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.

User: Joshua R. Tint

Joshua R. Tint's wiki page.

Joshua R. Tint has authored 2 sequences.

A354153 a(n) is the smallest value of a+b+c for nonnegative integers such that a^b + c = n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 5, 5, 6, 7, 8, 9, 10, 11, 6, 7, 8, 9, 10, 11, 12, 13, 14, 7, 8, 6, 7, 8, 9, 10, 7, 8, 9, 10, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23
Offset: 1

Author

Joshua R. Tint, May 27 2022

Keywords

Comments

An obvious upper bound for this sequence is a(n) <= n-1 because 0^0 + (n-1) = n.
Another upper bound can be defined recursively: a(n) <= a(n-1) + 1 because if n-1 = a^b + c, then n = a^b + c + 1, thus one possible sum is a+b+c+1 or a(n-1) + 1.

Examples

			a(1) = 0 because 0^0 + 0 = 1 and 0 + 0 + 0 = 0.
a(9) = 5 because 3^2 + 0 = 9 and 3 + 2 + 0 = 5 and there is no ordered triple (a,b,c) such that a^b + c = 9 with a+b+c < 5.
		

Programs

  • Python
    def a(n):
        minSum = n-1
        for a in range(n-1):
            for b in range(n-a-1):
                if a**b>n:
                    break
                c = n-a**b
                if a+b+c
    				

A307360 A sequence in which every divisor other than 1 is used at most three times.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229
Offset: 1

Author

Joshua R. Tint, Apr 04 2019

Keywords

Comments

In other words, for every k > 1, there are at most 3 multiples of k in the sequence. - Rémy Sigrist, Apr 08 2019
The sequence begins at 1. The smallest integer greater than the last term which is not divisible by a divisor already used three times (excluding one) is added to the sequence.
Contains all prime numbers (A000040), given that the prime numbers only have the divisors of themselves and one, by definition, therefore the only divisor which could exist in the sequence already to disqualify the number from inclusion in the sequence would be the prime number itself, but a number cannot have a divisor higher than itself (the prime numbers), so given that the sequence increases, the divisor could not exist in the sequence, and any prime number would be included.
Terms are {1} or primes or squares of primes (A000430) or numbers of the form prime(2k + 1) * prime(2k + 2) (A089581) where k >= 0. - David A. Corneth, Apr 09 2019

Examples

			For instance, 8 is not in the sequence because 2, 4, and 6 are all divisible by 2 and appear previously in the sequence. The sequence, then, skips to nine. After 9, no more numbers divisible by three appear in the sequence, given that after 3 and 6, it is the third number divisible by three to appear in the sequence.
		

Crossrefs

See A166684 for the variant in which every divisor other than one is used at most twice.
Union of {1}, A000430 and A089581.

Programs

  • Maple
    N:= 1000: # for terms <= N
    M:= Vector(N):
    Candidates:= {$2..N}:
    A[1]:= 1:
    for n from 2 while Candidates <> {} do
      A[n]:= min(Candidates):
      Candidates:= Candidates minus {A[n]};
      for d in numtheory:-divisors(A[n]) minus {1} do
         M[d]:= M[d]+1;
         if M[d] = 3 then Candidates:= Candidates minus {seq(i,i=2*d..N, d)} fi;
      od;
    od:
    seq(A[i],i=1..n-1); # Robert Israel, Apr 09 2019
  • Mathematica
    Select[Range@ 229, Or[# == 1, PrimeQ@ #, PrimeQ@ Sqrt@ #, And[SquareFreeQ@ #, If[PrimeNu@ # == 2, And[OddQ@ First@ #, Apply[SameQ, (# - {1, 2})/2]] &@ PrimePi[FactorInteger[#][[All, 1]]], False]]] &] (* Michael De Vlieger, Apr 11 2019 *)
  • PARI
    is(n) = if(n==1, return(1)); my(f=factor(n)); if(f[, 2] == [1]~ || f[, 2] ==[2]~, return(1)); if(f[,2] == [1,1]~ && nextprime(f[1,1]+1) == f[2,1] && primepi(f[1,1]) % 2 == 1, return(1)); 0 \\ David A. Corneth, Apr 09 2019

Extensions

More terms from Jinyuan Wang, Apr 07 2019