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Integers m such that the number of divisors whose last digit equals the last digit of m sets a new record.
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%I #27 Mar 26 2021 12:34:32

%S 1,11,40,60,120,240,360,480,600,1200,1800,2400,3600,7200,8400,12600,

%T 16800,25200,50400,75600,100800,151200,201600,252000,277200,453600,

%U 504000,554400,831600,1108800,1663200,2217600,2772000,3326400,4989600,5544000,6652800,7207200

%N Integers m such that the number of divisors whose last digit equals the last digit of m sets a new record.

%C Inspired by Project Euler, Problem 474 (see link).

%C The corresponding number of divisors whose last digit equals the last digit: 1, 2, 3, 4, 6, 8, 9, 10, 12, 16, 18, 20, ...

%H Project Euler, <a href="https://projecteuler.net/problem=474">Problem 474: Last digits of divisors</a>.

%F For n >= 3, a(n) = 10 * A002182(n) (conjectured).

%e a(5) = 120 is in the sequence because A330348(120) = 6, the six corresponding divisors are {10, 20, 30, 40, 60, 120} and 6 is larger than any earlier value in A330348.

%t d[n_] := DivisorSum[n, 1 &, Mod[# - n, 10] == 0 &]; dm = 0; s = {}; Do[d1 = d[n]; If[d1 > dm, dm = d1; AppendTo[s, n]], {n, 1, 10^7}]; s (* _Amiram Eldar_, Mar 23 2021 *)

%o (PARI) f(n) = my(dig = n%10); sumdiv(n, d, d%10 == dig); \\ A330348

%o lista(nn) = my(m, k=0, kk); for (n=1, nn, kk = f(n); if (kk>k, print1(n, ", "); k = kk)); \\ _Michel Marcus_, Mar 24 2021

%Y Cf. A002182, A002183, A330348, A335491.

%K nonn,base

%O 1,2

%A _Bernard Schott_, Mar 23 2021