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A355439 Smallest Lucas number beginning with n.

%I A355439 #11 Jul 08 2022 13:21:59
%S A355439 1,2,3,4,521,64079,7,843,9349,103682,11,123,1364,141422324,15127,
%T A355439 167761,17393796001,18,199,20633239,2139295485799,2207,23725150497407,
%U A355439 24476,2537720636,263115950957276,271443,28143753123,29,3010349,312119004989,322,33385282,3461452808002,3571
%N A355439 Smallest Lucas number beginning with n.
%C A355439 a(1382) is the first term with > 1000 digits (1156). - _Michael S. Branicky_, Jul 08 2022
%H A355439 Michael S. Branicky, <a href="/A355439/b355439.txt">Table of n, a(n) for n = 1..1381</a>
%H A355439 Ron Knott, <a href="https://r-knott.surrey.ac.uk/Fibonacci/fibmaths.html#section8">Every number starts some Fibonacci Number</a>, The Mathematical Magic of the Fibonacci Numbers.
%o A355439 (PARI) L(n) = real((2 + quadgen(5)) * quadgen(5)^n); \\ A000032
%o A355439 isok(k, dn) = my(dk=digits(L(k))); if (#dk >= #dn, Vec(dk, #dn) == dn);
%o A355439 a(n) = my(k=0, dn=digits(n)); while (!isok(k, dn), k++); L(k);
%o A355439 (Python)
%o A355439 def aupton(nn):
%o A355439     ans, f, g, k = dict(), 2, 1, 0
%o A355439     while len(ans) < nn:
%o A355439         sf = str(f)
%o A355439         for i in range(1, len(sf)+1):
%o A355439             if int(sf[:i]) > nn:
%o A355439                 break
%o A355439             if sf[:i] not in ans:
%o A355439                 ans[sf[:i]] = f
%o A355439         f, g, k = g, f+g, k+1
%o A355439     return [int(ans[str(i)]) for i in range(1, nn+1)]
%o A355439 print(aupton(35)) # _Michael S. Branicky_, Jul 08 2022
%Y A355439 Cf. A000032, A355438.
%Y A355439 Cf. A020344, A020345.
%K A355439 nonn,base
%O A355439 1,2
%A A355439 _Michel Marcus_, Jul 02 2022