A357300 - OEIS

%I #27 Sep 26 2022 06:17:45

%S 1,10,100,108,120,180,1040,1020,1170,1008,1260,1680,10010,10530,10200,

%T 10260,10560,10800,11340,10920,12600,10080,15840,18480,15120,102060,

%U 104400,101640,100320,102600,100980,117600,114660,107100,174240,113400,105840,100800,120120,143640

%N a(n) is the smallest number m with exactly n divisors whose first digit equals the first digit of m.

%C a(m) <= a(551) = 18681062400 for m < 555. All terms with values up to 2*10^10 start with 1. Do there exist a(n) starting with any other digit? - _Charles R Greathouse IV_, Sep 25 2022

%e Of the twelve divisors of 108, four have their first digit equals to the first digit of 108: 1, 12, 18 and 108, and there is no such smaller number, hence a(4) = 108.

%t f[n_] := IntegerDigits[n][[1]]; s[n_] := Module[{fn = f[n]}, DivisorSum[n, 1 &, f[#] == fn &]]; seq[len_, nmax_] := Module[{v = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = s[n]; If[i <= len && v[[i]] == 0, c++; v[[i]] = n]; n++]; v]; seq[40, 10^6] (* _Amiram Eldar_, Sep 23 2022 *)

%o (PARI) f(n) = my(fd=digits(n)[1]); sumdiv(n, d, digits(d)[1] == fd); \\ A357299

%o a(n) = my(k=1); while (f(k)!=n, k++); k; \\ _Michel Marcus_, Sep 23 2022

%o (PARI) v=vector(1000); v[1]=r=1; forfactored(n=2, 10^11, t=a(n[1],n[2],r); if(t>r && v[t]==0, v[t]=n[1]; print(t" "n[1]" = "n[2]); while(v[r],r++); r--)) \\ _Charles R Greathouse IV_, Sep 25 2022

%Y Cf. A335491 (with last digit), A206287, A355592, A357299.

%Y Similar, but with: A333456 (Niven numbers), A335038 (Zuckerman numbers).

%K nonn,base

%O 1,2

%A _Bernard Schott_, Sep 23 2022

%E More terms from _Michel Marcus_, Sep 23 2022