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.

A358890 a(n) is the first term of the first maximal run of n consecutive numbers with increasing greatest prime factors.

Original entry on oeis.org

14, 4, 1, 8, 90, 168, 9352, 46189, 2515371, 721970, 6449639, 565062156, 11336460025, 37151747513, 256994754033, 14037913234203
Offset: 1

Views

Author

Reinhard Zumkeller, Jan 10 2003

Keywords

Comments

a(16) > 10^13. - Giovanni Resta, Jul 25 2013
The convention gpf(1) = A006530(1) = 1 is used (otherwise we would have a(2) = 2 and a(3) = 24). - Pontus von Brömssen, Dec 05 2022
a(17) > 10^14. - Martin Ehrenstein, Dec 10 2022

Examples

			a(7) = 9352 because the first sequence of seven consecutive numbers with increasing greatest prime factors is 9352=167*7*2^3, 9353=199*47, 9354=1559*3*2, 9355=1871*5, 9356=2339*2^2, 9357=3119*3, and 9358=4679*2. [Corrected by _Jon E. Schoenfield_, Sep 21 2022]
		

Crossrefs

Cf. A006530, A070087, A079748, A079749 (erroneous version), A100384.

Programs

  • Maple
    V:= Vector(11): count:= 0:
    a:= 1: m:= 1: w:= 1:
    for k from 2 while count < 11 do
      v:= max(numtheory:-factorset(k));
      if v > m then m:= v
      else
        if V[k-a] = 0 then V[k-a]:= a; count:= count+1; fi;
        a:= k; m:= v;
      fi
    od:
    convert(V,list); # Robert Israel, Dec 05 2022
  • Python
    from sympy import factorint
    def A358890(n):
        m = 1
        gpf1 = 1
        k = 1
        while 1:
            while 1:
                gpf2 = max(factorint(m+k))
                if gpf2 < gpf1: break
                gpf1 = gpf2
                k += 1
            if k == n: return m
            m += k
            gpf1 = gpf2
            k = 1 # Pontus von Brömssen, Dec 05 2022

Formula

A079748(a(n)) = n-1.
From Pontus von Brömssen, Dec 05 2022: (Start)
A079748(a(n)-1) = 0 for n != 3.
For n != 3, a(n) = A070087(m)+1, where m is the smallest positive integer such that A070087(m+1) - A070087(m) = n.
(End)

Extensions

More terms from Don Reble, Jan 17 2003
Corrected by Jud McCranie, Feb 11 2003
a(14)-a(15) from Giovanni Resta, Jul 25 2013
Name edited, a(1) and a(2) corrected by Pontus von Brömssen, Dec 05 2022
a(16) from Martin Ehrenstein, Dec 07 2022