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 1-4 of 4 results.

A168262 Intersection of A003418 and A116998.

Original entry on oeis.org

1, 2, 6, 12, 60, 420, 840, 27720, 360360, 5354228880
Offset: 1

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Author

Matthew Vandermast, Nov 23 2009

Keywords

Comments

If, for some prime p, A045948(p) > p^2, then all members of the sequence are less than A003418(p). (Let p_(n) be a prime for which the inequality is satisfied, and let p_(n+1) be the smallest prime > (p_(n))^2. No number smaller than A003418(p_(n+1)) can belong to this sequence. However, for any p_(n) that satisfies the inequality, so does p_(n+1), leading to an endless cycle.) This inequality is first satisfied at p=53, as A045948(53)=5040 > 53^2=2809.
Proof: It follows from the definitions of p_(n) and p_(n+1), and from Bertrand's Postulate, that 2(A045948(p_(n))) > 2((p_(n))^2) > p_(n+1). Therefore 2((A045948(p_(n)))^2 > (p_(n+1))^2.
Since any prime that divides A003418(p_(n)) divides A003418(p_(n+1)) at least twice as often, A045948(p_(n+1)) cannot be less than the product of (A045948(p_n))^2 and A034386(p_(n)). (The latter term greatly exceeds 2 for any actual p_(n).)
Therefore A045948(p_(n+1)) > 2((A045948(p_n))^2 > (p_(n+1))^2, and p_(n+1) satisfies the inequality, implying that no number smaller than A003418(p_(n+2)) can belong to this sequence.

Links

Crossrefs

Also intersection of A003418 and A060735, and of A003418 and A168264. (A168264 is a subsequence of A060735, which is a subsequence of A116998.)
See also A001221, A168263.

A168264 For all sufficiently high values of k, d(n^k) > d(m^k) for all m < n. (Let k, m, and n represent positive integers only.)

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 30, 60, 120, 180, 210, 420, 840, 1260, 1680, 2310, 4620, 9240, 13860, 18480, 27720, 30030, 60060, 120120, 180180, 240240, 360360, 510510, 1021020, 2042040, 3063060, 4084080, 6126120, 9699690, 19399380, 38798760, 58198140
Offset: 1

Views

Author

Matthew Vandermast, Nov 23 2009

Keywords

Comments

d(n) is the number of divisors of n (A000005(n)).

Examples

			Since the exponents in 1680's prime factorization are (4,1,1,1), the k-th power of 1680 has (4k+1)(k+1)^3 = 4k^4 + 13k^3 + 15k^2 + 7k + 1 divisors. Comparison with the analogous formulas for all smaller members of A025487 shows the following:
a) No number smaller than 1680 has a positive coefficient in its "power formula" for any exponent larger than k^4.
b) The only power formula with a k^4 coefficient as high as 4 is that for 1260 (4k^4 + 12k^3 + 13k^2 + 6k + 1).
c) The k^3 coefficient for 1680 is higher than for 1260.
So for all sufficiently high values of k, d(1680^k) > d(m^k) for all m < 1680.

Links

Crossrefs

Subsequence of A025487, A060735, A116998. Includes A002110, A168262, A168263.
See also A168265, A168266, A168267.

Formula

If the canonical factorization of n into prime powers is Product p^e(p), then the formula for the number of divisors of the k-th power of n is Product_p (ek + 1). (See also A146289, A146290.)
For two positive integers m and n with different prime signatures, let j be the largest exponent of k for which m and n have different coefficients, after the above formula for each integer is expanded as a polynomial. Let m_j and n_j denote the corresponding coefficients. d(n^k) > d(m^k) for all sufficiently high values of k if and only if n_j > m_j.

A168263 For any m < n, and for all values of k, d(n^k) > d(m^k). (Let k, m, and n represent positive integers only.)

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 60, 120, 180, 840, 1260, 1680, 27720
Offset: 1

Views

Author

Matthew Vandermast, Nov 23 2009

Keywords

Comments

d(n) is the number of divisors of n (A000005(n)).
All members must be highly composite numbers (A002182) with at least as many distinct prime factors as any smaller positive integer (A116998). (See Formula and Example sections.) It turns out that these two conditions are jointly sufficient.
Ramanujan proved that a) for any prime p, there exist a finite number of highly composite numbers with p as its largest prime factor; and b) in the canonical prime factorization of a highly composite number with largest prime factor p, the exponents for all primes > p are never smaller than they are in the factorization of A003418(p). (See formula 54 of the Ramanujan paper.)
It follows that, if the intersection of A003418 and A116998 is finite, so is the intersection of A002182 and A116998. For proof that the former intersection is finite, see A168262.
By using the given formula for the number of divisors, it is possible to define a canonical polynomial p_n(k) for every natural number n. For example, because 60 = (2^2)(3^1)(5^1), we define p_60(k) = (1+2k)(1+k)(1+k). The present sequence is defined only by examining whether p_n(k) achieves a record for natural numbers k, but the question could also be asked whether p_n(k) achieves a record for all k > 0. This stricter requirement does not hold for a(7)-a(13) at various positive values of k < 1, but it does hold for a(1)-a(6). The present sequence is "full", so a(1)-a(6) are the only numbers to satisfy the stronger property. - Hal M. Switkay, Aug 17 2025

Examples

			1) 1680 has more divisors than any smaller positive integer; thus for all m < n, d(1680^1) > d(m^1).
2) Since the exponents in 1680's prime factorization are (4,1,1,1), the k-th power of 1680 has (4k+1)(k+1)^3 = 4k^4 + 13k^3 + 15k^2 + 7k + 1 divisors. Comparison with the analogous formulas for all smaller members of A025487 shows the following:
a) No number smaller than 1680 has a positive coefficient in its "power formula" for any exponent larger than k^4.
b) The only power formula with a k^4 coefficient as high as 4 is that for 1260 (4k^4 + 12k^3 + 13k^2 + 6k + 1).
c) The k^3 coefficient for 1680 is higher than for 1260.
So for all sufficiently high values of k, d(1680^k) > d(m^k) for all m < 1680.
3) Careful comparison of 1680's "power formula" with the analogous formulas for smaller members of A025487 shows that no intermediate value of k can exist for which d(m^k) >= d(1680^k) if m < 1680.

References

  • S. Ramanujan, Highly composite numbers, Proc. Lond. Math. Soc. 14 (1915), 347-409; reprinted in Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962.

Links

Crossrefs

Intersection of A002182 and A116998. Also, intersection of A002182 and A060735, and of A002182 and A168264. (A168264 is a subsequence of A060735, which is a subsequence of A116998.)

Formula

If the canonical factorization of n into prime powers is Product p^e(p), then the formula for the number of divisors of the k-th power of n is Product_p (ek + 1). (See also A146289, A146290.)
For two positive integers m and n with different prime signatures, let j be the largest exponent of k for which m and n have different coefficients, after the above formula for each integer is expanded as a polynomial. Let m_j and n_j denote the corresponding coefficients. d(n^k) > d(m^k) for all sufficiently high values of k if and only if n_j > m_j.

A322841 Number of positive integers less than n with more distinct prime factors than n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 2, 0, 3, 0, 0, 5, 5, 0, 6, 0, 0, 0, 9, 0, 10, 0, 11, 0, 12, 0, 13, 13, 1, 1, 1, 1, 17, 1, 1, 1, 20, 0, 21, 2, 2, 2, 24, 2, 25, 2, 2, 2, 28, 2, 2, 2, 2, 2, 33, 0, 34, 3, 3, 36, 3, 0, 38, 4, 4, 0, 41, 5, 42, 5, 5, 5, 5, 0, 47, 6, 48
Offset: 1

Views

Author

Gus Wiseman, Dec 28 2018

Keywords

Examples

			Column n lists the a(n) positive integers less than n with more distinct prime factors than n:
  1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20
  ---------------------------------------------------------------------
                    6  6  6      10      12          15  15      18
                                  6      10          14  14      15
                                          6          12  12      14
                                                     10  10      12
                                                      6   6      10
                                                                  6

Links

Crossrefs

Positions of zeros are A116998.
Cf. A000010, A000961, A001221, A006049, A058933, A067003, A294277, A302242, A322837, A322838.

Programs

  • Maple
    b:= proc(n) option remember; nops(numtheory[factorset](n)) end:
a:= proc(n) option remember;
      (t-> add(`if`(b(i)>t, 1, 0), i=1..n-1))(b(n))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Dec 28 2018
  • Mathematica
    Table[Length[Select[Range[n],PrimeNu[#]>PrimeNu[n]&]],{n,100}]
  • PARI
    a(n) = my(omegan=omega(n)); sum(k=1, n-1, omega(k) > omegan); \\ Michel Marcus, Dec 29 2018
  • PARI
    first(n) = {my(t = 1, pp = 1, res = vector(n)); forprime(p = 2, oo, pp*=p; if(pp > n, v = vector(t); break); t++); for(i = 1, n, o = omega(i); res[i] = v[o+1]; for(j = 1, o, v[j]++)); res} \\ David A. Corneth, Dec 29 2018
Showing 1-4 of 4 results.

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