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 1-8 of 8 results.

A316990 Smallest exponent m of n such that A289280(n) | n^m.

Original entry on oeis.org

2, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 3, 5, 2, 3, 2, 5, 2, 3, 2, 5, 2, 2, 4, 6, 2, 2, 2, 6, 4, 2, 2, 4, 2, 3, 2, 6, 2, 3, 2, 6, 4, 3, 2, 6, 2, 2, 4, 6, 2, 3, 2, 6, 2, 2, 3, 3, 2, 4, 4, 4, 2, 2, 2, 7, 4, 4, 2, 4, 2, 2, 2, 7, 2, 3, 3, 7, 5, 2, 2, 5, 2, 4, 5, 7, 3, 3, 2, 4, 2, 2, 2, 3, 2, 3, 3, 7
Offset: 2

Views

Author

Michael De Vlieger, Jul 28 2018

Keywords

Comments

Consider the least k > n such that k | n^m for m > 1. (We note that k cannot divide n if k exceeds n.) Values of k appear in A289280, while this sequence lists values of m.
If row n of A162306 were extended to include terms greater than n, A289280(n) would be the first term to follow those already in the row.
a(n) = 2 for n with omega(n) = 1. In other words, A289280(n) | n^2 for n = p^e with one distinct prime divisor, since A289280(p^e) = p^(e+1).
First indices of {2, 3, 4, 5, ..., m} are {2, 6, 10, 22, 34, 74, 134, 262, 514, 1042, 2062, 4106, 8198, 16418, 32822, 65542, ...}, i.e., the least even squarefree semiprime s > 2^(m - 1) for m > 2. This is because 2 is the smallest prime, and minimal multiplicity of 2 increases a(n) most efficiently. Let n = Product(p^e) and A289280(n) = Product(p^d), knowing there may be different values of p. a(n) = max(ceiling(d/e)) for d and e that pertain to the same prime p. Examples: for n = 10 = 2*5, A289280(10) = 16 = 2^4. Thus we are concerned with the ratio 4/1, and a(10) = 4. For n = 12 = 2^2*3 we have A289280(12) = 16; here we have the ratio 4/2 = 2. The greater multiplicity of 2 reduces a(n) for n = 12.

Examples

			For n = 2, A289280(n) = 4 = 2^2, the square of n = 2: thus a(2) = 2.
For n = 8, A289280(n) = 16 = 2^4; 2^4 | 8^2, thus a(8) = 2.
For n = 10, the least k > 10 that divides 10^e for e > 1 is 16. 16 | 10^4, thus a(n) = 4.
		

Crossrefs

Programs

  • Mathematica
    Table[If[PrimePowerQ@ n, 2, Block[{k = n + 1, m = 1}, While[PowerMod[n, k, k] != 0, k++]; While[PowerMod[n, m, k] != 0, m++]; m]], {n, 2, 106}]

A362041 a(0) = 1; for n > 0, a(n) is the largest k < A013929(n) such that rad(k) = rad(A013929(n)), where rad(n) = A007947(n).

Original entry on oeis.org

1, 2, 4, 3, 6, 8, 12, 10, 18, 5, 9, 14, 16, 24, 20, 22, 15, 36, 7, 40, 26, 48, 28, 30, 21, 32, 34, 54, 45, 38, 50, 27, 42, 44, 60, 46, 72, 56, 33, 80, 52, 96, 98, 58, 39, 90, 11, 62, 25, 84, 64, 66, 75, 68, 70, 108, 63, 74, 120, 76, 51, 78, 100, 144, 82, 126, 13, 57, 86, 35, 88, 150, 92, 94, 147, 162
Offset: 0

Views

Author

Michael De Vlieger, May 01 2023

Keywords

Comments

Permutation of natural numbers.
Let m = A013929(n) and let R_m be the sequence of numbers k such that rad(k) = rad(m). a(n) gives the predecessor of m in R_m.

Examples

			A013929(1) = 4; the smallest k < 4 such that rad(k) = rad(4) = 2 is a(1) = 2.
A013929(2) = 8; the smallest k < 8 such that rad(k) = rad(8) = 2 is a(2) = 4.
A013929(3) = 9; the smallest k < 9 such that rad(k) = rad(9) = 3 is a(3) = 3.
A013929(4) = 12; the smallest k < 12 such that rad(k) = 6 is a(4) = 6.
		

Crossrefs

Programs

  • Mathematica
    rad[x_] := Times @@ FactorInteger[x][[All, 1]]; {1}~Join~Table[Function[r, SelectFirst[Range[m - 1, 1, -1], r == rad[#] &] ][rad[m]], {m, Select[Range[225], Not @* SquareFreeQ]}]

Formula

A013929(n) = p^e, a prime power, e > 0, implies a(n) = p^(e-1).
A013929(n) = p^2 implies a(n) = p.

A362045 a(n) = smallest k such that k > m^2 and rad(k) | m, where rad(k) = A007947(k) and m = A120944(n).

Original entry on oeis.org

48, 125, 224, 243, 567, 512, 832, 960, 1331, 2048, 1715, 2048, 2187, 1792, 2944, 4131, 3125, 4617, 3712, 3968, 8125, 4374, 5589, 5000, 8192, 9317, 6144, 8192, 10625, 8192, 19683, 15379, 19683, 12032, 11875, 11016, 11907, 13568, 12500, 19683, 13122, 14375, 15104, 16807, 15616, 19683, 19683, 17576, 45619
Offset: 1

Views

Author

Michael De Vlieger, Apr 05 2023

Keywords

Comments

The smallest k such that k > p^2 such that p is prime and rad(k) | p is p^3.

Examples

			a(1) = 48 since m = 6 and the smallest k > m^2 such that rad(k) | 6 is 48. This is to say, the number that follows 6^2 in A003586 is 48.
a(2) = 80 since m = 10 and the smallest k > m^2 such that rad(k) | 10 is 125. This is to say, the number that precedes 10^2 in A003592 is 125.
Table of n = 1..12, m = A120944(n), m^2, and a(n).
   n    m    m^2   a(n)
  ---------------------
   1    6     36     48
   2   10    100    125
   3   14    196    224
   4   15    225    243
   5   21    441    567
   6   22    484    512
   7   26    676    832
   8   30    900    960
   9   33   1089   1331
  10   34   1156   2048
  11   35   1225   1715
  12   38   1444   2048
		

Crossrefs

Programs

  • Mathematica
    Table[m = k^2 + 1; While[! Divisible[k, Times @@ FactorInteger[m][[All, 1]]], m++]; m, {k, Select[Range[6, 133], And[CompositeQ[#], SquareFreeQ[#]] &]}]

A362044 a(n) = largest k such that k < m^2 and rad(k) | m, where rad(k) = A007947(k) and m = A120944(n).

Original entry on oeis.org

32, 80, 128, 135, 343, 352, 512, 864, 891, 1088, 875, 1216, 1053, 1728, 2048, 2187, 1375, 2187, 2048, 2048, 3125, 4224, 2187, 4802, 4736, 3773, 5832, 5248, 4913, 5504, 7047, 4459, 7533, 8192, 6859, 10368, 10935, 8192, 11264, 8991, 12312, 12167, 8192, 5831, 8192, 9963, 10449, 16640, 16807, 17152, 18432
Offset: 1

Views

Author

Michael De Vlieger, Apr 05 2023

Keywords

Comments

The largest k such that k < p^2 such that p is prime and rad(k) | p is p itself.

Examples

			a(1) = 32 since m = 6 and the largest k < m^2 such that rad(k) | 6 is 32. This is to say, the number that precedes 6^2 in A003586 is 32.
a(2) = 80 since m = 10 and the largest k < m^2 such that rad(k) | 10 is 80. This is to say, the number that precedes 10^2 in A003592 is 80.
Table of n = 1..12, m = A120944(n), a(n), and m^2.
   n    m    a(n)   m^2
  ---------------------
   1    6     32     36
   2   10     80    100
   3   14    128    196
   4   15    135    225
   5   21    343    441
   6   22    352    484
   7   26    512    676
   8   30    864    900
   9   33    891   1089
  10   34   1088   1156
  11   35    875   1225
  12   38   1216   1444
		

Crossrefs

Programs

  • Mathematica
    Table[m = k^2 - 1; While[! Divisible[k, Times @@ FactorInteger[m][[All, 1]]], m--]; m, {k, Select[Range[6, 133], And[CompositeQ[#], SquareFreeQ[#]] &]}]

A319605 a(1) = 1, and for n > 1, a(n) is the least prime power of the form p^k >= n where p is a prime factor of n.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 7, 8, 9, 16, 11, 16, 13, 16, 25, 16, 17, 27, 19, 25, 27, 32, 23, 27, 25, 32, 27, 32, 29, 32, 31, 32, 81, 64, 49, 64, 37, 64, 81, 64, 41, 49, 43, 64, 81, 64, 47, 64, 49, 64, 81, 64, 53, 64, 121, 64, 81, 64, 59, 64, 61, 64, 81, 64, 125, 81, 67
Offset: 1

Views

Author

Rémy Sigrist, Jan 07 2019

Keywords

Comments

This sequence has similarities with A289280.
Each power of a prime appears in the sequence.
Each prime number appears once in the sequence.

Examples

			For n = 42:
- 42 has 3 prime factors: 2, 3 and 7,
- the least power of 2 >= 42 is 64,
- the least power of 3 >= 42 is 81,
- the least power of 7 >= 42 is 49,
- hence a(42) = 49.
		

Crossrefs

Programs

  • PARI
    a(n) = my (pp=factor(n)[,1]~); if (#pp <= 1, n, vecmin(apply(p -> p^(1+logint(n,p)), pp)))

Formula

a(n) >= n with equality iff n belongs to A000961.

A364919 a(0) = 1; a(n) is the smallest number m not already in the sequence such that rad(m) divides A019565(n).

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 9, 6, 7, 14, 21, 12, 25, 10, 15, 16, 11, 22, 27, 18, 55, 20, 33, 24, 49, 28, 63, 32, 35, 40, 45, 30, 13, 26, 39, 36, 65, 50, 75, 48, 91, 52, 81, 42, 125, 56, 105, 54, 121, 44, 99, 64, 143, 80, 117, 60, 77, 88, 147, 66, 169, 70, 135, 72, 17, 34
Offset: 0

Views

Author

Michael De Vlieger, Aug 30 2023

Keywords

Comments

Let k be a squarefree number and define R_k to be the set of numbers m such that rad(m) | k.
For n > 0, a(n) is the smallest m in R_k such that a(j) != m, j < n.
Conjecture: permutation of natural numbers.

Examples

			Let b(n) = A019565(n).
a(1) = 2 since b(1) = 2. Since 2 is prime, we find the first number in the prime power range of 2 that is not in the sequence and that is 2.
a(3) = 4 since b(3) = 6, and the smallest number m such that rad(m) | 6 that has not already appeared is 4.
a(5) = 8 since b(5) = 10. R_10 begins {1, 2, 4, 5, 8, 10, 16, ...} and the smallest number m in that list that is not already in the sequence is 8.
a(6) = 9 since b(6) = 15. R_15 begins {1, 3, 5, 9, 15, 25, ...} and the smallest m in that list not already in the sequence is 9, etc.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    f[x_] := Times @@ Prime@ Position[Reverse@ IntegerDigits[x, 2], 1][[All, 1]];
    c[] := False; c[1] = True; q[] := 1; a[0] = 1; r[_] := 1;
    Do[If[PrimeQ[#],
      While[c[Set[k, #^q[#]]], q[#]++],
      While[Or[c[r[#]], ! Divisible[#, rad[r[#]]]], r[#]++]; k = r[#] ] &[f[i]]; Set[{a[i], c[k]}, {k, True}], {i, nn}];
    Array[a, nn + 1, 0]

Formula

a(2^k) = prime(k+1).

A365324 a(1) = 2, a(n) = k + 1, where k is the least number greater than a(n-1) such that rad(k) | a(n-1), where rad(n) = A007947(n).

Original entry on oeis.org

2, 5, 26, 33, 82, 129, 244, 257, 66050, 78126, 78733, 79508, 81797, 271442, 524289, 531442, 551369, 571788, 580609, 707282, 1048577, 1419858, 1431645, 1476226, 1620897, 1712422, 2097153, 2146690, 2151297, 2505890, 2560001, 11082242, 16777217
Offset: 1

Views

Author

Michael De Vlieger, Nov 15 2023

Keywords

Examples

			a(2) = 5 since the least k > a(1) such that rad(k) | a(1) is 4, and 4 + 1 = 5.
a(3) = 26 since the least k > a(2) such that rad(k) | a(2) is 25, and 25 + 1 = 26.
a(4) = 33 since the smallest k > 26 such that rad(k) | 26 is 32, and 32 + 1 = 33, etc.
		

Crossrefs

Programs

  • Mathematica
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    NestList[(k = # + 1; While[! Divisible[#, rad[k]], k++]; k + 1) &, 2, 20]

Formula

a(n) = A289280(a(n-1)) + 1 for n > 1.

A365413 a(1) = 2, a(n) = k - 1, where k is the least number greater than a(n-1) such that rad(k) | a(n-1), where rad(n) = A007947(n).

Original entry on oeis.org

2, 3, 8, 15, 24, 26, 31, 960, 971, 942840, 944783, 946728, 948675, 950624, 952575, 954528, 956483, 958440, 959999, 2229048, 2232035, 2235024, 2238015, 2241008, 2244003, 2247000, 2249999, 2253000, 2256003, 2259008, 2262015, 2265024, 2268035, 2271048, 2274063, 2277080, 2280099
Offset: 1

Views

Author

Michael De Vlieger, Nov 15 2023

Keywords

Examples

			a(2) = 3 since 4 is the smallest k > a(1) such that rad(k) | a(1), and 4 - 1 = 3.
a(3) = 8 since 9 is the least k > a(2) such that rad(k) | a(2), and 9 - 1 = 8.
a(4) = 15 since 16 is the least k > a(3) such that rad(k) | a(3), and 16 - 1 = 15, etc.
		

Crossrefs

Programs

  • Mathematica
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    NestList[(k = # + 1; While[! Divisible[#, rad[k]], k++]; k - 1) &, 2, 20]

Formula

a(n) = A289280(a(n-1)) - 1 for n > 1.
Showing 1-8 of 8 results.