A137799 Consider the first run of composites that contains at least two numbers whose largest prime factor is prime(n), n >= 2. a(n) is the first of these numbers.

24, 120, 140, 528, 2184, 2975, 3230, 50232, 11745, 15686, 62234, 265639, 171957, 34075, 1133405, 2313685, 1060790, 332320, 1334161, 404858, 1388504, 1357216, 15800704, 5516293, 66896037, 11962832, 6084983, 129761775, 43216511, 90972513

Offset: 2

Author

Enoch Haga, Feb 11 2008

Keywords

• nonn
• changed

Comments

For the second of these numbers see A137800. Sequences have offset 2 (prime(2) = 3) because prime(1) = 2 is never the largest prime factor for two numbers in a run of composites.

Suggested by Puzzle 430, Carlos Rivera's The Prime Puzzles & Problems Connection.

Examples

			The composites between 23 and 29 form the first run containing two numbers with largest prime factor prime(2) = 3, viz. 24 = 2*2*2*3 and 27 = 3*3*3. Hence a(2) = 24.
The composites between 2313679 and 2313767 form the first run containing two numbers with largest prime factor prime(17) = 59, viz. 2313685 = 5*11*23*31*59 and 2313744 = 2*2*2*2*3*19*43*59. Hence a(17) = 2313685.
		

Crossrefs

Cf. A137800.

Programs

PARI

{m=30; v=vector(m); w=v; p=3; c=0; while(cb&&f[matsize(f)[1], 1]<=p&&g[matsize(g)[1], 1]<=p); c++; v[c]=a; w[c]=b; p=nextprime(p+1)); print("A137799:"); print(v); print("A137800:"); print(w)} /* Klaus Brockhaus, Feb 15 2008 */

UBASIC

10 'puzzle 430 (duplicate prime factors) 20 N=2313680 30 A=1:S=N\2:print N; 40 B=N\A 50 if B*A=N and B=prmdiv(B) and B<=S then print B;:goto 80 60 A=A+1 70 if A<=N\2 then 40 80 C=C+1:print C: if B=59 then T=T+1 81 if N=2313700 then stop 90 if T=2 then T=0:stop 100 N=N+1: if N=prmdiv(N) then C=0:T=0:stop:print:goto 100:else 30

Extensions

Edited and a(18) through a(31) added by Klaus Brockhaus, Feb 15 2008