%I
%S 24,120,140,528,2184,2975,3230,50232,11745,15686,62234,265639,171957,
%T 34075,1133405,2313685,1060790,332320,1334161,404858,1388504,1357216,
%U 15800704,5516293,66896037,11962832,6084983,129761775,43216511,90972513
%N 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.
%C 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.
%C Suggested by Puzzle 430, Carlos Rivera's The Prime Puzzles & Problems Connection.
%e 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.
%e 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.
%o 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
%o (PARI) {m=30; v=vector(m); w=v; p=3; c=0; while(c<m, b=p; t=0; until(t, a=b; f=factor(a); b=a+p; g=factor(b); t=nextprime(a+1)>b&&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 */
%Y Cf. A137800.
%K nonn
%O 2,1
%A _Enoch Haga_, Feb 11 2008
%E Edited and a(18) through a(31) added by _Klaus Brockhaus_, Feb 15 2008
