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.

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A362600 a(1) = 1, a(2) = 6, a(3) = 10; for n > 3, a(n) is the smallest positive number that has not yet appeared that shares a factor with a(n-1) and a(n-2) and also contains as factors the smallest primes that are not factors of both a(n-1) and a(n-2).

Original entry on oeis.org

1, 6, 10, 15, 12, 20, 30, 42, 35, 40, 60, 84, 70, 45, 18, 50, 75, 24, 80, 90, 105, 14, 36, 120, 140, 21, 48, 150, 210, 154, 33, 54, 110, 135, 66, 100, 165, 72, 130, 180, 126, 175, 160, 168, 195, 170, 78, 225, 190, 96, 240, 280, 63, 102, 270, 315, 28, 108, 300, 350, 147, 114, 330, 420, 77, 22
Offset: 1

Views

Author

Scott R. Shannon, May 02 2023

Keywords

Comments

No term can be a prime power as each term must contain at least two distinct prime factors. This make the sequence similar to A362754, A360519 and A361606. Some small composite numbers take many terms to appear, e.g., a(354476) = 65. Such terms are usually preceded by a term that contains all the lower primes as factors. In the first 500000 terms, other than the first term, there are no fixed points, and it is unknown if any exist.

Examples

			a(4) = 15 as a(2) = 6 = 2*3 and a(3) = 10 = 2*5, and 15 is the smallest unused number that shares a factor with 6 and 10 while also containing 5 and 3 as prime factors, the smallest primes not factors of 6 and 10 respectively. This is the first term to differ from A362754.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[_] := False;
    f[x_] := If[OddQ[x], 2, y = 3; While[Divisible[x, y], y = NextPrime[y]]; y];
    MapIndexed[Set[{a[First[#2]], c[#1]}, {#1, First[#2]}] &, {1, 6, 10}];
    i = a[2]; j = a[3]; q = 5; u = 12;
    Do[qq = f[j]; k = Ceiling[u/#] &[q*qq];
      While[Or[c[#], CoprimeQ[i, #], CoprimeQ[i, j]] &[k*q*qq], k++];
      k *= q*qq;
      Set[{a[n], c[k], i, j, q}, {k, True, j, k, qq}];
      If[k == u, While[Or[c[u], PrimePowerQ[u]], u++]], {n, 4, nn}];
    Array[a, nn] (* Michael De Vlieger, May 09 2023 *)

A363504 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that is a multiple of prime(Omega(a(n-1))).

Original entry on oeis.org

1, 2, 4, 3, 6, 9, 12, 5, 8, 10, 15, 18, 20, 25, 21, 24, 7, 14, 27, 30, 35, 33, 36, 28, 40, 42, 45, 50, 55, 39, 48, 11, 16, 49, 51, 54, 56, 63, 60, 70, 65, 57, 66, 75, 80, 22, 69, 72, 44, 85, 78, 90, 77, 81, 84, 91, 87, 93, 96, 13, 26, 99, 95, 102, 100, 98, 105, 110, 115, 108, 88, 112, 121, 111
Offset: 1

Views

Author

Scott R. Shannon, Jun 06 2023

Keywords

Comments

The sequence is conjectured to be a permutation of the positive integers, although it takes many terms for the primes to appear, e.g., a(371221) = 61. After the first two terms the fixed points in the first 500000 terms are 10, 26076, 26151, 26223, 26307, 26379, 26406, although it is possible more exist.

Examples

			a(8) = 5 as prime(Omega(a(7))) = prime(A001222(12)) = prime(3) = 5, and 5 is the smallest unused number that is a multiple of 5.
		

Crossrefs

A363956 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that is a multiple of prime(omega(a(n-1))).

Original entry on oeis.org

1, 2, 4, 6, 3, 8, 10, 9, 12, 15, 18, 21, 24, 27, 14, 30, 5, 16, 20, 33, 36, 39, 42, 25, 22, 45, 48, 51, 54, 57, 60, 35, 63, 66, 40, 69, 72, 75, 78, 50, 81, 26, 84, 55, 87, 90, 65, 93, 96, 99, 102, 70, 80, 105, 85, 108, 111, 114, 95, 117, 120, 100, 123, 126, 110, 115, 129, 132, 125, 28, 135, 138
Offset: 1

Views

Author

Scott R. Shannon, Jun 29 2023

Keywords

Comments

The sequence is conjectured to be a permutation of the positive integers, although it takes many terms for the primes to appear, e.g., a(210667) = 17. After the first two terms the only fixed points in the first 500000 terms are 15777 and 86977, although it is possible more exist.

Examples

			a(10) = 15 as prime(omega(a(9))) = prime(A001221(12)) = prime(2) = 3, and 15 is the smallest unused number that is a multiple of 3.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[] := False; m[] := 1; f[x_] := Prime@ PrimeNu[x];
    Array[Set[{a[#], c[#], m[#]}, {#, True, 2}] &, 2]; j = 2;
    Do[k = f[j];
     While[c[m[k] k], m[k]++]; k *= m[k];
     Set[{a[n], c[k], j}, {k, True, k}], {n, 3, nn}];
    Array[a, nn] (* Michael De Vlieger, Jul 01 2023 *)

A369271 The smallest positive number that has not yet appeared that is a multiple of the smallest prime that does not divide n.

Original entry on oeis.org

2, 3, 4, 6, 8, 5, 10, 9, 12, 15, 14, 20, 16, 18, 22, 21, 24, 25, 26, 27, 28, 30, 32, 35, 34, 33, 36, 39, 38, 7, 40, 42, 44, 45, 46, 50, 48, 51, 52, 54, 56, 55, 58, 57, 60, 63, 62, 65, 64, 66, 68, 69, 70, 75, 72, 78, 74, 81, 76, 49, 80, 84, 82, 87, 86, 85, 88, 90, 92, 93, 94, 95, 96, 99, 98
Offset: 1

Views

Author

Scott R. Shannon, Jan 18 2024

Keywords

Comments

The sequence is conjectured to be a permutation of the positive integers.
The empty product 1 never appears - Michael De Vlieger, Jan 24 2024

Examples

			a(7) = 10 as 2 is the smallest prime that does not divide 7, and 10 = 2*5 is the smallest multiple of 2 that has not yet appeared.
		

Crossrefs

Programs

  • Mathematica
    nn = 2^10;
    c[] := False; m[] := 1;
    f[x_] := Block[{q = 2}, While[Divisible[x, q], q = NextPrime[q]]; q];
    Do[While[c[Set[k, #*m[#]]], m[#]++] &[f[i]];
      Set[{a[i], c[k]}, {k, True}], {i, nn}];
    Array[a, nn] (* Michael De Vlieger, Jan 24 2024 *)

Formula

a(A002110(k-1)) = the k-th prime.

A369272 a(1) = 1; for n > 1, a(n) is the smallest positive number that has not yet appeared that is a multiple of the smallest prime that does not divide n*a(n-1).

Original entry on oeis.org

1, 3, 2, 6, 7, 5, 4, 9, 8, 12, 10, 14, 15, 11, 16, 18, 20, 21, 22, 24, 25, 27, 26, 30, 28, 33, 32, 36, 35, 44, 39, 40, 42, 45, 34, 50, 48, 55, 38, 51, 46, 60, 49, 54, 56, 57, 52, 65, 58, 63, 62, 66, 70, 77, 64, 69, 68, 72, 75, 84, 80, 78, 85, 81, 74, 90, 91, 87, 76, 93, 82, 95, 86, 96, 98, 99
Offset: 1

Views

Author

Scott R. Shannon, Jan 18 2024

Keywords

Comments

The sequence is conjectured to be a permutation of the positive integers, although it takes many terms for the primes to appear, e.g., a(63784) = 23.

Examples

			a(7) = 4 as 2 is the smallest prime that does not divide a(6)*7 = 5*7 = 35, and 4 = 2*2 is the smallest multiple of 2 that has not yet appeared.
		

Crossrefs

A379746 a(1)=1. For n>1 if a(n-1)=A002110(k), a(n)=prime(k+1). Otherwise a(n) is the smallest novel number whose prime factors have already occurred as previous terms.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 7, 14, 21, 28, 32, 35, 36, 40, 42, 45, 48, 49, 50, 54, 56, 60, 63, 64, 70, 72, 75, 80, 81, 84, 90, 96, 98, 100, 105, 108, 112, 120, 125, 126, 128, 135, 140, 144, 147, 150, 160, 162, 168, 175, 180
Offset: 1

Views

Author

David James Sycamore, Jan 01 2025

Keywords

Comments

Equivalent definition: Lexicographically earliest infinite sequence of distinct positive integers such that a(n) is the smallest novel number having prime power factorization Product_p_i^e_i where p_i is the least nondivisor prime of at most e_i distinct terms a(j); 1<=j<=n-1.
A permutation of the positive integers with prime powers q^k appearing in order (k>=1), and whose underlying sequence of least nondivisor primes is a permutation of A053669. Also, for distinct x, y; x
No multiple m*p (m>1) of a prime p can occur before p itself is a term.
From Michael De Vlieger, Jan 02 2025: (Start)
Efficient method of generating the sequence:
Define row k to be a(A363061(k)+1..A363061(k+1)).
Define R(i) to be { m <= i : rad(m) | i } = tensor product of prime power factor ranges of i that do not exceed i.
Then row k contains R(A002110(k+1)) \ R(A002110(k)).
Row 0 is R(1) = {1}.
Row 1 is R(2)\R(1) = {1, 2} \ {1} = {2},
i.e., {row 2 of A162306} \ {row 1 of A162306}
= {first A363061(1) terms of A000079} \ {1}.
Row 2 is R(6)\R(2) = {1, 2, 3, 4, 6} \ {1, 2} = {3, 4, 6},
where R(6) = row 6 of A162306 = first A363061(2) terms of A003586.
Row 3 is R(30)\R(6)
= {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30} \ {1, 2, 3, 4, 6}
= {5, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30},
where R(30) = row 30 of A162306 = first A363061(3) terms of A051037, etc.
Therefore, for k > 1, within each row, terms strictly increase from prime(k) to primorial A002110(k).
Furthermore, a(1..A363061(k)) is a permutation of R(A002110(k)), hence the sequence is infinite and a permutation of natural numbers. (End)

Examples

			a(1) = 1 = A002110(0) therefore a(2) = A053669(1) = 2.
a(2) = 2 = A002110(1) therefore a(3) = A053669(2) = 3.
a(3) = 3 is not a primorial term so a(4)=4 = 2^2 is the smallest novel number whose prime factors do not exceed 3.
Using the second definition we have a(1,2,3,4)=1,2,3,4
                  with least nondivisor primes 2,3,2,3 respectively. Therefore a(5)=2^1*3^1=6, the smallest novel number whose prime factors (2,3) are nondivisor primes of the first 4 terms, and whose exponents do not exceed the number of times these primes have occurred in the underlying sequence of least nondivisor primes.
		

Programs

  • Mathematica
    nn = 120; kk = 12;
    c[] := False; m[] := 0; h = 0; q = j = 1; u = 2;
    f[x_] := f[x] = FactorInteger[x][[All, 1]];
    MapIndexed[Set[P[First[#2] - 1], #1] &, FoldList[Times, 1, Prime@ Range[kk]]];
    {1}~Join~Reap[Do[
        If[j == P[h],
          If[h == kk, Break[]]; k = Prime[h + 1]; h++; q = Prime[h],
          k = u; While[Or[c[k], ! AllTrue[f[k], # <= q &]], k++]];
        j = Sow[k]; c[k] = True; If[k == u, While[c[u], u++] ],
    {n, 2, nn}] ][[-1, 1]] (* Michael De Vlieger, Jan 02 2025 *)

Formula

From Michael De Vlieger, Jan 02 2025: (Start)
a(A363061(k)) = A002110(k).
a(A363061(k)+1) = prime(k).
Seen as a table T(j,k), k = 1..A363061(j)-A363061(j-1) for j > 0, row 0 = {1},
row j = {row A002110(j) of A162306} \ {row A002110(j-1) of A162306}. (End)

A361331 Smallest index of n-th prime in A361330, or -1 if it does not appear.

Original entry on oeis.org

1, 2, 5, 22, 160, 1770, 23022, 391390, 7436428
Offset: 1

Author

Keywords

Comments

It appears that a(n) is growing as constant*primorial(n). - Scott R. Shannon, Mar 20 2023

Crossrefs

Cf. A002110 (primorials), A351495, A361330.

Extensions

a(7)-a(8) from Winston de Greef, Mar 18 2023
a(9) from Scott R. Shannon, Mar 20 2023

A361825 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that is a multiple of the smallest prime that does not divide a(n-2) + a(n-1).

Original entry on oeis.org

1, 2, 4, 5, 6, 8, 3, 10, 12, 9, 14, 16, 7, 18, 20, 15, 22, 24, 21, 26, 28, 25, 30, 32, 27, 34, 36, 33, 38, 40, 35, 42, 44, 39, 46, 48, 45, 50, 52, 55, 54, 56, 51, 58, 60, 57, 62, 64, 65, 66, 68, 63, 70, 72, 69, 74, 76, 49, 78, 80, 75, 82, 84, 81, 86, 88, 85, 90, 92, 87, 94, 96, 93, 98, 100, 95
Offset: 1

Author

Scott R. Shannon, Mar 25 2023

Keywords

Comments

The sequence is conjectured to be a permutation of the positive integers, although it takes many terms for the primes to appear, e.g., a(191443) = 19.

Examples

			a(3) = 4 as a(1) + a(2) = 1 + 2 = 3 which does not contain 2 as a prime factor, and 4 is the smallest unused number that is a multiple of 2.
a(4) = 5 as a(2) + a(3) = 2 + 4 = 6 = 2*3 which does not contain 5 as a prime factor, and 5 is the smallest unused number that is a multiple of 5.
		

Crossrefs

A362015 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that, given the list of primes that form the factors of all previous terms a(1)..a(n-1), is a multiple of the prime in that list which is a factor of the fewest previous terms. If two or more such primes exist the smallest is chosen.

Original entry on oeis.org

1, 2, 4, 6, 3, 9, 8, 12, 15, 5, 10, 20, 25, 18, 30, 35, 7, 14, 21, 28, 42, 49, 40, 56, 45, 63, 50, 70, 77, 11, 22, 33, 44, 55, 66, 88, 99, 110, 121, 84, 132, 91, 13, 26, 39, 52, 65, 78, 104, 117, 130, 143, 156, 169, 98, 154, 182, 165, 195, 176, 208, 105, 187, 17, 34, 51, 68, 85, 102, 119, 136
Offset: 1

Author

Scott R. Shannon, Apr 03 2023

Keywords

Comments

After 5 million terms the lowest number not to have appeared is 16 = 2^4. In that range 2 is a factor of 2614180 terms while 3 is a factor of 1763610 terms. As these are the most and second-most common prime factors this suggest that 16, and other higher powers of 2, will never appear as that would require 2 to be the least common factor of all previous terms. This is also true for the powers of the other smaller primes.
In the first 5 million terms the only fixed point, other than the first two terms, is 4175, although it is probable that more exist.

Examples

			a(5) = 3 as the list of primes that divide all previous terms a(1)..a(4) is 2 and 3, with 2 being a factor of three terms and 3 being a factor of one term. Therefore a(5) is the lowest multiple of 3 that has not appeared, which is 3.
		

Crossrefs

A363159 a(1)=1. Thereafter, if a(n-1) is a novel term, a(n) is the smallest prime which does not divide a(n-1). If a(n-1) has been seen k (>1) times already then a(n) = k*a(n-1).

Original entry on oeis.org

1, 2, 3, 2, 4, 3, 6, 5, 2, 6, 12, 5, 10, 3, 9, 2, 8, 3, 12, 24, 5, 15, 2, 10, 20, 3, 15, 30, 7, 2, 12, 36, 5, 20, 40, 3, 18, 5, 25, 2, 14, 3, 21, 2, 16, 3, 24, 48, 5, 30, 60, 7, 14, 28, 3, 27, 2, 18, 36, 72, 5, 35, 2, 20, 60, 120, 7, 21, 42, 5, 40, 80, 3, 30, 90, 7, 28, 56, 3, 33, 2, 22, 3, 36, 108
Offset: 1

Author

David James Sycamore, Jul 08 2023

Keywords

Comments

It follows from the definition that the sequence is infinite. Every number appears multiple times according to its prime factorization. All primes p appear infinitely many times, prime powers p^k (k>1) appear once only, all squarefree semiprimes appear twice, and so on.
On the first occasion of A007947(a(n-1)) = A002110(k-1), a(n) is the first occasion of prime(k).

Examples

			a(2)=2 since 1 is a novel term and 2 is the least prime which does not divide 1, a(3)=3 since 3 is the smallest prime which does not divide 2.
a(4)=4 since 2 has appeared twice.
a(7) = 6, therefore a(8) = 5.
f(30) = A001221(30) + 1 since f(15)=2 and 2*15=30. No other divisor d of 30 has the property d*f(d) >= 30 thus f(30)=3+1=4.
		

Programs

  • Mathematica
    nn = 120; c[_] := 0; a[1] = c[1] = k = 1;
    Do[If[c[j] == 0,
         c[j]++; p = 2; While[Divisible[j, p], p = NextPrime[p]]; Set[k, p],
         c[j]++; Set[k, j c[j]] ];
      Set[{a[n], j}, {k, k}], {n, 2, nn}];
    Array[a, nn] (* Michael De Vlieger, Jul 08 2023 *)
  • PARI
    lista(nn) = my(c, p, v=vector(nn)); v[1]=1; for(k=2, nn, if(c=sum(i=1, k-2, v[i]==v[k-1]), v[k]=(c+1)*v[k-1], p=2; while(v[k-1]%p==0, p=nextprime(p+1)); v[k]=p)); v \\ Jinyuan Wang, Jul 11 2023

Formula

For integer m let f(m) be the number of times m appears in the sequence.
f(1)=1, f(p)->oo for all prime p, and for n composite the following recursion applies:
f(n) = A001221(n) + Sum_{i=1..A321516(n)} [k_i*f(k_i)>=n], where k_i is a composite divisor of n and [] is the Iverson bracket.
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