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 1-7 of 7 results.

A372111 Partial sums of A124652.

Original entry on oeis.org

1, 3, 6, 10, 15, 24, 30, 38, 54, 66, 77, 84, 98, 126, 144, 168, 189, 216, 248, 279, 360, 370, 390, 403, 572, 594, 627, 646, 663, 702, 728, 777, 814, 858, 894, 942, 996, 1060, 1085, 1120, 1160, 1189, 1230, 1245, 1290, 1320, 1370, 1450, 1508, 1560, 1620, 1692, 1739
Offset: 1

Views

Author

Michael De Vlieger, Apr 25 2024

Keywords

Comments

Analogous to A109735 with respect to A109890.
A007947(A124652(n+1)) | a(n) for n > 2.

Examples

			See A124652.
		

Crossrefs

Programs

  • Mathematica
    nn = 54; c[_] := False; a[1] = 1; a[2] = 2; s = u = 3; c[1] = c[2] = True;
    f[x_] := f[x] = Times @@ FactorInteger[x][[All, 1]];
    {1}~Join~Reap[Do[Sow[s]; k = u; While[Or[Mod[s, f[k]] != 0, c[k]], k++];
        Set[{a[n], c[k]}, {k, True}];
        s += k; If[k == u, While[c[u], u++]], {n, 3, nn}] ][[-1, 1]]

A372009 Indices k such that A124652(k) is prime.

Original entry on oeis.org

2, 3, 5, 11, 12, 20, 24, 28, 29, 33, 42, 43, 53, 58, 67, 78, 93, 98, 104, 105, 109, 112, 118, 125, 126, 137, 141, 145, 146, 162, 174, 182, 185, 187, 188, 195, 200, 223, 224, 231, 232, 239, 246, 249, 252, 255, 259, 264, 271, 275, 283, 286, 287, 296, 298, 300, 326
Offset: 1

Views

Author

Michael De Vlieger, Apr 29 2024

Keywords

Comments

Analogous to A111238, a sequence which instead pertains to A109890.

Examples

			Let b(x) = A124652(x).
Table of first terms.
   n  a(n)  b(a(n))
  -----------------
   1    2      2
   2    3      3
   3    5      5
   4   11     11
   5   12      7
   6   20     31
   7   24     13
   8   28     19
   9   29     17
  10   33     37
  11   42     29
  12   43     41
  ...
		

Crossrefs

Programs

  • Mathematica
    nn = 300; c[_] := False;
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    f[x_] := Select[Range[x], Divisible[x, rad[#]] &];
    Array[Set[{a[#], c[#]}, {#, True}] &, 2]; s = a[1] + a[2];
    {2}~Join~Reap[Do[r = f[s]; k = SelectFirst[r, ! c[#] &];
        If[PrimeQ[k], Sow[i]]; c[k] = True;
        s += k, {i, 3, nn}] ][[-1, 1]]

Formula

Proper subset of A372028.

A372028 Numbers k such that A124652(k) divides A372111(k-1).

Original entry on oeis.org

3, 5, 7, 11, 12, 13, 15, 16, 17, 18, 20, 22, 24, 26, 27, 28, 29, 30, 31, 33, 40, 41, 42, 43, 44, 46, 49, 50, 51, 53, 55, 58, 59, 60, 62, 63, 64, 66, 67, 68, 69, 70, 71, 72, 73, 78, 79, 80, 92, 93, 95, 98, 101, 102, 103, 104, 105, 107, 109, 111, 112, 115, 116, 117
Offset: 1

Views

Author

Michael De Vlieger, May 05 2024

Keywords

Comments

Contains A372009(m), m > 1.
For k in this sequence, A124652(k) has the same relationship with A372111(k-1) as A109890(i) has with A109735(i-1) for i > 2.

Examples

			Let b(x) = A124652(x) and s(x) = A372111(x), where A372111 contains partial sums of A124652.
a(1) = 3 since b(3) = 3, a divisor of s(2) = 3.
a(2) = 5 since b(5) = 5, a divisor of s(4) = 10.
a(3) = 7 since b(7) = 6, a divisor of s(6) = 24, etc.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[_] := False;
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    f[x_] := Select[Range[x], Divisible[x, rad[#]] &];
    Array[Set[{a[#], c[#]}, {#, True}] &, 2]; s = a[1] + a[2];
    Reap[Do[r = f[s]; k = SelectFirst[r, ! c[#] &];
      If[Divisible[s, k], Sow[i]]; c[k] = True;
      s += k, {i, 3, nn}] ][[-1, 1]]

Formula

A124652(a(n)) is a number in row A372111(a(n)-1) of A027750.

A372284 Primes in the order in which they appear in A124652.

Original entry on oeis.org

2, 3, 5, 11, 7, 31, 13, 19, 17, 37, 29, 41, 47, 59, 23, 97, 53, 829, 71, 149, 101, 167, 79, 151, 43, 103, 157, 113, 139, 109, 137, 197, 353, 89, 277, 293, 269, 229, 73, 61, 571, 83, 191, 691, 251, 179, 127, 257, 193, 173, 239, 163, 331, 613, 617, 311, 67, 523
Offset: 1

Views

Author

Michael De Vlieger, Apr 29 2024

Keywords

Comments

Analogous to A111239, a sequence which instead pertains to A109890.

Examples

			Let b(x) = A372009(x).
Table of first terms:
   n  b(n) a(n)
  -------------
   1    2    2
   2    3    3
   3    5    5
   4   11   11
   5   12    7
   6   20   31
   7   24   13
   8   28   19
   9   29   17
  10   33   37
  11   42   29
  12   43   41
  ...
		

Crossrefs

Programs

  • Mathematica
    nn = 300; c[_] := False;
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    f[x_] := Select[Range[x], Divisible[x, rad[#]] &];
    Array[Set[{a[#], c[#]}, {#, True}] &, 2]; s = a[1] + a[2];
    {2}~Join~Reap[Do[r = f[s]; k = SelectFirst[r, ! c[#] &];
        If[PrimeQ[k], Sow[k]]; c[k] = True;
        s += k, {i, 3, nn}] ][[-1, 1]]

A372323 A124652(n) is the a(n)-th term in row A372111(n-1) of irregular triangle A162306.

Original entry on oeis.org

2, 4, 4, 4, 5, 7, 5, 8, 8, 2, 10, 8, 12, 11, 13, 6, 13, 6, 6, 9, 8, 11, 4, 8, 16, 5, 6, 7, 13, 12, 7, 10, 19, 15, 16, 17, 9, 6, 15, 10, 3, 11, 8, 18, 28, 14, 14, 10, 30, 28, 15, 4, 20, 33, 13, 12, 6, 22, 18, 21, 12, 11, 29, 12, 11, 8, 24, 18, 8, 14, 17, 32, 33
Offset: 3

Views

Author

Michael De Vlieger, May 05 2024

Keywords

Comments

Let b(x) = A124652(x) and let s(x) = A372111(x), where A372111 contains partial sums of A124652.
Let r(x) = A010846(x), the number of m <= x such that rad(m) | x, where rad = A007947.
Let row k of A162306 contain { m : rad(m) | k, m <= k }. Thus r(k) is the length of row k of A162306.
Let T(k,j) represent the j-th term in row k of irregular triangle A162306.
a(n) = j is the position of b(n) in row s(n-1) of A162306.
b(n) = T(s(n-1), a(n)).
Analogous to A371910, which instead regards A109890 and A109735.

Examples

			Let b(x) = A124652(x) and let s(x) = A372111(x), where A372111 contains partial sums of A124652.
a(3) = 2 since b(3) = 3 is the 2nd term in row s(3) = 3 of A162306, {1, [3]}.
a(4) = 4 since b(4) = 4 is the 4th term in row s(4) = 6 of A162306, {1, 2, 3, [4], 6}.
a(5) = 4 since b(5) = 5 is T(s(n-1), 4) = T(10, 4), {1, 2, 4, [5], 8, 10}.
a(6) = 4 since b(6) = 9 is T(s(n-1), 4) = T(15, 4), {1, 3, 5, [9], 15}.
a(7) = 5 since b(7) = 6 is T(s(n-1), 5) = T(24, 5), {1, 2, 3, 4, [6], 8, 9, 12, 16, 18, 24}, etc.
Table relating this sequence to b = A124652, s = A372111, r = A372322, and A162306.
   n b(n) s(n-1) a(n) r(n) row s(n-1) of A162306
  ---------------------------------------------------------------------
   3    3    3    2    2   {1, [3]}
   4    4    6    4    5   {1, 2, 3, [4], 6}
   5    5   10    4    6   {1, 2, 4, [5], 8, 10}
   6    9   15    4    5   {1, 3, 5, [9], 15}
   7    6   24    5   11   {1, 2, 3, 4, [6], ..., 24}
   8    8   30    7   18   {1, 2, 3, 4, 5, 6, [8], ..., 30}
   9   16   38    5    8   {1, 2, 4, 8, [16], 19, 32, 38}
  10   12   54    8   16   {1, 2, 3, 4, 6, 8, 9, [12], ..., 54}
  11   11   66    8   22   {1, 2, 3, 4, 6, 8, 9, [11], ..., 66}
  12    7   77    2    5   {1, [7], 11, 49, 77}
  13   14   84   10   28   {1, 2, 3, 4, ..., 12, [14], ..., 84}
  14   28   98    8   13   {1, 2, 4, 7, ..., 16, [28], ..., 98}
		

Crossrefs

Programs

  • Mathematica
    nn = 75; c[_] := False;
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    f[x_] := Select[Range[x], Divisible[x, rad[#]] &];
    Array[Set[{a[#], c[#]}, {#, True}] &, 2]; s = a[1] + a[2];
    Reap[Do[r = f[s]; k = SelectFirst[r, ! c[#] &];
      Sow[FirstPosition[r, k][[1]]]; c[k] = True;
      s += k, {i, 3, nn}] ][[-1, 1]]

A372399 Numbers k such that A124652(k) does not divide A372111(k-1).

Original entry on oeis.org

2, 4, 6, 8, 9, 10, 14, 19, 21, 23, 25, 32, 34, 35, 36, 37, 38, 39, 45, 47, 48, 52, 54, 56, 57, 61, 65, 74, 75, 76, 77, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 94, 96, 97, 99, 100, 106, 108, 110, 113, 114, 122, 123, 130, 136, 142, 153, 157, 158, 159, 170, 171
Offset: 1

Views

Author

Michael De Vlieger, May 05 2024

Keywords

Comments

Let b(x) = A124652(x) and let s(x) = A372111(x), where A372111 contains partial sums of A124652.
For n > 2, 1 < gcd(b(a(n)), s(a(n)-1)) < b(a(n)).
For n > 2, both b(a(n)) and s(a(n)-1) are necessarily composite, since prime p either divides or is coprime to n. Furthermore, both b(a(n)) and s(a(n)-1) have at least 2 distinct prime factors.
Indices of records in A124652 except {1, 2, 3, 5} are in this sequence.

Examples

			a(1) = 2 since b(2) = 2 does not divide s(1) = 1.
a(2) = 4 since b(4) = 4 does not divide s(3) = 6.
a(3) = 6 since b(6) = 9 does not divide s(5) = 15.
a(4) = 8 since b(8) = 8 does not divide s(7) = 30.
a(5) = 9 since b(9) = 16 does not divide s(8) = 38, etc.
Table of b(k) and s(k-1), where k = a(n), n = 2..12. Asterisked k denote terms such that rad(b(k)) | rad(s(k-1)); k = 73 and k = 4316 are the only other known indices where the terms have this quality.
     k      b(k)                        s(k-1)
    ----------------------------------------------------------
     4      4 =  2^2                    6 =  2 * 3
     6      9 =  3^2                   15 =  3 * 5
     8      8 =  2^3                   30 =  2 * 3 * 5
     9     16 =  2^4                   38 =  2 * 19
    10*    12 =  2^2 * 3               54 =  2 * 3^3
    14*    28 =  2^2 * 7               98 =  2 * 7^2
    19     32 =  2^5                  216 =  2^3 * 3^3
    21     81 =  3^4                  279 =  3^2 * 31
    23     20 =  2^2 * 5              370 =  2 * 5 * 37
    25    169 = 13^2                  403 = 13 * 31
    32     49 =  7^2                  728 =  2^3 * 7 * 13
    ...
    73*   100 =  2^2 * 5^2           4800 =  2^6 * 3 * 5^2
    ...
  4316*  4720 =  2^4 * 5 * 59    30806850 =  2 * 3 * 5^2 * 59^3
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[_] := False;
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    f[x_] := Select[Range[x], Divisible[x, rad[#]] &];
    Array[Set[{a[#], c[#]}, {#, True}] &, 2]; s = a[1] + a[2];
    {2}~Join~Reap[Do[
      r = f[s]; k = SelectFirst[r, ! c[#] &];
      If[! Divisible[s, k], Sow[i]];
      c[k] = True;
      s += k, {i, 3, nn}] ][[-1, 1]]

Formula

A124652(a(n)) is a number in row A372111(a(n)-1) of A272618.

A372322 a(n) = A010846(A372111(n)).

Original entry on oeis.org

1, 2, 5, 6, 5, 11, 18, 8, 16, 22, 5, 28, 13, 33, 23, 38, 11, 26, 12, 9, 58, 28, 80, 5, 30, 55, 19, 27, 19, 56, 37, 21, 27, 87, 44, 44, 48, 38, 18, 58, 42, 5, 110, 26, 112, 140, 38, 45, 32, 144, 102, 59, 5, 139, 225, 39, 44, 22, 180, 86, 114, 34, 23, 133, 41, 115
Offset: 1

Views

Author

Michael De Vlieger, May 05 2024

Keywords

Comments

Let r(x) = A010846(x), the number of m <= x such that rad(m) | x, where rad = A007947.
Let row k of A162306 contain { m : rad(m) | k, m <= k }. Thus r(k) is the length of row k of A162306.
a(n) is the length of row A372111(n) of A162306.
Analogous to A371909, which instead regards A109890 and A109735.

Examples

			Let s(x) = A372111(x) and let r(x) = A010846(x).
a(1) = 1 since r(s(1)) = r(1) = 1.
a(2) = 2 since r(s(2)) = r(3) = 2. For prime p, r(p) = card({1, p}) = 2.
a(3) = 5 since r(s(3)) = r(6) = 5. r(6) = card({1, 2, 3, 4, 6}) = 5.
a(4) = 6 since r(s(4)) = r(10) = 6. r(10) = card({1, 2, 4, 5, 8, 10}) = 6.
a(5) = 5 since r(s(5)) = r(15) = 5. r(15) = card({1, 3, 5, 9, 15}) = 5.
a(6) = 11 since r(s(6)) = r(24) = 11. r(24) = card({1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24}) = 11, etc.
		

Crossrefs

Programs

  • Mathematica
    nn = 68; c[_] := False;
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    f[x_] := Select[Range[x], Divisible[x, rad[#]] &];
    Array[Set[{a[#], c[#]}, {#, True}] &, 2]; s = a[1] + a[2];
    {1}~Join~Reap[Do[r = f[s]; k = SelectFirst[r, ! c[#] &];
      Sow[Length[r]]; c[k] = True;
      s += k, {i, 3, nn}] ][[-1, 1]]
Showing 1-7 of 7 results.