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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.
1
27, 54, 270, 4158, 20790, 270270, 36506106, 464053590, 18621166410
OFFSET
1,1
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
EXAMPLE
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.
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:
PROG
(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
CROSSREFS
Cf. A007953.
Sequence in context: A033903 A175806 A232922 * A181874 A042444 A042442
KEYWORD
nonn,base,more
AUTHOR
Michel Lagneau, Mar 08 2019
EXTENSIONS
a(9) from Giovanni Resta, Mar 08 2019
STATUS
approved