OFFSET

1,1

COMMENTS

The sequence includes an infinite family of arithmetic progressions. Such AP's can be constructed to each term, with large differences [like squares of primorials, A061742(7)]. It is necessary to solve suitable systems of linear Diophantine equations. E.g.: arithmetic progression subsequences of starting 9-chains is {mk+69147868+j} where j=0..8, m=510510^2 because square prime factors of a(4)+j=68147868+j are 4, 49, 121, 169, 4, 9, 289, 25, 4 resp. for j=0..8; k goes to infinity; 7th primorial is sufficient, 9th is not necessary. Construction is provable for arbitrary long [>9] chains. - Labos Elemer, Nov 25 2002

More precisely, if in one run {a(n)+j, j=0..8} the maximum smallest square factor is p^2, then an infinite subsequence is given by {a(n)+(p#)^2*k, k=0..oo}, where p# = A034386(p). One may get a smaller step taking the least L^2 which has a square factor in common with each of the 9 consecutive terms. - M. F. Hasler, Feb 03 2016

LINKS

Donovan Johnson, Table of n, a(n) for n = 1..1000

FORMULA

A078143 = { A077647[k] | A077647[k+1] = A077647[k]+1 } = { A077640[k] | A077640[k+2] = A077640[k]+2 } = { A078144[k] | A078144[k+4] = A078144[k]+4 } etc. Note that A049535 is defined differently. - M. F. Hasler, Feb 01 2016

a(n) < 4666864390*n. With more work this bound can be decreased significantly. - Charles R Greathouse IV, Nov 05 2017

MATHEMATICA

s9[x_] := Apply[Plus, Table[Abs[MoebiusMu[x+j]], {j, 0, 8}]]; Do[If[Equal[s9[n], 0], Print[n]], {n, 8000000, 1000000000}]

PROG

(PARI) is(n)=for(i=n, n+8, if(!issquarefree(i), return(0))); 1 \\ Charles R Greathouse IV, Nov 05 2017

CROSSREFS

KEYWORD

nonn

AUTHOR

Labos Elemer, Nov 22 2002

EXTENSIONS

a(6)-a(21) from Donovan Johnson, Nov 26 2008

STATUS

approved