login
A063737
Numbers n such that sum of digits of n is equal to the sum of the prime factors of n, counted with multiplicity.
10
2, 3, 4, 5, 7, 27, 378, 576, 588, 648, 729, 2688, 17496, 19683, 49896, 69888, 3796875, 3857868, 4898880, 5878656, 7077888, 8957952, 2499898464, 34998578496, 49997969280, 2928898896840, 7625597484987, 184958866998359685
OFFSET
1,1
COMMENTS
a(29) > 10^30. - Giovanni Resta, Apr 23 2017
a(29) > 10^50 if it exists. - Bert Dobbelaere, Jun 16 2024
EXAMPLE
27=3*3*3, 2+7=9, 3+3+3=9. 49896 = 2*2*2*3*3*3*3*7*11, 4+9+8+9+6 = 36, 2+2+2+3+3+3+3+7+11 = 36.
MATHEMATICA
g@n_ := Cases[Union@(Times @@ # & /@Select[Flatten[Table[IntegerPartitions[k, All, Prime@Range@PrimePi@(9*n)], {k, 1, 9*n}], 1], Plus@@#==DigitSum@(Times @@ #) &]),
_?(#<10^n&)];
g@18 (*Requires Mathematica version 14 or later*) (* Hans Rudolf Widmer, Jan 20 2024 *)
PROG
(ARIBAS): var stk: stack; end; for n := 1 to 2000000 do s := itoa(n); for j := 0 to length(s) - 1 do stack_push(stk, atoi(s[j..j])); end; if sum(stack2array(stk)) = sum(factorlist(n)) then write(n, " "); end; end; .
(PARI) isok(m) = my(f=factor(m)); sumdigits(m) == f[, 1]~*f[, 2]; \\ Michel Marcus, Dec 18 2020
(Python) MAXDIGITS=20
maxsum, maxval = 9*MAXDIGITS, 10**MAXDIGITS-1
from sympy import primerange
primes=list(primerange(0, maxsum))
nprimes, results = len(primes), []
def lensumdigits(x):
s, t = str(x), 0
for c in s: t+= ord(c)-48
return len(s), t
def solve(startidx, sump, val):
for idx in range(startidx, nprimes):
p=primes[idx]
s2, v2 = sump+p, val*p
ld, sd = lensumdigits(v2)
if sd==s2: results.append(v2)
if (s2 > maxsum) or (v2 > maxval) or ((p>10) and (s2 > 9*ld)):
return
solve(idx, s2, v2)
solve(0, 0, 1) ; print(sorted(results)) # Bert Dobbelaere, Jun 16 2024
CROSSREFS
Sequence in context: A273932 A270831 A063738 * A338121 A278342 A276522
KEYWORD
nonn,base,more
AUTHOR
Felice Russo, Aug 13 2001
EXTENSIONS
More terms from Klaus Brockhaus, Aug 17 2001
More terms from David Wasserman, Jul 11 2002
STATUS
approved