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 1-3 of 3 results.

A361534 Let h,i,j be the latest 3 terms in the sequence, starting with a(1)=1, a(2)=2, a(3)=3. Let R = rad(h*i*j), where rad is A007947, and let S be the smallest number of terms in U = {h,i,j} which are divisible by any prime p dividing R. Then, a(n) is the least novel multiple of the greatest such prime p.

Original entry on oeis.org

1, 2, 3, 6, 9, 4, 12, 15, 5, 8, 18, 10, 20, 21, 7, 25, 30, 14, 28, 35, 40, 42, 24, 49, 56, 27, 33, 11, 22, 36, 39, 13, 16, 45, 26, 52, 50, 55, 65, 78, 44, 66, 91, 104, 77, 88, 117, 130, 99, 110, 143, 156, 60, 121, 169, 182, 132, 154, 195, 208, 165, 176, 221, 17, 234, 48, 34, 51, 68, 54, 85, 70, 102
Offset: 1

Views

Author

David James Sycamore, Mar 15 2023

Keywords

Comments

In the first 2^20 terms there are only 3 occasions of adjacent terms with same rad, (e.g., a(12,13) = 10, 20); only two occasions of rad(h) = rad(j) (e.g., a(3,4,5) = 3,6,9) and there is no occasion of rad(h) = rad(i) = rad(j) (possible iff S = 3), so it's reasonable to assume that it never happens. In any case the definition would ensure continuation of the sequence since a(n) would be the least novel multiple of the greatest prime divisor of R.
The sequence is conjectured to be a permutation of the positive integers with the primes appearing in their natural order.
The scatterplot resembles a "fine-toothed comb" similar to that of A361133.

Examples

			a(1,2,3) = 1,2,3 and every prime divisor in R divides only one term (S = 1). Since 3 is the greatest such prime, a(4) = 6, the least novel multiple of 3.
a(2,3,4) = 2,3,6 and S = 2 since both 2 and 3 divide two terms in U and no prime divides more than that. Since 3 is the greatest such prime a(5) must be 9.
a(3,4,5) = 3,6,9 and S = 1 because 2|6 but no other term in {3,6,9}, therefore a(6) = 4, the least novel multiple of 2.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[] = False; q[] = 1;
    Array[Set[{a[#], c[#]}, {#, True}] &, 3];
    Set[{h, i, j}, {a[1], a[2], a[3]}]; u = 4;
    w = Array[FactorInteger[#][[All, 1]] &, 3];
    Do[g = Reverse@ Union@ Flatten[w];
     p = First[
        Join @@ Map[ReverseSort,
          SplitBy[SortBy[
            Table[{p, Count[{h, i, j}, _?(Divisible[#, p] &)]}, {p, g}],
            Last], Last]]][[1]] ;
     (k = q[#]; While[c[k #], k++]; k *= #; While[c[# q[#]], q[#]++]) &[p];
     Set[{a[n], c[k], w, h, i, j}, {k, True,
       Append[w[[2 ;; -1]], FactorInteger[k][[All, 1]]], i, j, k}];
     If[k == u, While[c[u], u++]], {n, 4, nn}];
    Array[a, nn] (* Michael De Vlieger, Mar 15 2023 *)

A361629 For n <= 2, a(n) = n. Thereafter let p be the greatest prime which divides the least number of terms in U = {a(n-2), a(n-1)}, then a(n) is the smallest multiple of p that is not yet in the sequence.

Original entry on oeis.org

1, 2, 4, 6, 3, 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, 112, 136, 153, 57, 19
Offset: 1

Views

Author

David James Sycamore, Mar 18 2023

Keywords

Comments

In other words, if R = A007947(a(n-2)*a(n-1)), and S (= 1 or 2) is the smallest number of terms in U divisible by any p|R, then a(n) is the least novel multiple of the greatest p|R which divides S terms in U.
When for some m > 1, a(n-1) = m*p, is the first occurrence of a multiple of prime p, then a(n) = p. The sequence after a(1) = 1 can be represented as an irregular table in which the n-th row T(n,k) starts with p = prime(n), and ends with the first subsequent term divisible by q = prime(n+1). The next row begins with q, and so on. The range of k is not always known; see Formula and Example.
Some primes (2,7,11,13,47,79,113,127,...) exhibit fully developed initial trajectories T(n,k) dominated by multiples of p, until p*q is reached, whereas for others the trajectory ends with a term m*q where m < p.
The scatterplot resembles a fine-toothed comb (as in A361133, A361534), in which each "tooth" represents the initial trajectory of a prime. It is not known whether infinitely many primes exhibit fully developed initial trajectories.
Conjectured to be a permutation of the positive integers with primes in order.

Examples

			a(1,2) = 1,2 and 2 is the smallest p|R = 2 which divides the least number (1) of terms in U = {1,2}. Therefore a(3) = 4, is the least multiple of 2 which has not already occurred.
a(2,3) = 2,4 and since rad(2) = rad(4) = 2, S = 2 then a(4) = 6, the least multiple of 2 which has not occurred earlier.
a(3,4) = 4,6 so S = 1 since 3|6 but not 2, so a(5) = 3, since 3 has not appeared previously.
The sequence after a(1) can be shown as an irregular table of finite subsequences where each row T(n) starts with prime(n) and ends with the first subsequent multiple of prime(n+1). The table includes the fully developed trajectories of 2,7,11,13 with 3,15,24 terms respectively, each having final term in A006094. T(4) starts with 7, ends with 77 and has 3*(prime(5)-1)/2 = 15 terms.
Table T(n) for n = 1..10:
  2,4,6
  3,8,9,12,10
  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
  17,234,34,51,54,68,85,80,102,119,112,136,153,57
  19,69,
  23,72,46,92,138,81,115,161,126,184,207,87
  29,90,58,116,174,93
		

Crossrefs

Programs

  • Mathematica
    Block[{a, c, g, h, i, j, k, p, q, u, w, nn},
    nn = 120; c[] = False; q[] = 1;
    a[1] = i = 1; a[2] = j = 2; c[1] = c[2] = True; w = {1, 2}; u = 3;
    Do[If[Equal @@ w, p = Times @@ w[[1]],
      g = Reverse@ Union@ Flatten[w];
      p = First[
         Join @@ Map[ReverseSort,
           SplitBy[SortBy[
             Table[{p, Count[{i, j}, _?(Divisible[#, p] &)]}, {p, g}],
             Last], Last]] ][[1]] ];
      k = q[p]; While[c[k*p], k++]; k *= p; While[c[p*q[p]], q[p]++]
      Set[{a[n], c[k], w, i, j},
          {k, True, Append[w[[2 ;; -1]], FactorInteger[k][[All, 1]]], j, k}];
      If[k == u, While[c[u], u++]], {n, 3, nn}];
    Array[a, nn] ] (* Michael De Vlieger, Mar 18 2023 *)

Formula

The number of terms T(n,k) is <= 3*(prime(k+1) - 1)/2, with equality when the trajectory is fully developed. The number of terms in shorter trajectories is not readily known. Therefore the irregular table (see Example) is expressed as T(n), rather than T(n,k).

A362631 Lexicographically earliest infinite sequence of distinct positive integers with a(n) = n for n <= 3, and for n > 3 a(n) is the least novel multiple of the greatest prime divisor of a(n-2) which does not divide a(n-1).

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 9, 10, 12, 15, 8, 20, 7, 25, 14, 30, 21, 35, 18, 28, 24, 42, 11, 49, 22, 56, 33, 63, 44, 70, 55, 77, 40, 66, 45, 88, 50, 99, 60, 110, 27, 121, 36, 132, 13, 143, 16, 26, 17, 39, 34, 52, 51, 65, 68, 78, 85, 91, 102, 104, 119, 117, 136, 130, 153, 156, 170, 169, 187, 182, 204, 195, 221, 75, 238, 80, 255, 32, 272, 19
Offset: 1

Views

Author

David James Sycamore, May 06 2023

Keywords

Comments

The definition reflects that of A098550 in that it places a condition on a(n-2) which does not apply to a(n-1).
If there is no prime divisor of a(n-2) which does not divide a(n-1), then by empty product convention a(n) = u the least unused number.
Some primes (23,29,31,47,...) enter because of dividing a(n-2) but not a(n-1), whereas others (5,7,11,13,17,19,...) enter as least u; see Example.
With the exception of 16 all least u terms (up to a(2^28)) are primes, so it seems likely that a(47) = 16 is a one-off (fluke) term.
The scatterplot resembles a fine-toothed comb similar to those of A361629, A361133 and A361534, in which each "tooth" starts with a novel prime p and continues with a run of (mostly) alternate multiples of p and the greatest prime less than p until interrupted by the arrival of the next prime, and so forth.
The sequence, after a(1) = 1 can be represented as an irregular table in which the n-th row starts with prime(n), see Example.
Conjectured to be a permutation of the positive integers with the primes in order.

Examples

			a(4,5) = 4, 6 and since rad(4)|rad(6) a(6) = least u = 5.
a(11,12) = 8, 20 and since rad(8)|rad(20) a(13) = least u = 7.
a(44,45) = 132, 13 and gpd(132) = 11 does not divide 13, and since it is the 13th occurrence of p = 11, a(46) = 13*11 = 143.
a(45,46) = 13, 143 which forces a(47) = least u = 16 (see Comment).
a(90,91) = 69, 114 and 23 is the greatest prime dividing 69 which does not divide 114. Since 23 has not appeared earlier in the sequence a(92) = 23.
There is as yet no known formula for the row lengths of the table below. Whereas most rows terminate with a multiple of the prime they start with, there are exceptions, e.g., 47, 109. This behavior is open to explanation.
The table starts:
2;
3,4,6;
5,9,10,12,15,8,20;
7,25,14,30,21,35,18,28,24,42;
11,49,22,56,33,63,44,70,55,77,40,66,45,88,50,99,60,110,27,121,36,132;
13,143,16,26;
17,39,34,52,51,65,68,78,85,91,102,104,119,117,136,130,153,156,170,169,187,182,204,195,221,75,238,80,255,32,272;
19,....
		

Crossrefs

Programs

  • Mathematica
    mm = 3; nn = 10^4; c[] := False; q[] := 1;
    Array[Set[{a[#], c[#]}, {#, True}] &, mm]; q[2]++; u = mm + 1;
    i = a[mm - 1]; j = a[mm]; s = {1};
    Do[t = FactorInteger[i][[All, 1]];
      If[SubsetQ[s, t], q[u]++; Set[k, u],
       p = SelectFirst[Reverse@ t, CoprimeQ[j, #] &];
       If[! IntegerQ[p], Set[k, u], While[c[p*q[p]], q[p]++]; k = p*q[p]] ];
      Set[{a[n], c[k], i, j, s}, {k, True, j, k, t}];
      If[k == u, While[c[u], u++]], {n, mm + 1, nn}];
    Array[a, nn]
Showing 1-3 of 3 results.