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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A361722 Index of where prime(n) first appears as a divisor of any term in A359804.

Original entry on oeis.org

2, 3, 4, 8, 13, 31, 44, 47, 55, 66, 84, 96, 121, 125, 135, 143, 154, 161, 179, 192, 197, 218, 231, 242, 267, 270, 279, 293, 303, 308, 341, 352, 372, 379, 403, 412, 426, 440, 462, 476, 494, 501, 524, 530, 542, 545, 578, 617, 626, 639, 645, 665, 668, 697, 717, 730, 741, 748, 770, 786, 798, 822, 850
Offset: 1

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Author

Keywords

Crossrefs

A351495 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 a(n-1).

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 8, 9, 10, 12, 15, 14, 18, 20, 21, 16, 24, 25, 22, 27, 26, 30, 7, 28, 33, 32, 36, 35, 34, 39, 38, 42, 40, 45, 44, 48, 50, 51, 46, 54, 55, 52, 57, 56, 60, 49, 58, 63, 62, 66, 65, 64, 69, 68, 72, 70, 75, 74, 78, 80, 81, 76, 84, 85, 82, 87, 86, 90, 77, 88, 93, 92, 96, 95, 94
Offset: 1

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Author

Scott R. Shannon, May 03 2022

Keywords

Comments

The sequence is conjectured to be a permutation of the positive integers.
The k-th prime appears as the next term after A002110(k-1) appears.

Examples

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

Programs

A364154 Lexicographically earliest sequence of distinct positive integers such that a(n) is least novel multiple m of the product of all primes less than the greatest prime factor of a(n-1) which do not divide a(n-1); a(1) = 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 30, 8, 9, 10, 12, 11, 210, 13, 2310, 14, 15, 16, 17, 30030, 18, 19, 510510, 20, 21, 40, 24, 22, 105, 26, 1155, 28, 45, 32, 23, 9699690, 25, 36, 27, 34, 15015, 38, 255255, 42, 35, 48, 29, 223092870, 31, 6469693230, 33, 70, 39, 770, 51, 10010
Offset: 1

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Author

David James Sycamore, Jul 11 2023

Keywords

Comments

It follows from the definition that the sequence is infinite, and that the records (outside of the first 7 terms) are all primorial numbers, meaning that it grows very quickly.
When there are no primes less than the greatest prime factor of a(n-1) which do not divide a(n-1) then m is the least novel multiple of 1, the empty product, and therefore a(n) = u, the least unused number in the sequence so far. The only way a prime can enter the sequence is as u. When a(n-1) = prime(k), a(n) is A002110(k-1), and any primorial term is followed by u. Thus: prime —> primorial —> u.
Sequence is a permutation of the positive integers since by the definition no number appears more than once and m = 1 eventually introduces any number not already placed by the first part of the definition (m > 1).

Examples

			a(1) = 1 and there are no primes < 1 which divide 1 therefore m = 1 so a(2) = 2, the least unused number. Likewise a(3) = 3.
a(4) = 2*2 = 4 since 2 is the only prime < 3 which does not divide 3 and 2 has already occurred.
Since a(7) = 7, a(8) = 2*3*5 = 30.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[] := False; m[] := 1; a[1] = j = 1; c[1] = True;
    Do[k = Times @@ Complement[Prime@ Range[PrimePi@ Last[#] - 1], #] &[
       FactorInteger[j][[All, 1]] ];
     While[c[k m[k]], m[k]++]; k *= m[k];
     Set[{a[n], c[k], j}, {k, True, k}], {n, 2, nn}];
    Array[a, nn]
  • PARI
    lista(nn) = my(c, m, v=List([1, 2])); for(k=3, nn, c=m=1; forprime(p=2, vecmax(factor(v[k-1])[, 1]), if(v[k-1]%p, m*=p)); while(setsearch(Set(v), c*m), c++); listput(v, c*m)); Vec(v) \\ Jinyuan Wang, Jul 11 2023

Extensions

More terms from Jinyuan Wang, Jul 11 2023

A361640 a(0) = 0, a(1) = 1; thereafter let b be the least power of 2 that does not appear in the binary expansions of a(n-2) and a(n-1), then a(n) is the smallest multiple of b that is not yet in the sequence.

Original entry on oeis.org

0, 1, 2, 4, 3, 8, 12, 5, 6, 16, 7, 24, 32, 9, 10, 20, 11, 64, 28, 13, 14, 48, 15, 128, 80, 17, 18, 36, 19, 40, 44, 21, 22, 56, 23, 192, 72, 25, 26, 52, 27, 256, 60, 29, 30, 96, 31, 384, 160, 33, 34, 68, 35, 88, 76, 37, 38, 104, 39, 112, 120, 41, 42, 84, 43
Offset: 0

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Author

Rémy Sigrist, Mar 19 2023

Keywords

Comments

This sequence is a variant of A359804; here we consider binary expansions, there prime factorizations.
All powers of 2 appear in the sequence, in ascending order.
This sequence is a permutation of the nonnegative integers (with inverse A361641): an odd term is always followed by two even terms, and after two even terms we can choose the least value not yet in the sequence.

Examples

			The first terms, in decimal and in binary, alongside the corresponding b's, are:
  n   a(n)  bin(a(n))  b
  --  ----  ---------  ---
   0     0          0  N/A
   1     1          1  N/A
   2     2         10    2
   3     4        100    4
   4     3         11    1
   5     8       1000    8
   6    12       1100    4
   7     5        101    1
   8     6        110    2
   9    16      10000    8
  10     7        111    1
  11    24      11000    8
  12    32     100000   32
		

Crossrefs

Cf. A006519, A359804, A361641 (inverse).

Programs

  • Mathematica
    nn = 120; c[] = False; q[] = 1;
    f[n_] := f[n] = -1 + Position[Reverse@ IntegerDigits[n, 2], 1][[All, 1]];
    a[1] = 0; a[2] = 1; c[0] = c[1] = True; i = f[0]; j = f[1];
    Do[(k = q[#]; While[c[k #], k++]; q[#] = k; k *= #) &[
       2^First@ Complement[Range[0, Max[#] + 1], #] &[Union[i, j]]];
      Set[{a[n], c[k], i, j}, {k, True, j, f[k]}], {n, 3, nn}];
    Array[a, nn] (* Michael De Vlieger, Mar 20 2023 *)
  • PARI
    See Links section.

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

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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

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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

Views

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.
		

Crossrefs

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.

A363444 a(n) = n for n <= 3; for n > 3, a(n) is the smallest positive number that has not yet appeared that includes as factors the distinct prime factors of a(n-2) and a(n-1) that are not shared between a(n-2) and a(n-1).

Original entry on oeis.org

1, 2, 3, 6, 4, 9, 12, 8, 15, 30, 10, 18, 45, 20, 24, 60, 5, 36, 90, 25, 42, 210, 35, 48, 420, 70, 21, 120, 140, 63, 150, 280, 84, 75, 350, 126, 105, 40, 168, 315, 50, 252, 525, 80, 294, 630, 55, 462, 840, 110, 231, 1050, 220, 693, 1260, 330, 77, 1470, 660, 154, 735, 990, 308, 945, 1320, 616, 1155
Offset: 1

Views

Author

Scott R. Shannon, Jun 02 2023

Keywords

Comments

The last prime to appear in the first 10000 terms is a(17) = 5, and it is unknown if more appear. The largest terms increase rapidly in size, e.g., a(8924) = 2233642178577810, although subsequent terms can be significantly smaller. It is unknown is all numbers eventually appear.

Examples

			a(4) = 6 as a(2) = 2 and a(3) = 3 contain the distinct prime factors 2 and 3 respectively, both of which only appear in one term. Therefore a(4) is the smallest unused number that contains both 2 and 3 as factors, which is 6.
a(6) = 9 as a(4) = 6 = 2*3 and a(5) = 4 = 2*2, so 3 is the only prime factor that is not shared between these terms. Therefore a(6) is the smallest unused number that contains 3 as a factor, which is 9.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[] := False; m[] := 1; Array[Set[{a[#], c[#]}, {#, True}] &, 3];
    i = {a[2]}; j = {a[3]}; Do[q = Times @@ SymmetricDifference[i, j]; While[c[Set[k, q m[q]]], m[q]++]; Set[{a[n], c[k], i, j}, {k, True, j, FactorInteger[k][[All, 1]]}], {n, 4, nn}]; Array[a, nn] (* Michael De Vlieger, Jun 05 2023 *)

A364280 Lexicographically earliest sequence of distinct positive integers such that a(n) is the least novel multiple of m, the product of all primes q < gpf(a(n-2)*a(n-1)) which do not divide a(n-2)*a(n-1); a(1) = 1, a(2) = 2.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 10, 9, 8, 11, 105, 12, 13, 385, 18, 14, 15, 16, 17, 15015, 20, 19, 51051, 30, 21, 22, 25, 42, 23, 230945, 84, 24, 35, 26, 33, 70, 27, 28, 40, 36, 29, 37182145, 48, 31, 1078282205, 54, 32, 34, 30030, 37, 6678671, 60060, 38, 51, 5005, 44, 39
Offset: 1

Views

Author

David James Sycamore, Jul 17 2023

Keywords

Comments

It follows from the definition that the sequence is infinite.
Let r(n) = a(n-2)*a(n-1).
If rad(r(n)) is a primorial, then every prime q < gpf(r(n)) divides r(n), so m = 1, the empty product, and a(n) = u, the smallest missing number in the sequence so far.
If rad(r(n)) is not a primorial, then m > 1, and significant spikes can occur in scatterplot when there are multiple primes < gpf(r(n)) which do not divide r(n) (e.g., a(12) = 105, a(15) = 385, a(21) = 15015).
The only way a prime can occur is as u.
The sequence is a permutation of the positive integers since no number appears more than once and m = 1 eventually introduces any number not already placed consequent to terms arising from m > 1.

Examples

			a(4) = 4, a(5) = 5, and 3 is the only prime < 5 which does not divide 20, therefore m = 3 and a(6) = 6 since 3 has occurred once already.
a(10) = 8, a(11) = 11 and the product of all primes < 11 which do not divide 8*11 = 88 is 3*5*7 = 105, which has not occurred earlier, so a(12) = 105.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[] := False; m[] := 1; a[1] = i = 1; a[2] = j = 2; c[1] = c[2] = True;
      f[x_] := Times @@ Complement[Prime@ Range[PrimePi@ #[[-1]] - 1], #] &[
        FactorInteger[x][[All, 1]]];
      Do[While[Set[k, f[i j]]; c[k m[k]], m[k]++]; k *= m[k];
        Set[{a[n], c[k], i, j}, {k, True, j, k}], {n, 3, nn}];
    Array[a, nn] (* Michael De Vlieger, Jul 17 2023 *)

Extensions

More terms from Michael De Vlieger, Jul 17 2023

A367082 a(1), a(2) = 2. Thereafter a(n) is the least novel multiple of the greatest prime which divides precisely one of a(n-1), a(n-2) but not the other. If no such prime exists a(n) is the least novel multiple of the smallest prime dividing neither a(n-1) nor a(n-2).

Original entry on oeis.org

1, 2, 4, 3, 6, 8, 9, 12, 10, 5, 14, 7, 16, 21, 28, 15, 35, 42, 20, 49, 56, 18, 63, 70, 25, 77, 11, 84, 22, 33, 24, 44, 55, 30, 66, 88, 27, 99, 110, 40, 121, 132, 36, 143, 13, 154, 26, 39, 45, 52, 65, 50, 78, 91, 98, 104, 117, 48, 130, 156, 60, 169, 182, 105, 195, 208, 75, 221, 17, 234, 34, 51, 54, 68, 85, 80, 102, 119
Offset: 1

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Author

Keywords

Comments

The second condition of the definition applies iff rad(a(n-1)) = rad(a(n-2)). This occurs when n = 3 (a(2) = 2, a(3) = 4), and seems never to occur again.
Conjecture: This is a permutation of the positive integers, with primes in order. Each prime p generates a trajectory T(p), dominated by multiples of p, see graph. Typically T(prime(k)) is initiated following the earliest term divisible by prime(k). Some trajectories are more prominent in the graph than others e.g. T(7),T(11),T(13) extend from p = prime(k) to prime(k+1)*prime(k). Others, e.g. T(3) terminate earlier but include prime(k)^2, whilst some do not reach as far as prime(k)^2 (T(5) has just two terms). Thus three categories of prime trajectory can be distinguished in the graph (respectively "full", "medium" and "small"). The graph resembles an inclined comb whose teeth correspond to full and medium trajectories; see Example.

Examples

			a(1,2) = 1,2 so a(3) = 4, the least novel multiple of 2 (which divides 2 but not 1). Since rad(2) = rad(4) = 2 there is no prime which divides one of a(2), a(3) but not the other so by the second condition of the definition a(4) = 3, the least novel multiple of the smallest prime (3) which divides neither a(2) = 2 nor a(3) = 4.
The sequence can be presented as an irregular table where row(k) starts with A008578(k), and with the exception of rows 1 and 2, ends with the earliest multiple of A008578(k+1).
The table starts:
  1;
  2,4;
  3,6,8,9,12,10;
  5,14;
  7,16,21,28,15,35,42,20,49,56,18,63,70,25,77;
  11,84,22,33,24,44,55,30,66,88,27,110,40,121,132,36,143;
  13,154.....
T(3) is a medium trajectory, includes 3^2 but not 3*5 = 15, which appears later, in T(7). T(5) is a short trajectory, stopped by 14, does not include 25 which is delayed until T(7); T(7) is the first full trajectory, including 49, and ending with 7*11 = 77. In full and medium trajectories T(p) we see pairs of consecutive multiples of p separated by a multiple of a smaller prime. If T(prime(k)) is full it contains (prime(k+1) - 1) multiples of prime(k) and (prime(k+1) - 1)/2 multiples of smaller primes, thus T(7) contains 15 terms; see Formula.
		

Crossrefs

Programs

  • Mathematica
    nn = 120;
      c[] := False; m[] := 1;
      f[x_] := f[x] = FactorInteger[x][[All, 1]];
      g[x_] :=
       Block[{q = 2},
        If[OddQ[x], q, While[Divisible[x, q], q = NextPrime[q] ]; q] ];
      Array[Set[{a[#], c[#]}, {#, True}] &, 2];
      i = a[1]; j = a[2]; u = 3; ri = {}; rj = {j};
      Do[Set[r, Times @@ Union[ri, rj]]; If[ri == rj,
        While[c[Set[k, # m[#] ] ], m[#]++] &[g[r]],
        While[c[Set[k, # m[#] ] ], m[#]++] &[FactorInteger[r][[-1, 1]] ] ];
       Set[{a[n], c[k], i, j, ri, rj}, {k, True, j, k, rj, f[k]}], {n, 3, nn}];
    Array[a, nn] (* Michael De Vlieger, Nov 06 2023 *)

Formula

If T(prime(k)) is a fully developed trajectory is contains (prime(k+1) - 1)*3/2 terms.
Previous Showing 11-20 of 21 results. Next