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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A319676 Numerator of A047994(n)/n where A047994 is the unitary totient function.

Original entry on oeis.org

1, 1, 2, 3, 4, 1, 6, 7, 8, 2, 10, 1, 12, 3, 8, 15, 16, 4, 18, 3, 4, 5, 22, 7, 24, 6, 26, 9, 28, 4, 30, 31, 20, 8, 24, 2, 36, 9, 8, 7, 40, 2, 42, 15, 32, 11, 46, 5, 48, 12, 32, 9, 52, 13, 8, 3, 12, 14, 58, 2, 60, 15, 16, 63, 48, 10, 66, 12, 44, 12, 70, 7, 72, 18, 16
Offset: 1

Views

Author

Michel Marcus, Sep 26 2018

Keywords

Crossrefs

Programs

  • Mathematica
    uphi[n_] := Product[{p, e} = pe; p^e - 1, {pe, FactorInteger[n]}];
    a[n_] := If[n == 1, 1, Numerator[uphi[n]/n]];
    Array[a, 100] (* Jean-François Alcover, Jan 10 2022 *)
  • PARI
    a(n)=my(f=factor(n)~); numerator(prod(i=1, #f, f[1, i]^f[2, i]-1)/n);

Formula

a(p) = p-1, for p prime (see A006093).
a(A002110(n)) = A038110(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/A319677(k) = Product_{p prime} (1 - 1/(p*(p+1))) = 0.7044422... (A065463). - Amiram Eldar, Nov 21 2022

A378720 a(n) is the numerator of the asymptotic density of numbers whose third smallest prime divisor is prime(n).

Original entry on oeis.org

0, 0, 1, 1, 4, 326, 628, 992, 98304, 125568, 733440, 281163264, 386427322368, 3178249003008, 12454223855616, 6450728943845376, 342348724735967232, 20218431581110665216, 39814891891080560640, 82739188294287768944640, 15336676441718784000, 61298453882755419734016000
Offset: 1

Views

Author

Robert G. Wilson v and Amiram Eldar, Dec 05 2024

Keywords

Comments

The third smallest prime divisor of a number k is the third member in the ordered list of the distinct prime divisors of k. Only numbers in A000977 have a third smallest prime divisor.
The partial sums of the fractions first exceed 1/2 after summing 4467 terms. Therefore, the median value of the distribution of the third prime divisor is prime(4467) = 42719 = A284411(3).

Examples

			The fractions begin with 0/1, 0/1, 1/30, 1/30, 4/165, 326/15015, 628/36465, 992/62985, 98304/7436429, 125568/11849255, ..., .
a(1) = a(2) = 0 since there are no numbers whose third prime divisor is 2 or 3.
a(3)/A378721(3) = 1/30 since the numbers whose third prime divisor is 5 are the numbers that are divisible by 2, 3 and 5, and their density if (1/2)*(1/3)*(1/5) = 1/30.
a(4)/A378721(4) = 1/30 since the numbers whose third prime divisor is 7 are the union of the numbers that are divisible by 2, 3 and 7 and not by 5 whose density is (1/2)*(1/3)*(1-1/5)*(1/7) = 2/105, the numbers that are divisible by 2, 5 and 7 and not by 3 whose density is (1/2)*(1-1/3)*(1/5)*(1/7) = 1/105, and the numbers that are divisible by 3, 5 and 7 and not by 2 whose density is (1-1/2)*(1/3)*(1/5)*(1/7) = 1/210, and 2/105 + 1/105 + 1/210 = 1/30.
		

References

  • József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, pp. 337-341.

Crossrefs

Programs

  • Mathematica
    a[n_] := Block[{p, q = Prime@ Range@ n}, p = Fold[Times, 1, q]; q = Most@ q; Plus @@ Times @@@ Subsets[q -1, {n -3}]/p]; a[1] = 0; Numerator@ Array[a, 22]
  • PARI
    a(n) = {my(v = primes(n), q = vecextract(apply(x -> x-1, v),"^-1"), p = vecprod(v), prd = vecprod(q)/p, sm = 0, sb); forsubset([#q, 2], s, sb = vecextract(q, s); sm += 1/vecprod(sb)); numerator(prd * sm);}

Formula

a(n)/A378721(n) = (1/prime(n)#) * (Product_{k=1..n-1} (prime(k) - 1)) * Sum_{j=1..n-1, i=1..j-1} 1/((prime(i)-1)*(prime(j)-1)), where prime(n)# = A002110(n) is the n-th primorial number.
Sum_{n>=1} a(n)/A378721(n) = 1.
Sum_{n=1..m} a(n)/A378721(n) > 1/2 for m >= 4467 = primepi(A284411(3)).

A378721 a(n) is the denominator of the asymptotic density of numbers whose third smallest prime divisor is prime(n).

Original entry on oeis.org

1, 1, 30, 30, 165, 15015, 36465, 62985, 7436429, 11849255, 73465381, 33426748355, 50708377254535, 436092044389001, 1863302371480277, 1086305282573001491, 64092011671807087969, 3909612711980232366109, 8449808119441147371913, 18598027670889965365580513, 3543193335582015099413
Offset: 1

Views

Author

Robert G. Wilson v and Amiram Eldar, Dec 05 2024

Keywords

Comments

See A378720 for more details.

Crossrefs

Cf. A000040, A038110, A038111, A342479, A342480, A378720 (numerators).

Programs

  • Mathematica
    a[n_] := Block[{p, q = Prime@ Range@ n}, p = Fold[Times, 1, q]; q = Most@ q; Plus @@ Times @@@ Subsets[q -1, {n -3}]/p]; a[1] = 0; Denominator@ Array[a, 21]
  • PARI
    a(n) = {my(v = primes(n), q = vecextract(apply(x -> x-1, v),"^-1"), p = vecprod(v), prd = vecprod(q)/p, sm = 0, sb); forsubset([#q, 2], s, sb = vecextract(q, s); sm += 1/vecprod(sb)); denominator(prd * sm);}

A076359 a(n) = denominator(n!/phi(n!)).

Original entry on oeis.org

1, 1, 1, 1, 4, 4, 8, 8, 8, 8, 16, 16, 192, 192, 192, 192, 3072, 3072, 55296, 55296, 55296, 55296, 110592, 110592, 110592, 110592, 110592, 110592, 442368, 442368, 13271040, 13271040, 13271040, 13271040, 13271040, 13271040, 477757440
Offset: 1

Views

Author

Labos Elemer, Oct 08 2002

Keywords

Comments

Numerator of Product_{p<=n, p prime} (1 - 1/p). - Franz Vrabec, Jan 28 2014

Crossrefs

Programs

  • Maple
    P:= 1: p:= 1:  v:= 1:
    while p < 100 do q:= nextprime(p);
       for i from p to q-1 do A[i]:= v od;
       P:= P * (1-1/q);
       v:= numer(P);
       p:= q;
    od:
    seq(A[i],i=1..q-1); # Robert Israel, Oct 18 2018
  • Mathematica
    dnf[n_]:=With[{nn=n!},Denominator[nn/EulerPhi[nn]]]; Array[dnf,40] (* Harvey P. Dale, Feb 21 2015 *)
  • PARI
    a(n) = numerator(prod(p=1, n, if (isprime(p),(1-1/p), 1))); \\ Michel Marcus, Jan 28 2014

Formula

a(n) = denominator(A000142(n)/A048855(n)).
a(n) = A038110(A036234(n)). - Robert Israel, Oct 18 2018

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.
		

Crossrefs

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.
		

Crossrefs

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]]

A335334 Sum of the integers in the reduced residue system of A002110(n).

Original entry on oeis.org

1, 6, 120, 5040, 554400, 86486400, 23524300800, 8045310873600, 4070927302041600, 3305592969257779200, 3074201461409734656000, 4094836346597766561792000, 6715531608420337161338880000, 12128250084807128913378017280000
Offset: 1

Views

Author

Jamie Morken, Jun 02 2020

Keywords

Comments

Sum of the integers up to A002110(n) and coprime to A002110(n).
The sequence gives the sum of row n of A286941(n).

Examples

			For n = 3: A002110(3) = 30, the reduced residue system of 30 is {1, 7, 11, 13, 17, 19, 23, 29}. The sum is a(3) = 120.
		

Crossrefs

Programs

  • Mathematica
    n = 15;
    A002110 = Drop[FoldList[Times, 1, Prime[Range[n]]], 1];
    A005867 = Drop[EulerPhi@FoldList[Times, 1, Prime@Range@n], 1];
    A002110*A005867/2
    (* Second program: *)
    Map[# EulerPhi[#]/2 &, FoldList[Times, Prime@ Range@ 14]] (* Michael De Vlieger, Apr 07 2021 *)
  • PARI
    a(n) = my(P=factorback(primes(n))); P*eulerphi(P)/2; \\ Michel Marcus, Jun 02 2020

Formula

a(n) = A023896(A002110(n)).
a(n) = A002110(n)*A005867(n)/2 = A070826(n)*A005867(n).
a(n) = (A002110(n)*A038110(n+1)/2)*A058250(n).

A356093 a(n) = numerator((prime(n)-1)/prime(n)#), where prime(n)# = A002110(n) is the n-th primorial.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 8, 3, 1, 2, 1, 6, 4, 1, 1, 2, 1, 2, 1, 1, 12, 1, 1, 4, 16, 10, 1, 1, 18, 8, 3, 1, 4, 1, 2, 5, 2, 27, 1, 2, 1, 6, 1, 32, 14, 3, 1, 1, 1, 2, 4, 1, 8, 25, 128, 1, 2, 9, 2, 4, 1, 2, 3, 1, 4, 2, 1, 8, 1, 2, 16, 1, 1, 2, 9, 1, 2, 6, 40, 4, 1, 2, 1
Offset: 1

Views

Author

Amiram Eldar, Jul 26 2022

Keywords

Comments

f(n) = a(n)/A356094(n) is the asymptotic density of numbers k such that prime(n) = A053669(k) is the smallest prime not dividing k.
The asymptotic mean of A053669 is 2.9200509773... (A249270) which is the weighted mean of the primes with f(n) as weights. The corresponding asymptotic standard deviation, which can be evaluated from the second raw moment Sum_{n>=1} f(n) * prime(n)^2, is 2.8013781465... .

Examples

			Fractions begin with 1/2, 1/3, 2/15, 1/35, 1/231, 2/5005, 8/255255, 3/1616615, 1/10140585, 2/462120945, ...
		

Crossrefs

Cf. A002110, A006093, A039787, A053669, A249270, A356094 (denominators).
Similar sequences: A038110, A338559, A340818, A341431, A342450, A342479.

Programs

  • Mathematica
    primorial[n_] := Product[Prime[i], {i, 1, n}]; Numerator[Table[(Prime[i] - 1)/primorial[i], {i, 1, 100}]]
  • PARI
    a(n) = numerator((prime(n)-1)/factorback(primes(n))); \\ Michel Marcus, Jul 26 2022
    
  • Python
    from math import gcd
    from sympy import prime, primorial
    def A356093(n): return (p:=prime(n)-1)//gcd(p,primorial(n)) # Chai Wah Wu, Jul 26 2022

Formula

a(n) = 1 iff prime(n) is in A039787.
Let f(n) = a(n)/A356094(n):
f(n) = A006093(n)/A002110(n).
Sum_{n>=1} f(n) = 1.
Sum_{n>=1} f(n) * prime(n) = A249270.

A377734 Number of integers less than n that have the same smallest prime factor as n.

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 3, 1, 4, 0, 5, 0, 6, 2, 7, 0, 8, 0, 9, 3, 10, 0, 11, 1, 12, 4, 13, 0, 14, 0, 15, 5, 16, 2, 17, 0, 18, 6, 19, 0, 20, 0, 21, 7, 22, 0, 23, 1, 24, 8, 25, 0, 26, 3, 27, 9, 28, 0, 29, 0, 30, 10, 31, 4, 32, 0, 33, 11, 34, 0, 35, 0, 36, 12, 37, 2, 38, 0, 39
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 05 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Length[Select[Range[n - 1], If[# == 1, 1, FactorInteger[#][[1, 1]]] == If[n == 1, 1, FactorInteger[n][[1, 1]]] &]], {n, 80}]
    seq[len_] := Module[{t = Table[FactorInteger[n][[1,1]], {n, 1, len}], s = Table[0, {len}]}, Do[s[[i]] = Count[t[[1;;i-1]], t[[i]]], {i, 1, len}]; s]; seq[80] (* Amiram Eldar, Nov 21 2024 *)
  • PARI
    a(n) = if (n>1, my(p=vecmin(factor(n)[,1])); sum(k=2, n-1, p == vecmin(factor(k)[,1])), 0); \\ Michel Marcus, Nov 16 2024

Formula

a(n) = |{j < n : lpf(j) = lpf(n)}|.
a(n) = A078898(n) - 1.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Sum_{k>=1} (A038110(k)/A038111(k))^2 = 0.2847976823663... . - Amiram Eldar, Nov 21 2024

A306353 Number of composites among the first n composite numbers whose least prime factor p is that of the n-th composite number.

Original entry on oeis.org

1, 2, 3, 1, 4, 5, 6, 2, 7, 8, 9, 3, 10, 11, 1, 12, 4, 13, 14, 15, 5, 16, 2, 17, 18, 6, 19, 20, 21, 7, 22, 23, 1, 24, 8, 25, 26, 3, 27, 9, 28, 29, 30, 10, 31, 4, 32, 33, 11, 34, 35, 36, 12, 37, 2, 38, 39, 13, 40, 41, 5, 42, 14, 43, 44, 3, 45, 15, 46, 6, 47, 48, 16, 49, 50, 51, 17, 52, 53, 54, 18, 55, 56, 7
Offset: 1

Views

Author

Jamie Morken and Vincenzo Librandi, Feb 09 2019

Keywords

Comments

Composites with least prime factor p are on that row of A083140 which begins with p
Sequence with similar values: A122005.
Sequence written as a jagged array A with new row when a(n) > a(n+1):
1, 2, 3,
1, 4, 5, 6,
2, 7, 8, 9,
3, 10, 11,
1, 12,
4, 13, 14, 15,
5, 16,
2, 17, 18,
6, 19, 20, 21,
7, 22, 23,
1, 24,
8, 25, 26,
3, 27,
9, 28, 29, 30.
A153196 is the list B of the first values in successive rows with length 4.
B is given by the formula for A002808(x)=A256388(n+3), an(x)=A153196(n+2)
For example: A002808(26)=A256388(3+3), an(26)=A153196(3+2).
A243811 is the list of the second values in successive rows with length 4.
A047845 is the list of values in the second column and A104279 is the list of values in the third column of the jagged array starting on the second row.
Sequence written as an irregular triangle C with new row when a(n)=1:
1,2,3,
1,4,5,6,2,7,8,9,3,10,11,
1,12,4,13,14,15,5,16,2,17,18,6,19,20,21,7,22,23,
1,24,8,25,26,3,27,9,28,29,30,10,31,4,32,33,11,34,35,36,12,37,2,38,39,13,40,41,5,42,14,43,44,3,45,15,46,6,47,48,16,49,50,51,17,52,53,54,18,55,56,7,57,19,58,4,59.
A243887 is the last value in each row of C.
The second value D on the row n > 1 of the irregular triangle C is a(A053683(n)) or equivalently A084921(n). For example for row 3 of the irregular triangle:
D = a(A053683(3)) = a(16) = 12 or D = A084921(3) = 12. This is the number of composites < A066872(3) with the same least prime factor p as the A053683(3) = 16th composite, A066872(3) = 26.
The number of values in each row of the irregular triangle C begins: 3,11,18,57,39,98,61,141,265,104,351,268,...
The second row of the irregular triangle C is A117385(b) for 3 < b < 15.
The third row of the irregular triangle C has similar values as A117385 in different order.

Examples

			First composite 4, least prime factor is 2, first case for 2 so a(1)=1.
Next composite 6, least prime factor is 2, second case for 2 so a(2)=2.
Next composite 8, least prime factor is 2, third case for 2 so a(3)=3.
Next composite 9, least prime factor is 3, first case for 3 so a(4)=1.
Next composite 10, least prime factor is 2, fourth case for 2 so a(5)=4.
		

Crossrefs

Programs

  • Mathematica
    counts = {}
    values = {}
    For[i = 2, i < 130, i = i + 1,
    If[PrimeQ[i], ,
    x = PrimePi[FactorInteger[i][[1, 1]]];
      If[Length[counts] >= x,
       counts[[x]] = counts[[x]] + 1;
       AppendTo[values, counts[[x]]], AppendTo[counts, 1];
       AppendTo[values, 1]]]]
       (* Print[counts] *)
       Print[values]
  • PARI
    c(n) = for(k=0, primepi(n), isprime(n++)&&k--); n; \\ A002808
    a(n) = my(c=c(n), lpf = vecmin(factor(c)[,1]), nb=0); for(k=2, c, if (!isprime(k) && vecmin(factor(k)[,1])==lpf, nb++)); nb; \\ Michel Marcus, Feb 10 2019

Formula

a(n) is approximately equal to A002808(n)*(A038110(x)/A038111(x)), with A000040(x)=A020639(A002808(n)).
For example if n=325, a(325)~= A002808(325)*(A038110(2)/A038111(2)) with A000040(2)=A020639(A002808(325)).
This gives an estimate of 67.499... and the actual value of a(n)=67.
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