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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A241920 After a(1)=1, each n appears A061395(n) times, where A061395 gives the index of the largest prime factor of n.

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 5, 5, 6, 6, 7, 7, 7, 7, 8, 9, 9, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 16, 17, 17, 17, 17, 17, 17, 17, 18, 18, 19, 19, 19, 19, 19, 19, 19, 19, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 22
Offset: 1

Views

Author

Antti Karttunen, May 01 2014

Keywords

Comments

Only numbers that occur just once are the powers of two (A000079).

Links

Crossrefs

Cf. A082288, A061395, A203623, A067255, A241918, A241914, A241915.

A307056 Row n = digits of A025487(n) in primorial base.

Original entry on oeis.org

1, 1, 0, 2, 0, 1, 0, 0, 1, 1, 0, 2, 0, 0, 2, 2, 0, 4, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 3, 0, 0, 2, 0, 0, 0, 2, 0, 2, 0, 2, 2, 0, 0, 3, 1, 0, 0, 4, 0, 0, 0, 4, 1, 1, 0, 4, 4, 0, 0, 6, 0, 0, 0, 6, 2, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 2, 2, 0, 1, 2, 3, 0, 0, 1, 5, 0, 0, 0, 1, 5, 4
Offset: 1

Views

Author

Michael De Vlieger, Mar 21 2019

Keywords

Comments

A025487 is the sequence of products of primorials (A002110).

Examples

			First rows of this sequence:
   n  A025487(n)  Row n
-------------------------------
   1          1   1
   2          2   1, 0
   3          4   2, 0
   4          6   1, 0, 0
   5          8   1, 1, 0
   6         12   2, 0, 0
   7         16   2, 2, 0
   8         24   4, 0, 0
   9         30   1, 0, 0, 0
  10         32   1, 0, 1, 0
  11         36   1, 1, 0, 0
  12         48   1, 3, 0, 0
  13         60   2, 0, 0, 0
  14         64   2, 0, 2, 0
  15         72   2, 2, 0, 0
  16         96   3, 1, 0, 0
  17        120   4, 0, 0, 0
  18        128   4, 1, 1, 0
  19        144   4, 4, 0, 0
  20        180   6, 0, 0, 0
  21        192   6, 2, 0, 0
  22        210   1, 0, 0, 0, 0
  ...

Links

Crossrefs

Cf. A002110, A002182, A025487.

Programs

  • Mathematica
    f[n_] := {{1}}~Join~Block[{lim = Product[Prime@ i, {i, n}], ww = NestList[Append[#, 1] &, {1}, n - 1], dec}, dec[x_] := Apply[Times, MapIndexed[Prime[First@ #2]^#1 &, x]]; Map[Block[{w = #, k = 1}, Sort@ Prepend[If[Length@ # == 0, #, #[[1]]], Product[Prime@ i, {i, Length@ w}] ] &@ Reap[Do[If[# < lim, Sow[#]; k = 1, If[k >= Length@ w, Break[], k++]] &@ dec Set[w, If[k == 1, MapAt[# + 1 &, w, k], PadLeft[#, Length@ w, First@ #] &@ Drop[MapAt[# + Boole[i > 1] &, w, k], k - 1] ]], {i, Infinity}] ][[-1]] ] &, ww]]; Block[{nn = 10, b}, b = MixedRadix[Reverse@ Prime@ Range@ nn]; Map[IntegerDigits[#, b] &, Union@ Flatten@ f@ nn]]

A325609 Unsorted q-signature of n!. Irregular triangle read by rows where T(n,k) is the multiplicity of q(k) in the factorization of n! into factors q(i) = prime(i)/i.

Original entry on oeis.org

1, 2, 1, 4, 1, 5, 2, 1, 7, 3, 1, 9, 3, 1, 1, 12, 3, 1, 1, 14, 5, 1, 1, 16, 6, 2, 1, 17, 7, 3, 1, 1, 20, 8, 3, 1, 1, 22, 9, 3, 1, 1, 1, 25, 9, 3, 2, 1, 1, 27, 11, 4, 2, 1, 1, 31, 11, 4, 2, 1, 1, 33, 11, 4, 3, 1, 1, 1, 36, 13, 4, 3, 1, 1, 1, 39, 13, 4, 3, 1, 1, 1, 1
Offset: 1

Views

Author

Gus Wiseman, May 12 2019

Keywords

Comments

Every positive integer has a unique q-factorization (encoded by A324924) into factors q(i) = prime(i)/i, i > 0. For example:
11 = q(1) q(2) q(3) q(5)
50 = q(1)^3 q(2)^2 q(3)^2
360 = q(1)^6 q(2)^3 q(3)
Row n is the sequence of nonzero exponents in the q-factorization of n!.
Also the number of terminal subtrees with Matula-Goebel number k of the rooted tree with Matula-Goebel number n!.

Examples

			We have 10! = q(1)^16 q(2)^6 q(3)^2 q(4), so row n = 10 is (16,6,2,1).
Triangle begins:
  {}
   1
   2  1
   4  1
   5  2  1
   7  3  1
   9  3  1  1
  12  3  1  1
  14  5  1  1
  16  6  2  1
  17  7  3  1  1
  20  8  3  1  1
  22  9  3  1  1  1
  25  9  3  2  1  1
  27 11  4  2  1  1
  31 11  4  2  1  1
  33 11  4  3  1  1  1
  36 13  4  3  1  1  1
  39 13  4  3  1  1  1  1
  42 14  5  3  1  1  1  1

Crossrefs

Row lengths are A000720.
Row sums are A325544(n) - 1.
Column k = 1 is A325543.
Cf. A056239, A067255, A112798, A118914, A124010.
Matula-Goebel numbers: A007097, A061775, A109129, A196050, A317713, A324935.
Factorial numbers: A000142, A011371, A022559, A071626, A115627, A325276.
q-factorization: A324922, A324923, A324924, A325614, A325615, A325660.

Programs

  • Mathematica
    difac[n_]:=If[n==1,{},With[{i=PrimePi[FactorInteger[n][[1,1]]]},Sort[Prepend[difac[n*i/Prime[i]],i]]]];
Table[Length/@Split[difac[n!]],{n,20}]

A356008 A variant of Look and Say sequence (A005150) based on exponents in prime factorization of n (see Comments section for precise definition).

Original entry on oeis.org

1, 6, 105, 12, 315, 18, 945, 24, 525, 6006, 2835, 420, 8505, 42042, 735, 48, 25515, 1050, 76545, 12012, 440895, 294294, 229635, 840, 1575, 2060058, 2625, 84084, 688905, 54, 2066715, 96, 5731635, 14420406, 2205, 36, 6200145, 100942842, 74511255, 24024, 18600435
Offset: 1

Views

Author

Rémy Sigrist, Jul 23 2022

Keywords

Comments

To compute a(n):
- a(1) = 1,
- for n > 1:
- consider the prime factorization of n:
n = Product_{i = 1..k} prime(i)^e_i
(where e_k > 0 and prime(i) denotes the i-th prime number),
- apply the Look and Say procedure to the list (e_k, ..., e_1),
- the result, say (f_m, ..., f_1), gives the prime exponents for a(n):
a(n) = Product_{i = 1..m} prime(i)^f_i.
There are only two fixed points: a(1) = 1 and a(36) = 36.
All terms are distinct and belong to A244990 (but some terms of A244990, like 210 = 7*5*3*2, do not appear here).

Examples

			For n = 99:
- 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
- the list of exponents is: 1 0 0 2 0,
- applying the Look and Say procedure, we obtain: 1 1 2 0 1 2 1 0,
- so a(99) = 19^1 * 17^1 * 13^2 * 11^0 * 7^1 * 5^2 * 3^1 * 2^0 = 28658175.

Links

Crossrefs

Cf. A002110, A005150, A007814, A008776, A067255, A244990, A356014, A356016.

Programs

  • PARI
    See Links section.

Formula

a(n) = n mod 2.
A007814(a(n)) = A007814(n).
a(prime(n)) = 7*5*3^(n-1) for any n > 1.
a(A002110(n)) = 2*3^n = A008776(n) for any n > 0.

A356014 Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique e; the resulting list corresponds to the exponents in the prime factorization of a(n).

Original entry on oeis.org

1, 2, 3, 4, 3, 2, 3, 8, 9, 10, 3, 12, 3, 10, 3, 16, 3, 18, 3, 20, 21, 10, 3, 24, 9, 10, 27, 20, 3, 2, 3, 32, 21, 10, 3, 4, 3, 10, 21, 40, 3, 10, 3, 20, 45, 10, 3, 48, 9, 50, 21, 20, 3, 54, 21, 40, 21, 10, 3, 12, 3, 10, 63, 64, 21, 10, 3, 20, 21, 10, 3, 72, 3
Offset: 1

Views

Author

Rémy Sigrist, Jul 23 2022

Keywords

Comments

We ignore the exponents (all 0's) for the prime numbers beyond the greatest prime factor of n.
This sequence operates on prime exponents as A090079 and A337864 operate on binary and decimal digits, respectively.

Examples

			For n = 99:
- 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
- the list of exponents is: 1 0 0 2 0,
- compressing consecutive values, we obtain: 1 0 2 0,
- so a(99) = 7^1 * 5^0 * 3^2 * 2^0 = 63.

Crossrefs

Cf. A007814, A061742, A067255, A090079, A115964, A294674, A319521, A337864, A356008, A356021 (fixed points).

Programs

  • PARI
    a(n) = { my (v=1, e=-1, k=0); forprime (p=2, oo, if (n==1, return (v), if (e!=e=valuation(n,p), v*=prime(k++)^e); n/=p^e)) }

Formula

a(a(n)) = a(n).
a(n^k) = a(n)^k for any k >= 0.
a(n) = A319521(A356008(n)).
A007814(a(n)) = A007814(n).
a(n) = 3 iff n belongs to A294674 \ {1}.
a(n) = 4 iff n belongs to A061742 \ {1}.
a(n) = 8 iff n belongs to A115964.

A362227 a(n) = Product_{k=1..w(n)} p(k)^(S(n,k)-1), where set S(n,k) = row n of A272011 and w(n) = A000120(n) is the binary weight of n.

Original entry on oeis.org

1, 2, 4, 12, 8, 24, 72, 360, 16, 48, 144, 720, 432, 2160, 10800, 75600, 32, 96, 288, 1440, 864, 4320, 21600, 151200, 2592, 12960, 64800, 453600, 324000, 2268000, 15876000, 174636000, 64, 192, 576, 2880, 1728, 8640, 43200, 302400, 5184, 25920, 129600, 907200, 648000, 4536000, 31752000, 349272000, 15552
Offset: 0

Views

Author

Michael De Vlieger, Jun 08 2023

Keywords

Comments

In other words, let S(n) contain place values of 1's in the binary expansion of n, ordered greatest to least, where S(n,1) = floor(log_2(n+1)) = A000523(n+1) and the remaining terms in S strictly decrease. This sequence reads S(n,k)+1 instead as a multiplicity of prime(k) so as to produce a number with strictly decreasing prime exponents.

Examples

			a(0) = 1 since 1 is the empty product.
a(1) = 2 since 1 = 2^0, s = {0}, hence a(1) = prime(1)^(0+1) = 2^1 = 2.
a(2) = 4 since 2 = 2^1, s = {1}, hence a(2) = 2^(1+1) = 4.
a(3) = 12 since 3 = 2^1+2^0, s = {1,0}, hence a(3) = 2^2*3^1 = 12, etc.
The table below relates first terms of this sequence greater than 1 to A272011 and A067255:
   n   A272011(n)  a(n)  A067255(a(n))
  ------------------------------------
   1   0             2   1
   2   1             4   2
   3   1,0          12   2,1
   4   2             8   3
   5   2,0          24   3,1
   6   2,1          72   3,2
   7   2,1,0       360   3,2,1
   8   3            16   4
   9   3,0          48   4,1
  10   3,1         144   4,2
  11   3,1,0       720   4,2,1
  12   3,2         432   4,3
  13   3,2,0      2160   4,3,1
  14   3,2,1     10800   4,3,2
  15   3,2,1,0   75600   4,3,2,1
  16   4            32   5
  ...
This sequence appears below, seen as an irregular triangle T(m,j) delimited by 2^m where j = 1..2^(m-1) for m > 0:
   1;
   2;
   4, 12;
   8, 24,  72, 360;
  16, 48, 144, 720, 432, 2160, 10800, 75600;
  ...
T(m,1) = 2^m.
T(m,2^(m-1)) = A006939(m) for m > 0.

Links

Crossrefs

Cf. A000120, A006939, A067255, A087980, A272011.

Programs

  • Mathematica
    Array[Times @@ MapIndexed[Prime[First[#2]]^(#1 + 1) &, Length[#] - Position[#, 1][[All, 1]] ] &[IntegerDigits[#, 2]] &, 48, 0]

Formula

This sequence, sorted, is A087980.
a(2^k) = 2^(k+1).
a(2^k-1) = A006939(k-1).

A072941 Least multiple of n having no prime gaps.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 30, 11, 12, 13, 210, 15, 16, 17, 18, 19, 60, 105, 2310, 23, 24, 25, 30030, 27, 420, 29, 30, 31, 32, 1155, 510510, 35, 36, 37, 9699690, 15015, 120, 41, 210, 43, 4620, 45, 223092870, 47, 48, 49, 150, 255255, 60060, 53, 54, 385
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 12 2002

Keywords

Comments

a(n) = smallest m such that m is a multiple of n and in the prime factorization of m every prime between the smallest prime factor of n and the largest appears at least once.
A073490(a(n))=0; a(n)=n iff A073490(A007947(n))=0; A006530(a(n)) = A006530(n); A020639(a(n)) = A020639(n); A001221(n) <= A001221(a(n)); A001221(a(n))=A049084(A006530(n))-A049084(A020639(n))+1; A001222(n) <= A001222(a(n)); A001222(a(n)) + A001221(n) = A001221(a(n)) + A001222(n).

Examples

			a(99)=a(3*3*11)=3*3*5*7*11=3465.

Links

Crossrefs

Cf. A073490, A074044, A067255.

Programs

  • Haskell
    a072941 n = product $ zipWith (^) ps $ map (max 1) es where
            (ps, es) = unzip $ dropWhile ((== 0) . snd) $
                       zip a000040_list $ a067255_row n
-- Reinhard Zumkeller, Dec 21 2013

Formula

a(n)=A002110(A049084(A006530(n)))*A003557(n)/A002110(A049084(A020639(n))-1).

Extensions

Example corrected by Nadia Heninger, Jul 06 2005

A306737 Irregular triangle where row n is a list of indices in A002110 with multiplicity whose product is A002182(n).

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 3, 1, 1, 3, 2, 3, 1, 1, 1, 3, 1, 2, 3, 1, 1, 2, 3, 1, 1, 4, 2, 4, 1, 1, 1, 4, 1, 2, 4, 1, 1, 2, 4, 2, 2, 4, 1, 1, 1, 2, 4, 1, 2, 2, 4, 1, 1, 1, 1, 2, 4, 1, 1, 3, 4, 1, 2, 5, 2, 2, 2, 4, 1, 1, 1, 3, 4, 1, 1, 2, 5, 2, 2, 5, 1, 1, 1, 2, 5, 1, 2, 2, 5, 1, 1, 1, 1, 2, 5
Offset: 1

Views

Author

Michael De Vlieger, Mar 06 2019

Keywords

Comments

Each highly composite number A002182(n) can be expressed as a product of primorials in A002110.
Row 1 = {0} by convention.
Maximum value in row n is given by A001221(A002182(n)).
Row n in reverse order is the conjugate of A067255(A002182(n)), a list of the multiplicities of the prime divisors of A002182(n).

Examples

			Terms in the first rows n of this sequence, followed by the corresponding primorials whose product = A002182(n):
   n    T(n,k)       A002110(T(n,k))   A002182(n)
  -----------------------------------------------
   1:   0;              1                =     1
   2:   1;              2                =     2
   3:   1, 1;           2 * 2            =     4
   4:   2;              6                =     6
   5:   1, 2;           2 * 6            =    12
   6:   1, 1, 2;        2 * 2 * 6        =    24
   7:   2, 2;           6 * 6            =    36
   8:   1, 1, 1, 2;     2 * 2 * 2 * 6    =    48
   9:   1, 3;           2 * 30           =    60
  10:   1, 1, 3;        2 * 2 * 30       =   120
  11:   2, 3;           6 * 30           =   180
  12:   1, 1, 1, 3;     2 * 2 * 2 * 30   =   240
  13:   1, 2, 3;        2 * 6 * 30       =   360
  14:   1, 1, 2, 3;     2 * 2 * 6 * 30   =   720
  15:   1, 1, 4;        2 * 2 * 210      =   840
  ...
Row 6 = {1,1,2} since A002110(1)*A002110(1)*A002110(2) = 2*2*6 = 24 and A002182(6) = 24. The conjugate of {2,1,1} = {3,1} and 24 = 2^3 * 3^1.
Row 10 = {1,1,3} since A002110(1)*A002110(1)*A002110(3) = 2*2*30 = 120 and A002182(10) = 120. The conjugate of {3,1,1} = {3,1,1} and 120 = 2^3 * 3^1 * 5^1.

Links

Crossrefs

Cf. A001221, A002110, A002182, A067255, A304886.

Programs

  • Mathematica
    With[{s = DivisorSigma[0, Range[250000]]}, Map[Reverse@ Table[LengthWhile[#, # >= i &], {i, Max@ #}] &@ If[# == 1, {0}, Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> Last@ # &, f]]@ FactorInteger@ #] &@ FirstPosition[s, #][[1]] &, Union@ FoldList[Max, s]] /. {} -> {0}] // Flatten

A325236 Squarefree k such that phi(k)/k - 1/2 is positive and minimal for k with gpf(k) = prime(n).

Original entry on oeis.org

1, 2, 3, 15, 21, 231, 273, 255, 285, 167739, 56751695, 7599867, 3829070245, 567641679, 510795753, 39169969059, 704463969, 3717740976339, 42917990271, 547701649495, 45484457928390429, 59701280265935165
Offset: 0

Views

Author

Michael De Vlieger, Apr 19 2019

Keywords

Comments

Let gpf(k) = A006530(k) and let phi(n) = A000010(n) for k in A005117.
There are 2^(n-1) numbers k with gpf(k) = prime(n), since we can only either have p_i^0 or p_i^1 where p_i | k and i <= n. For example, for n = 2, there are only 2 squarefree numbers k with prime(2) = 3 as greatest prime factor. These are 3 = 2^0 * 3^1, and 6 = 2^1 * 3^1. We observe that we can write multiplicities of the primes as A067255(k), and thus for the example derive 3 = "0,1" and 6 = "1,1". Thus for n = 3, we have 5 = "0,0,1", 15 = "0,1,1", 10 = "1,0,1", and 30 = "1,1,1". This establishes the possible values of k with respect to n. We choose the value of k in n for which phi(k)/k - 1/2 is positive and minimal.
We know that prime k (in A000040) have phi(k)/k = A006093(n)/A000040(n) and represent maxima in n. We likewise know primorials k (in A002110) have phi(k)/k = A038110(n)/A060753(n) and represent minima in n. This sequence shows squarefree numbers k with gpf(k) = n such that their value phi(k)/k is closest to but more than 1/2.
Apart from a(1) = 2, all terms are odd. For n > 1 and k even, phi(k)/k - 1/2 is negative.

Examples

			First terms of this sequence appear in the chart below between asterisks.
The values of n appear in the header, values of k followed parenthetically by phi(k)/k appear in column n. The x axis plots k according to primepi(gpf(k)), while the y axis plots k according to phi(k)/k:
    0       1          2             3             4
    .       .          .             .             .
-- *1* -----------------------------------------------
  (1/1)     .          .             .             .
    .       .          .             .             .
    .       .          .             .             .
    .       .          .             .             7
    .       .          .             5           (6/7)
    .       .          .           (4/5)           .
    .       .          .             .             .
    .       .          .             .            35
    .       .         *3*            .          (24/35)
    .       .        (2/3)           .             .
    .       .          .             .             .
    .       .          .             .             .
    .       .          .             .           *21*
    .       .          .             .           (4/7)
    .       .          .           *15*            .
    .       .          .          (8/15)           .
    .      *2*         .             .             .
----------(1/2)---------------------------------------
    .       .          .             .             .
    .       .          .             .            105
    .       .          .             .          (16/35)
    .       .          .             .            14
    .       .          .            10           (3/7)
    .       .          .           (2/5)           .
    .       .          .             .             .
    .       .          .             .            70
    .       .          6             .          (12/35)
    .       .        (1/3)           .             .
    .       .          .             .            42
    .       .          .            30           (2/7)
    .       .          .          (4/15)           .
    .       .          .             .            210
    .       .          .             .           (8/35)
...
a(3) = 15 for the following reasons. There are 4 possible values of k with n = 3. These are 5, 15, 10, and 30 with phi(k)/k = 4/5, 8/15, 2/5, and 4/15, respectively. Subtracting 1/2 from each of the latter values, we derive 3/10, 1/30, -1/10, and -7/30 respectively. Since the smallest of these differences is 3/10 pertaining to k = 15, a(3) = 15.

Links

Crossrefs

Cf. A000010, A000040, A002110, A005117, A006093, A006530, A038110, A060753, A307540, A325237.

Programs

  • Mathematica
    With[{e = 15}, Map[MinimalBy[#, If[# < 0, # + 1, #] &[#[[2]] - 1/2] &] &, SplitBy[#, Last]] &@ Array[{#2, EulerPhi[#2]/#2, If[! IntegerQ@ #, 0, #] &[1 + Floor@ Log2@ #1]} & @@ {#, Times @@ MapIndexed[Prime[First@ #2]^#1 &, Reverse@ IntegerDigits[#, 2]]} &, 2^(e + 1), 0]][[All, 1, 1]]

A325237 Squarefree k such that 1/2 - phi(k)/k is positive and minimal for k with gpf(k) = prime(n).

Original entry on oeis.org

2, 6, 10, 105, 165, 195, 4641, 5187, 5313, 266133, 8870433, 3068957045, 11063481, 10164297, 667797009, 909411789, 32221169781185, 1963007211216415, 421522466365, 3012887561310445
Offset: 1

Views

Author

Michael De Vlieger, Apr 19 2019

Keywords

Comments

Let gpf(k) = A006530(k) and let phi(n) = A000010(n) for k in A005117. There are 2^(n-1) numbers k with gpf(k) = n, since we can only either have p_i^0 or p_i^1 where p_i | k and i <= n. For example, for n = 2, there are only 2 squarefree numbers k with prime(2) = 3 as greatest prime factor. These are 3 = 2^0 * 3^1, and 6 = 2^1 * 3^1. We observe that we can write multiplicities of the primes as A067255(k), and thus for the example derive 3 = "0,1" and 6 = "1,1". Thus for n = 3, we have 5 = "0,0,1", 15 = "0,1,1", 10 = "1,0,1", and 30 = "1,1,1". This establishes the possible values of k with respect to n. We choose the value of k in n for which 1/2 - phi(k)/k is positive and minimal.
We know that prime k (in A000040) have phi(k)/k = A006093(n)/A000040(n) and represent maxima in n. We likewise know primorials k (in A002110) have phi(k)/k = A038110(n)/A060753(n) and represent minima in n. This sequence shows squarefree numbers k with gpf(k) = n such that their value phi(k)/k is closest to but less than 1/2.
Conjecture: for n > 3, k is always odd. This assertion is reliant upon phi(2 prime(n))/2 prime(n) = phi(2)/2 * phi(prime(n))/prime(n) = 1/2 * (prime(n) - 1)/prime(n), and it is clear that 1/2 is an asymptote for even k.

Examples

			First terms of this sequence appear in the chart below between asterisks.
The values of n appear in the header, values of k followed parenthetically by phi(k)/k appear in column n. The x axis plots k according to primepi(gpf(k)), while the y axis plots k according to phi(k)/k:
    0       1          2             3             4
    .       .          .             .             .
--- 1 ------------------------------------------------
  (1/1)     .          .             .             .
    .       .          .             .             .
    .       .          .             .             .
    .       .          .             .             7
    .       .          .             5           (6/7)
    .       .          .           (4/5)           .
    .       .          .             .             .
    .       .          .             .            35
    .       .          3             .          (24/35)
    .       .        (2/3)           .             .
    .       .          .             .             .
    .       .          .             .             .
    .       .          .             .            21
    .       .          .             .           (4/7)
    .       .          .            15             .
    .       .          .          (8/15)           .
    .      *2*         .             .             .
----------(1/2)---------------------------------------
    .       .          .             .             .
    .       .          .             .           *105*
    .       .          .             .          (16/35)
    .       .          .             .            14
    .       .          .           *10*          (3/7)
    .       .          .           (2/5)           .
    .       .          .             .             .
    .       .          .             .            70
    .       .         *6*            .          (12/35)
    .       .        (1/3)           .             .
    .       .          .             .            42
    .       .          .            30           (2/7)
    .       .          .          (4/15)           .
    .       .          .             .            210
    .       .          .             .           (8/35)
...
a(3) = 10 for the following reasons. There are 4 possible values of k with n = 3. These are 5, 15, 10, and 30 with phi(k)/k = 4/5, 8/15, 2/5, and 4/15, respectively. Subtracting each of the latter values from 1/2, we derive -3/10, -1/30, 1/10, and 7/30 respectively. Since the smallest of these differences is 1/10 pertaining to k = 10, a(3) = 10.

Links

Crossrefs

Cf. A000010, A000040, A002110, A005117, A006093, A006530, A038110, A060753, A307540, A325236.

Programs

  • Mathematica
    With[{e = 20}, Map[MinimalBy[#, If[# > 0, # + 1, Abs@ #] &[#[[2]] - 1/2] &] &, SplitBy[#, Last]] &@ Array[{#2, EulerPhi[#2]/#2, If[! IntegerQ@ #, 0, #] &[1 + Floor@ Log2@ #1]} & @@ {#, Times @@ MapIndexed[Prime[First@ #2]^#1 &, Reverse@ IntegerDigits[#, 2]]} &, 2^e - 1]][[All, 1, 1]]
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