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First odd semiprime > 10^n.
1

%I #19 Jul 09 2023 09:40:19

%S 9,15,111,1003,10001,100001,1000001,10000001,100000001,1000000013,

%T 10000000003,100000000007,1000000000007,10000000000015,

%U 100000000000013,1000000000000003,10000000000000003,100000000000000015

%N First odd semiprime > 10^n.

%H Dario Alpern, <a href="https://www.alpertron.com.ar/ECM.HTM">Factorization using the Elliptic Curve Method.</a>

%e a(0)=9 because 9=3*3 is the first odd semiprime following 10^0=1.

%e a(13) = 10000000000015 = 5*2000000000003.

%t osp[n_]:=Module[{k=1},While[PrimeOmega[n+k]!=2,k=k+2];n+k]; Join[{9}, Table[osp[10^i],{i,20}]] (* _Harvey P. Dale_, Jan 17 2012 *)

%o (PARI) print1(9,","); for(n=1,10,forstep(i=10^n+1,10^(n+1)-1,2,f=factor(i); ms=matsize(f); if((ms[1]==1&&f[1,2]==2)||(ms[1]==2&&f[1,2]==1&&f[2,2]==1),print1(i,","); break))) /* Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 21 2006 */

%o (Python)

%o from sympy import factorint, nextprime

%o def is_semiprime(n): return sum(e for e in factorint(n).values()) == 2

%o def next_odd_semiprime(n):

%o nxt = n + 1 + n%2

%o while not is_semiprime(nxt): nxt += 2

%o return nxt

%o def a(n): return next_odd_semiprime(10**n)

%o print([a(n) for n in range(20)]) # _Michael S. Branicky_, Sep 15 2021

%Y Cf. A046315 (odd semiprimes), A098147(n)=a(n)-10^n continuation of this sequence, A003717 (smallest n-digit prime).

%K nonn

%O 0,1

%A _Hugo Pfoertner_, Aug 28 2004