A321024 - OEIS

%I #19 Nov 01 2018 15:59:27

%S 3,5,9,14,15,20,21,27,33,35,39,45,49,50,51,55,57,63,65,69,75,80,81,84,

%T 87,93,95,99,105,110,111,117,119,123,125,129,132,135,140,141,147,152,

%U 153,154,155,159,165,170,171,177,183,185,189,195,200,201,207,208,209

%N Let j be one of the prime factors of k. Sequence lists numbers k such that the prime before j is a prime factor of k+1.

%C Contains arbitrarily long strings of consecutive integers. Here are the shortest ones arranged by increasing numbers of terms:

%C {3}

%C {14,15}

%C {49,50,51}

%C {152,153,154,155}

%C {10217,10218,10219,10220,10221}

%C {634842, 634843, 634844, 634845, 634846, 634847}

%C {123945, 123946, 123947, 123948, 123949, 123950, 123951}

%C {2852055, 2852056, 2852057, 2852058, 2852059, 2852060, 2852061, 2852062}

%C {49057063, 49057064, 49057065, 49057066, 49057067, 49057068, 49057069, 49057070, 49057071}, etc.

%H Paolo P. Lava, <a href="/A321024/b321024.txt">Table of n, a(n) for n = 1..5000</a>

%e 152 is divisible by 19 and 153 by 17;

%e 153 is divisible by 3 and 154 by 2;

%e 154 is divisible by 7 and 155 by 5;

%e 155 is divisible by 5 and 156 by 3.

%p with(numtheory): P:=proc(n) local a,k;

%p a:=factorset(n) minus {2};

%p for k from 1 to nops(a) do if frac((n+1)/prevprime(a[k]))=0

%p then RETURN(n); fi; od; end: seq(P(i),i=2..300);

%t Select[Range[210], Function[k, AnyTrue[DeleteCases[NextPrime[ FactorInteger[k][[All, 1]], -1 ], p_ /; p < 0], Mod[k + 1, #] == 0 &]]] (* _Michael De Vlieger_, Oct 31 2018 *)

%o (PARI) is(n) = my(f = factor(n>>valuation(n, 2))[,1]); n++; for(i = 1, #f~, if(n % precprime(f[i]-1) == 0, return(1))); 0 \\ _David A. Corneth_, Oct 30 2018

%Y Cf. A072562, A073606.

%K nonn

%O 1,1

%A _Paolo P. Lava_, Oct 26 2018