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Numbers with all divisors starting with digit 1.
+10
3
1, 11, 13, 17, 19, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1111
COMMENTS
Equivalently, integers m with all divisors starting with the same first digit of m; in fact, as 1 divides all the integers, this digit is necessarily 1; also, for these terms m: A357299(m) = A000005(m). - Bernard Schott, Sep 25 2022
EXAMPLE
All divisors of 187 (1, 11, 17, 187) start with digit 1.
MATHEMATICA
fQ[n_] := Module[{d = Divisors[n]}, Union[IntegerDigits[#][[1]] & /@ d] == {1}]; Select[Range[1111], fQ] (* T. D. Noe, Feb 13 2012 *)
a(n) is the smallest number m with exactly n divisors whose first digit equals the first digit of m.
+10
3
1, 10, 100, 108, 120, 180, 1040, 1020, 1170, 1008, 1260, 1680, 10010, 10530, 10200, 10260, 10560, 10800, 11340, 10920, 12600, 10080, 15840, 18480, 15120, 102060, 104400, 101640, 100320, 102600, 100980, 117600, 114660, 107100, 174240, 113400, 105840, 100800, 120120, 143640
COMMENTS
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
EXAMPLE
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.
MATHEMATICA
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 *)
PROG
(PARI) f(n) = my(fd=digits(n)[1]); sumdiv(n, d, digits(d)[1] == fd); \\ A357299a(n) = my(k=1); while (f(k)!=n, k++); k; \\ Michel Marcus, Sep 23 2022 (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
CROSSREFS
Similar, but with: A333456 (Niven numbers), A335038 (Zuckerman numbers).
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