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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A243103 Product of numbers m with 2 <= m <= n whose prime divisors all divide n.

Original entry on oeis.org

1, 2, 3, 8, 5, 144, 7, 64, 27, 3200, 11, 124416, 13, 6272, 2025, 1024, 17, 35831808, 19, 1024000, 3969, 247808, 23, 859963392, 125, 346112, 729, 2809856, 29, 261213880320000000, 31, 32768, 264627, 18939904, 30625, 26748301344768, 37, 23658496, 369603, 32768000000, 41
Offset: 1

Views

Author

Michael De Vlieger, Aug 19 2014

Keywords

Comments

This sequence is the product of n-regular numbers.
A number m is said to be "regular" to n or "n-regular" if all the prime factors p of m also divide n.
The divisor is a special case of a regular m such that m also divides n in addition to all of its prime factors p | n.
Analogous to A007955 (Product of divisors of n).
If n is 1 or prime, a(n) = n.
If n is a prime power, a(n) = A007955(n).
Note: b-file ends at n = 4619, because a(4620) has more than 1000 decimal digits.
Product of the numbers 1 <= k <= n such that (floor(n^k/k) - floor((n^k - 1)/k)) = 1. - Michael De Vlieger, May 26 2016

Examples

			a(12) = 124416 since 1 * 2 * 3 * 4 * 6 * 8 * 9 * 12 = 124416. These numbers are products of prime factors that are the distinct prime divisors of 12 = {2, 3}.
From _David A. Corneth_, Feb 09 2015: (Start)
Let p# be the product of primes up to p, A002110. Then
a(13#) ~= 8.3069582 * 10 ^ 4133
a(17#) ~= 1.3953000 * 10 ^ 22689
a(19#) ~= 3.8258936 * 10 ^ 117373
a(23#) ~= 6.7960327 * 10 ^ 594048
a(29#) ~= 1.3276817 * 10 ^ 2983168
a(31#) ~= 2.8152792 * 10 ^ 14493041
a(37#) ~= 1.9753840 * 10 ^ 69927040
Up to n = 11# already in the table.
(End)
		

Crossrefs

Cf. A162306 (irregular triangle of regular numbers of n), A010846 (number of regular numbers of n), A244974 (sum of regular numbers of n), A007955, A244052 (record transform of regular numbers of n).

Programs

  • Maple
    A:= proc(n) local F, S, s, j, p;
      F:= numtheory:-factorset(n);
      S:= {1};
      for p in F do
        S:= {seq(seq(s*p^j, j=0..floor(log[p](n/s))), s=S)}
      od;
      convert(S,`*`)
    end proc:
    seq(A(n), n=1..100); # Robert Israel, Feb 09 2015
  • Mathematica
    regularQ[m_Integer, n_Integer] := Module[{omega = First /@ FactorInteger @ m }, If[Length[Select[omega, Divisible[n, #] &]] == Length[omega], True, False]]; a20140819[n_Integer] := Times @@ Flatten[Position[Thread[regularQ[Range[1, n], n]], True]]; a20140819 /@ Range[41]
    regulars[n_] := Block[{f, a}, f[x_] := First /@ FactorInteger@ x; a = f[n];{1}~Join~Select[Range@ n, SubsetQ[a, f@ #] &]]; Array[Times @@ regulars@ # &, 12] (* Michael De Vlieger, Feb 09 2015 *)
    Table[Times @@ Select[Range@ n, (Floor[n^#/#] - Floor[(n^# - 1)/#]) == 1 &], {n, 41}] (* Michael De Vlieger, May 26 2016 *)
  • PARI
    lista(nn) = {vf = vector(nn, n, Set(factor(n)[,1])); vector(nn, n, prod(i=1, n, if (setintersect(vf[i], vf[n]) == vf[i], i, 1)));} \\ Michel Marcus, Aug 23 2014
    
  • PARI
    for(n=1, 100, print1(prod(k=1, n, k^(floor(n^k/k) - floor((n^k - 1)/k))),", ")) \\ Indranil Ghosh, Mar 22 2017
    
  • Python
    from sympy import primefactors
    def A243103(n):
        y, pf = 1, set(primefactors(n))
        for m in range(2,n+1):
            if set(primefactors(m)) <= pf:
                y *= m
        return y # Chai Wah Wu, Aug 28 2014
    
  • Scheme
    ;; A naive implementation, code for A123275bi given under A123275:
    (define (A243103 n) (let loop ((k n) (m 1)) (cond ((= 1 k) m) ((= 1 (A123275bi n k)) (loop (- k 1) (* m k))) (else (loop (- k 1) m)))))
    ;; Antti Karttunen, Mar 22 2017

Formula

a(n) = product of terms of n-th row of irregular triangle A162306(n,k).
a(n) = Product_{k=1..n} k^( floor(n^k/k)-floor((n^k -1)/k) ). - Anthony Browne, Jul 06 2016
From Antti Karttunen, Mar 22 2017: (Start)
a(n) = Product_{k=2..n, A123275(n,k)=1} k.
For n >= 1, A046523(a(n)) = A283990(n).
(End)

A244974 Sum of numbers m <= n whose set of prime divisors is a subset of the set of prime divisors of n.

Original entry on oeis.org

1, 3, 4, 7, 6, 16, 8, 15, 13, 30, 12, 45, 14, 36, 33, 31, 18, 79, 20, 66, 41, 64, 24, 103, 31, 70, 40, 80, 30, 235, 32, 63, 84, 114, 73, 198, 38, 120, 92, 163, 42, 310, 44, 140, 130, 132, 48, 246, 57, 213, 108, 154, 54, 300, 97, 217, 116, 150, 60, 600, 62, 156, 180, 127, 109, 540, 68, 246
Offset: 1

Views

Author

Michael De Vlieger, Jul 08 2014

Keywords

Comments

a(n) = A000203(n) when n is prime or a perfect prime power (A000961). This is because all products of the prime divisor p in such numbers produce divisors.
a(n) > A000203(n) when n is composite and not a perfect prime power.

Examples

			For n = 4, A162306(4) = {1, 2, 4} and a(4) = 7.
For n = 5, A162306(5) = {1, 5} and a(5) = 6.
For n = 6, A162306(6) = {1, 2, 3, 4, 6} and a(6) = 16.
		

Crossrefs

a(n) = sum of terms of n-th row of triangle A162306(n,k).

Programs

  • Mathematica
    Table[Total@ Union[{1}, Function[d, Select[Range@ n, Union[d, First /@ FactorInteger@ #] == d &]][First /@ FactorInteger@ n]], {n, 68}] (* or *)
    Table[Sum[k (Floor[n^k/k] - Floor[(n^k - 1)/k]), {k, n}], {n, 68}] (* Michael De Vlieger, May 26 2016 *)
  • PARI
    a(n) = {summ = 0; spn = factor(n)[,1]~; for (m=1, n, spm = factor(m)[,1]~; if (setintersect(spm, spn) == spm, summ += m);); summ;} \\ Michel Marcus, Jul 17 2014

Formula

a(n) = Sum_{k=1..n} k*( floor(n^k/k)-floor((n^k - 1)/k) ). - Anthony Browne, May 25 2016
a(n) = Sum_{j=1..n} Sum_{i=j..gcd(n^j,j)} i. - Wesley Ivan Hurt, Apr 05 2021

A376248 Irregular triangle where row n lists m such that rad(m) | n and bigomega(m) <= bigomega(n), where rad = A007947 and bigomega = A001222.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 2, 4, 1, 5, 1, 2, 3, 4, 6, 9, 1, 7, 1, 2, 4, 8, 1, 3, 9, 1, 2, 4, 5, 10, 25, 1, 11, 1, 2, 3, 4, 6, 8, 9, 12, 18, 27, 1, 13, 1, 2, 4, 7, 14, 49, 1, 3, 5, 9, 15, 25, 1, 2, 4, 8, 16, 1, 17, 1, 2, 3, 4, 6, 8, 9, 12, 18, 27, 1, 19, 1, 2, 4, 5, 8, 10, 20, 25, 50, 125
Offset: 1

Views

Author

Michael De Vlieger, Oct 09 2024

Keywords

Comments

Analogous to A162306 regarding m such that rad(m) | n, but instead of taking m <= n, we take m such that bigomega(m) <= bigomega(n).
Row n is a finite set of products of prime power factors p^k (i.e., p^k | n) such that Sum_{p|n} k <= bigomega(n).
For prime power n = p^k, k >= 0 (i.e., n in A000961), row p^k of this sequence is the same as row p^k of A027750 and A162306. Therefore, for prime p, row p of this sequence is the same as row p of A027750 and A162306: {1, p}.
For n in A024619, row n of this sequence does not match row n of A162306, since the former contains gpf(n)^bigomega(n) = A006530(n)^A001222(n), which is larger than n.

Examples

			Triangle begins:
   n    row n of this sequence:
  -------------------------------------------
   1:   1;
   2:   1,  2;
   3:   1,  3;
   4:   1,  2   4;
   5:   1,  5;
   6:   1,  2,  3,  4,  6,  9;
   7:   1,  7;
   8:   1,  2,  4,  8;
   9:   1,  3,  9;
  10:   1,  2,  4,  5, 10, 25;
  11:   1, 11;
  12:   1,  2,  3,  4,  6,  8, 9, 12, 18, 27;
        ...
Row n = 10 of this sequence, presented according to 2^k, k = 0..bigomega(n) by columns, 5^i, i = 0..bigomega(n) by rows, showing terms m > n with an asterisk. The remaining m and the parenthetic 8 are in row 10 of A162306:
   1   2   4  (8)
   5  10
  25*
Row n = 12 of this sequence, presented according to 2^k, k = 0..bigomega(n) by columns, 3^i, i = 0..bigomega(n) by rows, showing terms m > n with an asterisk. The remaining m are in row 12 of A162306:
   1   2   4   8
   3   6  12
   9  18*
  27*
		

Crossrefs

Programs

  • Mathematica
    Table[Clear[p]; MapIndexed[Set[p[First[#2]], #1] &, FactorInteger[n][[All, 1]]]; k = PrimeOmega[n]; w = PrimeNu[n]; Union@ Map[Times @@ MapIndexed[p[First[#2]]^#1 &, #] &, Select[Tuples[Range[0, k], w], Total[#] <= k &] ], {n, 120}]

Formula

Row n of this sequence is { m : rad(m) | n, bigomega(m) <= bigomega(n) }.
A376567(n) = binomial(bigomega(n) + omega(n)) = Length of row n, where omega = A001221.

A286563 Triangular table T(n,k) read by rows: T(n,1) = 1, and for 1 < k <= n, T(n,k) = the highest exponent e such that k^e divides n.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 2, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 3, 0, 1, 0, 0, 0, 1, 1, 0, 2, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 1

Views

Author

Antti Karttunen, May 20 2017

Keywords

Comments

T(n,k) > 0 for k in row n of A027750. - Michael De Vlieger, May 20 2017
Compare rows to those of triangle A279907, smallest exponent e of n divisible by k. The values of k > -1 in row n of A279907 pertain to k in row n of A162306 rather than k in row n of A027750. - Michael De Vlieger, May 21 2017

Examples

			The first fifteen rows of this triangular table:
  1,
  1, 1,
  1, 0, 1,
  1, 2, 0, 1,
  1, 0, 0, 0, 1,
  1, 1, 1, 0, 0, 1,
  1, 0, 0, 0, 0, 0, 1,
  1, 3, 0, 1, 0, 0, 0, 1,
  1, 0, 2, 0, 0, 0, 0, 0, 1,
  1, 1, 0, 0, 1, 0, 0, 0, 0, 1,
  1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
  1, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1,
  1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
  1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1,
  1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
		

Crossrefs

Lower triangular region of A286561.
Cf. A286564 (same triangle reversed).
Cf. A169594 (row sums).
Cf. also arrays A051731, A286158, A027750, A279907, A280269.

Programs

  • Maple
    T := (n, k) -> ifelse(k = 1, 1, padic:-ordp(n, k)):
    for n from 1 to 12 do seq(T(n, k), k = 1..n) od;  # Peter Luschny, Apr 07 2025
  • Mathematica
    Table[If[k == 1, 1, IntegerExponent[n, k]], {n, 15}, {k, n}] // Flatten (* Michael De Vlieger, May 20 2017 *)
  • Python
    def T(n, k):
        i=1
        if k==1: return 1
        while n%(k**i)==0:
            i+=1
        return i-1
    for n in range(1, 21): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, May 20 2017
  • Scheme
    (define (A286563 n) (A286561bi (A002024 n) (A002260 n))) ;; For A286561bi see A286561.
    

Formula

T(n,k) = A286561(n,k) listed row by row for n >= 1, k = 1 .. n.

A138109 Positive integers k whose smallest prime factor is greater than the cube root of k and strictly less than the square root of k.

Original entry on oeis.org

6, 15, 21, 35, 55, 65, 77, 85, 91, 95, 115, 119, 133, 143, 161, 187, 203, 209, 217, 221, 247, 253, 259, 287, 299, 301, 319, 323, 329, 341, 377, 391, 403, 407, 437, 451, 473, 481, 493, 517, 527, 533, 551, 559, 583, 589, 611, 629, 649, 667, 671, 689, 697, 703
Offset: 1

Views

Author

David S. Newman, May 04 2008

Keywords

Comments

This sequence was suggested by Moshe Shmuel Newman.
A020639(n)^2 < a(n) < A020639(n)^3. - Reinhard Zumkeller, Dec 17 2014
In other words, k = p*q with primes p, q satisfying p < q < p^2. - Charles R Greathouse IV, Apr 03 2017
If "strictly less than" in the definition were changed to "less than or equal to" then this sequence would also include the squares of primes (A001248), resulting in A251728. - Jon E. Schoenfield, Dec 27 2022

Examples

			6 is a term because the smallest prime factor of 6 is 2 and 6^(1/3) = 1.817... < 2 < 2.449... = sqrt(6).
From _Michael De Vlieger_, Apr 27 2024: (Start):
Table of p*q where p = prime(n) and q = prime(n+k):
n\k   1     2     3     4     5     6     7     8     9    10    11
-------------------------------------------------------------------
1:    6;
2:   15,   21;
3:   35,   55,   65,   85,   95,  115;
4:   77,   91,  119,  133,  161,  203,  217,  259,  287,  301,  329;
     ... (End)
		

Crossrefs

Subsequence of A251728 and of A006881. A006094 is a proper subset.

Programs

  • Haskell
    a138109 n = a138109_list !! (n-1)
    a138109_list = filter f [1..] where
       f x = p ^ 2 < x && x < p ^ 3 where p = a020639 x
    -- Reinhard Zumkeller, Dec 17 2014
    
  • Mathematica
    s = {}; Do[f = FactorInteger[i]; test = f[[1]][[1]]; If [test < N[i^(1/2)] && test > N[i^(1/3)], s = Union[s, {i}]], {i, 2, 2000}]; Print[s]
    Select[Range[1000],Surd[#,3]Harvey P. Dale, May 10 2015 *)
  • PARI
    is(n)=my(f=factor(n)); f[,2]==[1,1]~ && f[1,1]^3 > n \\ Charles R Greathouse IV, Mar 28 2017
    
  • PARI
    list(lim)=if(lim<6, return([])); my(v=List([6])); forprime(p=3,sqrtint(1+lim\=1)-1, forprime(q=p+2, min(p^2-2,lim\p), listput(v,p*q))); Set(v) \\ Charles R Greathouse IV, Mar 28 2017
    
  • Python
    from math import isqrt
    from sympy import primepi, primerange
    def A138109(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return int(n+x+(t:=primepi(s:=isqrt(x)))+(t*(t-1)>>1)-sum(primepi(min(x//p,p**2)) for p in primerange(s+1)))
        return bisection(f,n,n) # Chai Wah Wu, Mar 05 2025

Formula

From Michael De Vlieger, Apr 27 2024: (Start)
Let k = a(n); row k of A162306 = {1, p, q, p^2, p*q}, therefore A010846(k) = 5.
A079047(n) = card({ q : p < q < p^2 }), p and q primes. (End)

A361373 Number of prime powers p^m <= n such that p | n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 3, 2, 4, 1, 5, 1, 4, 3, 4, 1, 6, 1, 5, 3, 5, 1, 6, 2, 5, 3, 5, 1, 9, 1, 5, 4, 6, 3, 8, 1, 6, 4, 7, 1, 9, 1, 6, 5, 6, 1, 8, 2, 7, 4, 6, 1, 8, 3, 7, 4, 6, 1, 10, 1, 6, 5, 6, 3, 10, 1, 7, 4, 10, 1, 9, 1, 7, 5, 7, 3, 10, 1, 8, 4, 7, 1, 12, 3, 7
Offset: 1

Views

Author

Michael De Vlieger, Jun 17 2024

Keywords

Comments

Let p be prime. The term "prime power" p^m, m > 0, used here is that of A246655 = A000040 U A246547, the union of primes and perfect prime powers. Essentially, 1 is not considered a prime power.

Examples

			Let S = {k <= n : rad(k) | n} = row n of A162306
a(1) = 0 since S = {1} has 0 prime powers.
a(2) = 1 since S = {1, [2]} has 1 prime power.
a(4) = 2 since S = {1, [2, 4]} has 2 prime powers.
a(6) = 3 since S = {1, [2, 3, 4], 6} has 3 prime powers.
a(10) = 4 since S = {1, [2, 4, 5, 8], 10} has 4 prime powers.
a(12) = 5 since S = {1, [2, 3, 4], 6, [8, 9], 12} has 5 prime powers, etc.
		

Crossrefs

Programs

  • Maple
    a := n -> add(ilog[p](n), p in NumberTheory:-PrimeFactors(n)):
    seq(a(n), n = 1..92); # Peter Luschny, Jun 20 2024
  • Mathematica
    {0}~Join~Table[Total@ Map[Floor@ Log[#, n] &, FactorInteger[n][[All, 1]]], {n, 2, 120}]
  • PARI
    a(n) = if (n==1, 0, my(f=factor(n)[,1]); sum(k=1, #f, logint(n, f[k]))); \\ Michel Marcus, Jun 20 2024
    
  • Python
    from sympy import integer_log, primefactors
    def A361373(n): return sum(integer_log(n,p)[0] for p in primefactors(n)) # Chai Wah Wu, Sep 20 2024

Formula

a(n) = Sum_{p | n} floor(log n / log p).
a(n) = number of prime powers in row n of A162306.
a(n) < A000005(n), since A000005 counts 1.
a(n) < A010846(n), since A010846 counts 1.
Let tau = A000005, rad = A007947, rcf = A010846, and lpf = A020639.
a(p) = tau(p) - 1 = rcf(p) - 1 = 1 since S = row p of both A027750 and A162306 = {1, p} contains the prime power p.
a(p^m) = tau(p^m) - 1 = rcf(p^m) = 1 = m since S = row p^m of both A027750 and A162306 = {1, p, p^2, ..., p^m} contains the prime powers {p, p^2, ..., p^m}.
a(k) = tau(k) - 1 = 3 for squarefree composite k = p*q, p < q < p^2 in A138109 since S = row k of A162306 = {1, p, q, p^2, p*q} contains 3 prime powers {p, q, p^2}.
a(k) < tau(k) for k in A138511 and k in A126706 since m = lpf(k)^(-1 + floor(log k / log lpf(k))) is such that m < k but m does not divide k.

A381801 Irregular triangle read by rows: row n lists the residues r mod n of numbers k such that rad(k) | n, where rad = A007947.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 3, 4, 0, 1, 0, 1, 2, 4, 0, 1, 3, 0, 1, 2, 4, 5, 6, 8, 0, 1, 0, 1, 2, 3, 4, 6, 8, 9, 0, 1, 0, 1, 2, 4, 7, 8, 0, 1, 3, 5, 6, 9, 10, 12, 0, 1, 2, 4, 8, 0, 1, 0, 1, 2, 3, 4, 6, 8, 9, 10, 12, 14, 16, 0, 1, 0, 1, 2, 4, 5, 8, 10, 12, 16
Offset: 1

Views

Author

Michael De Vlieger, Mar 07 2025

Keywords

Comments

Define S(p,n) to be the set of residues r (mod n) taken by the power range of prime divisor p, i.e., {p^m, m >= 1}.
Define T(n) to be the union of the tensor product of distinct terms in S(p,n) for all p|n, where the products are expressed mod n.
Row n of this triangle is T(n), a superset of row n of A381799.
For n > 1, the intersection of row n of this triangle and row n of A038566 is {1}.

Examples

			Table of c(n) = A381800(n) and T(n) for select n:
 n  c(n)  T(n)
-----------------------------------------------------------------------------
 1    1   {0}
 2    2   {0, 1}
 3    2   {0, 1}
 4    3   {0, 1, 2}
 5    2   {0, 1}
 6    5   {0, 1, 2, 3, 4}
 8    4   {0, 1, 2, 4}
 9    3   {0, 1, 3}
10    7   {0, 1, 2, 4, 5, 6, 8}
11    2   {0, 1}
12    8   {0, 1, 2, 3, 4, 6, 8, 9}
14    6   {0, 1, 2, 4, 7, 8}
15    8   {0, 1, 3, 5, 6, 9, 10, 12}
16    5   {0, 1, 2, 4, 8}
18   12   {0, 1, 2, 3, 4, 6, 8, 9, 10, 12, 14, 16}
20    9   {0, 1, 2, 4, 5, 8, 10, 12, 16}
24   11   {0, 1, 2, 3, 4, 6, 8, 9, 12, 16, 18}
28    9   {0, 1, 2, 4, 7, 8, 14, 16, 21}
30   19   {0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 21, 24, 25, 27}
36   16   {0, 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 20, 24, 27, 28, 32}
For n = 10, we have S(2,10) = {1, 2, 4, 6, 8} and S(5,10) = {1, 5}. Therefore we have the following distinct products:
   1  2  4  8  6
   5  0
Hence T(10) = {0, 1, 2, 4, 5, 6, 8}; terms in A003592 belong to these residues (mod 10).
For n = 12, we have S(2,12) = {1, 2, 4, 8} and S(3,12) = {1, 3, 9}. Therefore we have the following distinct products:
   1  2  4  8
   3  6  0
   9
Thus T(12) = {0, 1, 2, 3, 4, 6, 8, 9}, terms in A003586 belong to these residues (mod 12).
For n = 30, we have {1, 2, 4, 8, 16}, {1, 3, 9, 21, 27}, and {1, 5, 25}. Therefore we have the following distinct products:
   1  2  4  8  16         5  10  20         25
   3  6 12 24            15   0
   9 18
  27
Thus T(30) = {0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 21, 24, 25, 27}; terms in A051037 belong to these residues (mod 30).
		

Crossrefs

Programs

  • Mathematica
    Table[Union@ Flatten@ Mod[TensorProduct @@ Map[(p = #; NestWhileList[Mod[p*#, n] &, 1, UnsameQ, All]) &, FactorInteger[n][[All, 1]] ], n], {n, 30}]

Formula

Row 1 is {0} since 1 is the empty product and the only number that has zero prime factors is 1, congruent to 0 (mod 1).
Row n begins with {0,1} for n > 1.
For prime p, row p = {0,1}.
For prime power p^m, m > 0, row p = union of {0} and {p^i, i < m}.
Row n is a subset of row n of A121998, considering n in A121998 instead as n mod n = 0.
Row n is a superset of row n of A162306, considering n in A162306 instead as n mod n = 0.

A299991 Numbers n for which A243822(n) > A000005(n).

Original entry on oeis.org

30, 42, 60, 66, 70, 74, 78, 82, 84, 86, 90, 94, 98, 102, 106, 110, 114, 118, 120, 122, 126, 130, 132, 134, 138, 140, 142, 146, 150, 154, 156, 158, 162, 165, 166, 168, 170, 174, 178, 180, 182, 186, 190, 194, 195, 198, 202, 204, 206, 210, 214, 218, 220, 222, 226
Offset: 1

Views

Author

Michael De Vlieger, Feb 25 2018

Keywords

Comments

Composite numbers m have nondivisors k in the cototient such that k | n^e with e > 1. These k appear in row n of A272618 and are enumerated by A243822(n). These nondivisors k are a kind of "regular" number along with divisors d of n; both are listed in row n of A162306 and are together enumerated by A010846(n).
This sequence lists numbers that have more nondivisors k in the cototient of n than divisors d.
This sequence contains all n for which A299990(n) is positive.
The smallest odd term is 165.
For m >= 3, A002110(m) is in a(n).
For m >= 9, A244052(m) is in a(n).

Examples

			30 is the first term since it is the smallest number for which A243822(n) > A000005(n), alternatively, for which A010846(n) > 2*A000005(n).
		

Crossrefs

Programs

  • Mathematica
    Select[Range@ 226, Function[n, Count[Range[n], _?(PowerMod[n, Floor@ Log2@ n, #] == 0 &)] > 2 DivisorSigma[0, n]]]

A372720 a(n) = A000005(n) - A008479(n).

Original entry on oeis.org

0, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 4, 1, 3, 3, 1, 1, 3, 1, 4, 3, 3, 1, 4, 1, 3, 1, 4, 1, 7, 1, 1, 3, 3, 3, 4, 1, 3, 3, 5, 1, 7, 1, 4, 4, 3, 1, 4, 1, 2, 3, 4, 1, 1, 3, 5, 3, 3, 1, 10, 1, 3, 4, 1, 3, 7, 1, 4, 3, 7, 1, 4, 1, 3, 3, 4, 3, 7, 1, 5, 1, 3, 1, 10, 3, 3, 3
Offset: 1

Views

Author

Michael De Vlieger, May 13 2024

Keywords

Comments

A095960(50) = 3, a(50) = 2.
a(162) = -2 is the first negative term.

Examples

			Table of a(n), b(n) = A000005(n), and c(n) = A008479(n) for n <= 12:
  n  b(n) c(n) a(n)
 ------------------
  1    1    1    0
  2    2    1    1
  3    2    1    1
  4    3    2    1
  5    2    1    1
  6    4    1    3
  7    2    1    1
  8    4    3    1
  9    3    2    1
 10    4    1    3
 11    2    1    1
 12    6    2    4
a(12) = 4 since 12 has 6 divisors {1, 2, 3, 4, 6, 12}, and row 12 of A369609 has 2 terms {6, 12}.
a(18) = 3 since 18 has 6 divisors {1, 2, 3, 6, 9, 18}, and row 18 of A369609 has 3 terms {6, 12, 18}.
a(50) = 2 since 50 has 6 divisors {1, 2, 5, 10, 25, 50}, and row 50 of A369609 has 4 terms {10, 20, 40, 50}
a(162) = -2 since 162 has 10 divisors {1,2,3,6,9,18,27,54,81,162} but row 162 of A369609 has 12 terms {6,12,18,24,36,48,54,72,96,108,144,162}.
a(500) = 0 since 500 has as many divisors {1,2,4,5,10,20,25,50,100,125,250,500} as terms in row 500 of A369609 {10,20,40,50,80,100,160,200,250,320,400,500}.
		

Crossrefs

Programs

  • Mathematica
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]]; Table[r = rad[n]; DivisorSigma[0, n] - Count[Range[n/r], _?(Divisible[r, rad[#]] &)], {n, 120}]
  • PARI
    a(n) = my(f=factor(n)[, 1], s); forvec(v=vector(#f, i, [1, logint(n, f[i])]), if(prod(i=1, #f, f[i]^v[i])<=n, s++)); numdiv(n) - s; \\ after A008479 \\ Michel Marcus, Jun 03 2024

Formula

a(n) = A095960(n) for n in A303554, i.e., for squarefree n or prime powers n.
a(n) = A095960(n) for n in A360767, i.e., for nonsquarefree composite n such that omega(n) > 1 and A003557(n) < A119288(n), since A008479(n) is the number of terms k in row n of A010846 such that k <= A003557(n).
a(n) = A183093(n) - A355432(n).

A376846 Number of m <= n such that rad(m) | n and Omega(m) > Omega(n), where rad = A007947 and Omega = A001222.

Original entry on oeis.org

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

Views

Author

Michael De Vlieger, Oct 06 2024

Keywords

Comments

Number of m not exceeding n such that the squarefree kernel of m divides n, and m has more prime factors with repetition than does n.
Number of m in row n of A162306 such that Omega(m) > Omega(n).

Examples

			Table of select n such that a(n) > 0:
   n  a(n)  List of m such that Omega(m) > Omega(n).
  -------------------------------------------------
  10   1    {8}
  14   1    {8}
  18   1    {16}
  20   1    {16}
  22   2    {8, 16}
  26   2    {8, 16}
  28   1    {16}
  30   2    {16, 24}
  33   1    {27}
  34   3    {8, 16, 32}
  36   1    {32}
  38   3    {8, 16, 32}
  39   1    {27}
  40   1    {32}
  42   4    {16, 24, 32, 36}
		

Crossrefs

Programs

  • Mathematica
    rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    {0}~Join~Table[With[{k = PrimeOmega[n]}, Count[Range[n], _?(And[Divisible[n, rad[#]], PrimeOmega[#] > k] &)]], {n, 2, 120}]

Formula

a(n) = card({m <= n : rad(m) | n, Omega(m) > Omega(n) }).
a(n) = 0 for prime power n (in A000961).
a(n) < A010846(n).
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