%I #22 Apr 09 2019 04:02:25
%S 1,27,54,81,108,135,162,216,243,270,324,351,361,405,432,513,540,621,
%T 702,703,810,1026,1053,1080,1215,1242,1458,1620,1728,1944,2071,2079,
%U 2106,2133,2160,2187,2403,2413,2592,2700,2701,2916,3024,3051,3105,3267,3321,4023,4033,4050,4158
%N Integer k such that digsum(k) = digsum (k/p(1)) = digsum (k/p(2)) = ... for all prime factors p(i) of k, where digsum(k) = A007953(k) is the digital sum of k.
%C a(n) == 0 or 1 (mod 9). If a(n) == 0 (mod 9), a(n) == 0 (mod 27).
%e 4158 = 2*3^3*7*11 is in the sequence because 4 + 1 + 5 + 8 = 18, and:
%e 4158/2 = 2079 and digsum(2079) = 18;
%e 4158/3 = 1386 and digsum(1386) = 18;
%e 4158/7 = 594 and digsum(594) = 18;
%e 4158/11 = 378 and digsum(378) = 18.
%p with(numtheory):nn:=4200:
%p for k from 1 to nn do:
%p d:=factorset(k):n1:=nops(d):it:=0:
%p b:=convert(k, base, 10):n2:=nops(b):s:=sum(‘b[i]’, ‘i’=1..n2):
%p for i from 1 to n1 do:
%p x:=k/d[i]:b1:=convert(x, base, 10):n3:=nops(b1):
%p s1:=sum(‘b1[i]’, ‘i’=1..n3):
%p if s1=s
%p then
%p it:=it+1:
%p else
%p fi:
%p od:
%p if it=n1
%p then
%p printf(`%d, `,k):
%p else
%p fi:
%p od:
%t sod[n_] := Total@IntegerDigits@n; Select[Range[1, 5000], {sod[#]} == Union[sod /@ (#/First /@ FactorInteger[#])] &] (* _Giovanni Resta_, Mar 12 2019 *)
%o (PARI) isok(k) = {my(pf = factor(k)[,1]~, sd = sumdigits(k)); for (i=1, #pf, if (sumdigits(k/pf[i]) != sd, return (0));); return (1);} \\ _Michel Marcus_, Mar 12 2019
%Y Cf. A007953, A305548, A306761 (a subsequence).
%K nonn,base
%O 1,2
%A _Michel Lagneau_, Mar 12 2019