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.

27, 54, 270, 4158, 20790, 270270, 36506106, 464053590, 18621166410

Offset: 1

Author

Michel Lagneau, Mar 08 2019

Keywords

• nonn
• base
• more

Comments

Conjecture 1: a(n) == 0 (mod 54) for n > 1.

Conjecture 2: a(n)/27 is a squarefree number for n > 1.

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

Examples

			a(4) = 4158 = 2*3^3*7*11 because 4 + 1 + 5 + 8 = 18, and:
4158/2 = 2079 and digsum(2079) = 18;
4158/3 = 1386 and digsum(1386) = 18;
4158/7 = 594 and digsum(594) = 18;
4158/11 = 378 and digsum(378) = 18.
		

Crossrefs

Cf. A007953.

Programs

Maple

with(numtheory):nn:=10^6:
for n from 1 to 10 do:
ii:=0:
  for k from 1 to nn while(ii=0) do:
     d:=factorset(k):n1:=nops(d):it:=0:
     b:=convert(k, base, 10):n2:=nops(b):s:=sum(‘b[i]’, ‘i’=1..n2):
      for i from 1 to n1 do:
        x:=n/d[i]:b1:=convert(x, base, 10):n3:=nops(b1):
        s1:=sum(‘b1[i]’, ‘i’=1..n3):
        if s1=s
         then
         it:=it+1:
         else
        fi:
      od:
       if it=n
        then
        ii:=1:printf(`%d %d \n`,n,it):
       else
      fi:
od:

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);}
a(n) = {my(k=2); while (!isok(k, n), k++); k;} \\ Michel Marcus, Mar 09 2019

Extensions

a(9) from Giovanni Resta, Mar 08 2019