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 1-5 of 5 results.

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.

Links

Crossrefs

Cf. A007947, A027750, A109735, A109890, A124652, A372009, A372111, A372399.

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.

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.

Links

Crossrefs

Cf. A007947, A010846, A124652, A162306, A371909, A372111, A372323.

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]]

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}

Links

Crossrefs

Cf. A007947, A010846, A124652, A162306, A371910, A372111, A372322.

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

Links

Crossrefs

Cf. A007947, A124652, A272618, A372111.

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.

A373534 Decimal expansion of Pi^(1/2)*Gamma(1/20)/(10*Gamma(11/20)).

Original entry on oeis.org

2, 1, 3, 5, 3, 4, 4, 9, 3, 3, 2, 4, 8, 0, 0, 4, 2, 2, 8, 0, 4, 6, 4, 7, 5, 2, 7, 9, 6, 8, 3, 7, 0, 6, 7, 7, 8, 8, 1, 0, 8, 7, 9, 3, 6, 6, 0, 1, 6, 4, 9, 4, 0, 0, 4, 0, 7, 7, 3, 1, 4, 4, 2, 9, 1, 0, 8, 7, 0, 3, 3, 0, 0, 1, 4, 9, 6, 8, 8, 3, 7, 8, 0, 6, 6, 5, 8, 3, 6, 5, 1, 2, 2, 2, 2, 2, 0, 5, 9, 6, 5
Offset: 1

Views

Author

Takayuki Tatekawa, Jun 08 2024

Keywords

Comments

Constants from generalized Pi integrals: the case of n=20.

Examples

			2.135344933248004228046475279683...

Links

Crossrefs

Cf. A085565, A113477, A262427, A371824, A371881, A371930, A372111, A372327.

Programs

  • Maple
    (2*sqrt(Pi)*GAMMA(21/20))/GAMMA(11/20): evalf(%, 102); # Peter Luschny, Jun 17 2024
  • Mathematica
    RealDigits[2*Sqrt[Pi]/20*Gamma[1/20]/Gamma[11/20], 10, 5001][[1]]

Formula

Equals 2*Integral_{x=0..1} dx/sqrt(1-x^20).
Equals (2*sqrt(Pi)*Gamma(21/20))/Gamma(11/20). - Peter Luschny, Jun 17 2024
Showing 1-5 of 5 results.

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