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.

Showing 1-6 of 6 results.

A323730 Table read by rows: row n lists every number j whose n-th power has exactly j divisors.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 28, 40, 1, 5, 9, 45, 1, 1, 7, 1, 225, 1, 153, 1, 640, 1, 11, 441, 2541, 4851, 1, 6348, 1, 13, 25, 325, 1, 19474560, 1, 1, 976, 1, 17, 1089, 9537, 18513, 1, 1225, 1, 19, 1, 1521, 70840000, 107747640000, 1, 81, 1, 1, 23, 1, 343000, 3763008, 245790720
Offset: 0

Views

Author

Jon E. Schoenfield, Jan 25 2019

Keywords

Comments

Row n lists every j such that tau(j^n) = j.
Since tau(1^n) = tau(1) = 1 for all n, every row of the table includes 1 as a term.
Each prime p appears as a term in row p-1 since, for n=p-1, tau(p^n) = tau(p^(p-1)) = p.

Examples

			Row n=3 includes 28 as a term because tau(28^3) = tau((2^2 * 7)^3) = tau(2^6 * 7^3) = (6+1)*(3+1) = 7*4 = 28.
Row n=3 includes 40 as a term because tau(40^3) = tau((2^3 * 5)^3) = tau(2^9 * 5^3) = (9+1)*(3+1) = 10*4 = 40.
Row n=5 includes no terms other than 1 because there exists no number j > 1 such that tau(j^5) = j.
Row n=23 includes 245790720 as a term because tau(245790720^23) = tau((2^11 * 3^3 * 5 * 7 * 127)^23) = tau(2^253 * 3^69 * 5^23 * 7^23 * 127^23) = (253+1)*(69+1)(23+1)*(23+1)*(23+1) = 254*70*24^3 = 245790720.
Table begins as follows:
   n | row n
  ---+---------------------------------
   0 | 1;
   1 | 1, 2;
   2 | 1, 3;
   3 | 1, 28, 40;
   4 | 1, 5, 9, 45;
   5 | 1;
   6 | 1, 7;
   7 | 1, 225;
   8 | 1, 153;
   9 | 1, 640;
  10 | 1, 11, 441, 2541, 4851;
  11 | 1, 6348;
  12 | 1, 13, 25, 325;
  13 | 1, 19474560;
  14 | 1;
  15 | 1, 976;
  16 | 1, 17, 1089, 9537, 18513;
  17 | 1, 1225;
  18 | 1, 19;
  19 | 1, 1521, 70840000, 107747640000;
  20 | 1, 81;
  21 | 1;
  22 | 1, 23;
  23 | 1, 343000, 3763008, 245790720;

Links

Crossrefs

Cf. A073049 (Least m > 1 such that m^n has m divisors, or 0 if no such m exists).
Cf. A000005, A323731, A323732, A323733, A323734.

Formula

A073049(n) = T(n,2) if row n contains more than 1 term, 0 otherwise.
A323731(n) is the number of terms in row n.
A323732 lists the numbers n such that row n contains only the single term 1.
A323733 lists the numbers n such that row n contains more than one term; i.e., A323733 is the complement of A323732.
A323734(n) = T(n, A323731(n)) is the largest term in row n.

A323731 a(n) is the number of numbers k whose n-th power has exactly k divisors.

Original entry on oeis.org

1, 2, 2, 3, 4, 1, 2, 2, 2, 2, 5, 2, 4, 2, 1, 2, 5, 2, 2, 4, 2, 1, 2, 4, 2, 2, 2, 2, 5, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 3, 1, 5, 10, 3, 2, 5, 2, 2, 2, 1, 4, 2, 3, 1, 6, 2, 2, 2, 6, 4, 4, 3, 4, 2, 2, 5, 1, 2, 2, 5, 4, 5, 2, 3, 3, 1, 4, 2, 5, 2, 2, 2, 2, 2, 2, 1
Offset: 0

Views

Author

Jon E. Schoenfield, Jan 26 2019

Keywords

Comments

a(n) is the number of terms in row n of A323730.
Since 1^n = 1 has exactly 1 divisor for all n, a(n) >= 1.
A323732 lists the numbers j such that a(j) = 1 (i.e., such that A073049(j) = 0); for each such j, the only number k whose j-th power has exactly k divisors is 1.
A323733 lists the numbers j such that a(j) > 1 (i.e., such that A073049(j) > 0).

Examples

			a(0) = 1 because there is only one number k whose 0th power (k^0 = 1) has exactly k divisors (namely, k=1).
a(2) = 2 because there are two numbers k such that tau(k^2) = k: tau(1^2) = tau(1) = 1 and tau(3^2) = tau(9) = 3.
a(43) = 10 because there are 10 numbers k such that tau(k^43) = k: 1, 7569, 2197000, 4296680960, 11128700700, 16629093000, 223705109760, 19462344549120, 32521578186240, and 5580197619796800.

Links

Crossrefs

Cf. A000005, A073049, A323730, A323732, A323733, A323734.

A323734 a(n) is the largest number j whose n-th power has exactly j divisors.

Original entry on oeis.org

1, 2, 3, 40, 45, 1, 7, 225, 153, 640, 4851, 6348, 325, 19474560, 1, 976, 18513, 1225, 19, 107747640000, 81, 1, 23, 245790720, 49, 2601, 2133, 3025, 94221, 56241820800, 31, 20063232, 4225, 15262600, 4761, 19236456, 37, 25462407801600, 5929, 2952832000, 21921921
Offset: 0

Views

Author

Jon E. Schoenfield, Jan 26 2019

Keywords

Comments

a(n) is the largest (and last) of the A323731(n) numbers in row n of A323730.
If a(n)=1 then n is a term in A323732; otherwise, n is a term in A323733.

Examples

			The numbers j whose 3rd powers have exactly j divisors are 1, 28, and 40; the largest of these is 40, so a(3) = 40.
The only number j whose 5th power has exactly j divisors is 1, so a(1) = 1.

Links

Crossrefs

Cf. A000005, A073049, A323730, A323731, A323732, A323733.

A323732 Numbers k for which there exists no j > 1 such that j^k has exactly j divisors.

Original entry on oeis.org

5, 14, 21, 41, 50, 54, 67, 76, 86, 90, 111, 113, 119, 131, 142, 153, 165, 175, 186, 202, 204, 216, 224, 230
Offset: 1

Views

Author

Jon E. Schoenfield, Jan 26 2019

Keywords

Comments

This sequence lists the numbers k such that A073049(k) = 0.
Equivalently:
numbers k for which the only number j such that j^k has exactly j divisors is 1;
numbers k such that A323731(k)=1;
numbers k such that A323734(k)=1.
The complement of this sequence is A323733.
The next terms after a(24)=230 appear to be 233, 253, 269, 273, 285, 293, 303, 307, 318, 321, 328, 345, 354, 357, 369, 370, 373, 384, 393, 402, 410, 412, 414, 426, 429, 431, 440, 441, 445, 468, ...

Examples

			There exists no j > 1 such that j^5 has exactly j divisors, so 5 is a term.
For k=15 and j=976, j^k = 976^15 = (2^4 * 61)^15 = 2^60 * 61^15, which has exactly (60+1)*(15+1) = 61*16 = 976 = j divisors, so k=15 is not a term.

Crossrefs

Cf. A000005, A073049, A323730, A323731, A323733, A323734.

A323733 Numbers k for which there exists at least one number j > 1 such that j^k has exactly j divisors.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 73
Offset: 1

Views

Author

Jon E. Schoenfield, Jan 26 2019

Keywords

Comments

Complement of A323732.
This sequence lists the numbers k such that A073049(k) > 0.
Equivalently:
numbers k for which 1 is not the only number j such that j^k has exactly j divisors;
numbers k such that A323731(k) > 1;
numbers k such that A323734(k) > 1.

Examples

			For k=9 and j=640, j^k = 640^9 = (2^7 * 5)^9 = 2^63 * 5^9, which has exactly (63+1)*(9+1) = 64*10 = 640 = j divisors, so k=9 is a term.
There exists no j > 1 such that j^14 has exactly j divisors, so 14 is not a term.

Crossrefs

Cf. A000005, A073049, A323730, A323731, A323732, A323734.

A270337 Composite numbers equal to the number of divisors of one of their powers.

Original entry on oeis.org

9, 25, 28, 40, 45, 49, 81, 121, 153, 169, 225, 289, 325, 343, 361, 441, 529, 625, 640, 841, 961, 976, 1089, 1225, 1369, 1521, 1681, 1849, 2133, 2197, 2209, 2401, 2541, 2601, 2809, 3025, 3249, 3481, 3721, 4225, 4489, 4753, 4761, 4851, 5041, 5329, 5929, 6241, 6348, 6561, 6859, 6889
Offset: 1

Views

Author

Paolo P. Lava, Mar 15 2016

Keywords

Comments

Prime numbers are not considered since every prime p satisfies p = d(p^(p-1)), where d() represents the number of divisors.
In general, p^k = d((p^k)^((p^k-1)/k)) for any prime p and for any power k such that (p^k-1)/k is an integer.

Examples

			9 = d(9^4); 28 = d(28^3); 153 = d(153^8); etc.

Links

Crossrefs

Cf. A000005, A073049, A270389.

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,k,n;
for n from 2 to q do if not isprime(n) then a:=tau(n); k:=0;
while a
  • Mathematica
    nn = 2000; Select[Select[Range@ nn, CompositeQ], Function[k, (SelectFirst[k^Range[nn/2], DivisorSigma[0, #] == k &] /. n_ /; MissingQ@ n -> 0) > 0]] (* Michael De Vlieger, Mar 17 2016, Version 10.2 *)
Showing 1-6 of 6 results.

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