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A137799
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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.
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3
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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
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OFFSET
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2,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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PROG
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(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
(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 */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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