A384247(a(n))
  --+------+---------------------
  1 |    1 | 1 / 1 = 1
  2 |    2 | 2 / 1 = 2
  3 |    6 | 6 / 2 = 3
  4 |    8 | 8 / 4 = 2
  5 |   12 | 12 / 6 = 2

Links

Crossrefs

Cf. A019434, A092506, A384247.
Similar sequences: A007694, A298759, A319481, A335327, A373057.

Programs

  • Mathematica
    f[p_, e_] := p^e*(1 - 1/p^(2^(IntegerExponent[e, 2]))); iphi[1] = 1; iphi[n_] := iphi[n] = Times @@ f @@@ FactorInteger[n]; q[n_] := Divisible[n, iphi[n]]; Select[Range[70000], q]
  • PARI
    iphi(n) = {my(f = factor(n)); n * prod(i = 1, #f~, (1 - 1/f[i, 1]^(1 << valuation(f[i, 2], 2))));}
isok(k) = !( k % iphi(k));

A067583 Integers of the form phi(n^n)/phi(n)^n where phi is the Euler totient function A000010(n).

Original entry on oeis.org

1, 2, 8, 243, 128, 177147, 32768, 129140163, 94143178827, 2147483648, 50031545098999707, 26588814358957503287787, 19383245667680019896796723, 9223372036854775808, 7509466514979724803946715958257547
Offset: 1

Views

Author

Benoit Cloitre, Jan 30 2002

Keywords

Programs

  • Mathematica
    Select[Table[EulerPhi[n^n]/EulerPhi[n]^n,{n,100}],IntegerQ] (* Harvey P. Dale, Dec 19 2021 *)

Formula

If A007694(n)==0 (mod 3) (or equivalently if A007694(n) = 2^u*3^v with u, v >=1) then a(n) =3^(A007694(n) -1); if A007694(n)=2^u a(n)=2^(A007694(n) -1). Remark : if n=3^v phi(n^n)/phi(n)^n=1/2^(n-1)
Previous Showing 41-47 of 47 results.

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