A066906 -id:A066906 - cp's OEIS frontend

A174538 The smallest k such that the number of steps in the n Collatz sequences starting at k+i, i=0..n-1, is always prime.

3, 3, 3, 60, 246, 560, 560, 560, 4722, 4722, 6032, 6666, 13956, 13956, 13956, 13956, 13956, 13956, 13956, 13956, 13956, 13956, 13956, 13956, 81488, 81488, 81488, 83840, 89535, 89535, 89535, 282880, 282984, 282984, 282984, 282984

Offset: 1

Michel Lagneau, Mar 21 2010

nonn

Indices of A006577 that start a run of at least n primes.

We find long sequences of primes, for example there are 55 primes starting at index 282984, namely 83, 101, 127, 251,...

			A006577(3)=7, A006577(4)=2 and A006577(5)=5 are all prime, so a(1)=a(2)=a(3) =3 mark the start of that run.

Donovan Johnson, Table of n, a(n) for n = 1..619 (terms <= 10^10)
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A008908, A066906, A070165, A033491.

A174538 := proc(n)
        local k,allp;
        for k from 1 do
                allp := true;
                for i from 0 to n-1 do
                        if not isprime(A006577(k+i)) then
                                allp := false;
                                break;
                        end if;
                end do:
                if allp then
                        return k;
                end if;
        end do:
end proc: # R. J. Mathar, Jul 08 2012

Edited by R. J. Mathar, Jul 08 2012

A174550 Run lengths of 2 or larger for consecutive prime numbers in A006577.

Original entry on oeis.org

3, 2, 2, 2, 2, 2, 4, 2, 3, 3, 2, 2, 3, 3, 4, 2, 5, 4, 3, 2, 3, 4, 2, 2, 2, 2, 3, 3, 2, 5, 2, 2, 3, 2, 3, 2, 2, 3, 8, 2, 4, 2, 2, 2, 2, 2, 3, 3, 3, 6, 3, 4, 2, 2, 3, 3, 2, 2, 4, 2, 2, 3, 6, 2, 3, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 2, 5, 3, 2, 4, 3, 5, 3, 3, 2, 8, 2, 2, 2, 2, 3, 8, 4, 3, 3, 3, 4, 2, 3, 8, 2, 3, 3, 5, 3

Offset: 1

Michel Lagneau, Mar 22 2010

This sequence is given only for n <=5000 with max(s(n)) = 10. But we can find long sequences of primes, for example,length(s(12956))= 55, and corresponding to A006577(282984 + k), k = 0,1,...,54. We obtain a sequence of 55 consecutive prime numbers given in the example below.

			a(1) = 3 represents the run (7, 2, 5).
a(2) = 2 represents the run (3, 19).
a(3) = 2 represents the run (17, 17).
a(7) = 4 represents the run (19, 19, 107, 107).
a(12956) = 55 represents the run (83, 251, 83, 251, 127, 127, 127, 251, 83, 83, 83, 83, 83, 83, 83, 83, 83, 251, 83, 83, 83, 83, 83, 83, 101, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 251, 251, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83)

Cf. A174538, A006577, A066906.

 nn:=2000:T:=array(1..nn):for n from 1 to nn do: m:=n:for p from 0 to 1000 while (m<>1) do: if irem(m,2)=1 then m:=3*m+1:else m:=m/2:fi:od:T[n]:=p:od:ii:=1:for i from 1 to nn do:if type(T[i],prime)=true and type(T[i+1],prime)=true then ii:=ii+1:else if ii<>1 then printf(`%d, `, ii):ii:=1:else fi:fi:od:

cp's OEIS Frontend

A174538 The smallest k such that the number of steps in the n Collatz sequences starting at k+i, i=0..n-1, is always prime.

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