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A306761 Let digsum(k) = A007953(k) denote the digital sum of k. The sequence lists the smallest integer k such that digsum(k) = digsum (k/d(1)) = digsum (k/d(2)) = ... = digsum (k/d(n)) where d(i) are the n distinct prime factors of k.

%I A306761 #19 Mar 11 2019 20:17:02
%S A306761 27,54,270,4158,20790,270270,36506106,464053590,18621166410
%N A306761 Let digsum(k) = A007953(k) denote the digital sum of k. The sequence lists the smallest integer k such that digsum(k) = digsum (k/d(1)) = digsum (k/d(2)) = ... = digsum (k/d(n)) where d(i) are the n distinct prime factors of k.
%C A306761 Conjecture 1: a(n) == 0 (mod 54) for n > 1.
%C A306761 Conjecture 2: a(n)/27 is a squarefree number for n > 1.
%C A306761 The smallest multiple of 54 that can be a(10) is 1069808930190, which is also a multiple of 7^2, so the two conjectures above cannot be both true. - _Giovanni Resta_, Mar 08 2019
%e A306761 a(4) = 4158 = 2*3^3*7*11 because 4 + 1 + 5 + 8 = 18, and:
%e A306761 4158/2 = 2079 and digsum(2079) = 18;
%e A306761 4158/3 = 1386 and digsum(1386) = 18;
%e A306761 4158/7 = 594 and digsum(594) = 18;
%e A306761 4158/11 = 378 and digsum(378) = 18.
%p A306761 with(numtheory):nn:=10^6:
%p A306761 for n from 1 to 10 do:
%p A306761 ii:=0:
%p A306761   for k from 1 to nn while(ii=0) do:
%p A306761      d:=factorset(k):n1:=nops(d):it:=0:
%p A306761      b:=convert(k, base, 10):n2:=nops(b):s:=sum(‘b[i]’, ‘i’=1..n2):
%p A306761       for i from 1 to n1 do:
%p A306761         x:=n/d[i]:b1:=convert(x, base, 10):n3:=nops(b1):
%p A306761         s1:=sum(‘b1[i]’, ‘i’=1..n3):
%p A306761         if s1=s
%p A306761          then
%p A306761          it:=it+1:
%p A306761          else
%p A306761         fi:
%p A306761       od:
%p A306761        if it=n
%p A306761         then
%p A306761         ii:=1:printf(`%d %d \n`,n,it):
%p A306761        else
%p A306761       fi:
%p A306761 od:
%o A306761 (PARI) isok(k, n) = {if (omega(k) != n, return(0)); my(pf = factor(k)[,1]~, sd = sumdigits(k)); for (i=1, n, if (sumdigits(k/pf[i]) != sd, return (0));); return (1);}
%o A306761 a(n) = {my(k=2); while (!isok(k, n), k++); k;} \\ _Michel Marcus_, Mar 09 2019
%Y A306761 Cf. A007953.
%K A306761 nonn,base,more
%O A306761 1,1
%A A306761 _Michel Lagneau_, Mar 08 2019
%E A306761 a(9) from _Giovanni Resta_, Mar 08 2019