A063737 Numbers n such that sum of digits of n is equal to the sum of the prime factors of n, counted with multiplicity.

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

Author

Felice Russo, Aug 13 2001

Keywords

• nonn
• base
• more

Comments

a(29) > 10^30. - Giovanni Resta, Apr 23 2017

a(29) > 10^50 if it exists. - Bert Dobbelaere, Jun 16 2024

Examples

			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.
		

Crossrefs

Cf. A001414, A007953.

Programs

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;

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 *)

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

Extensions

More terms from Klaus Brockhaus, Aug 17 2001

More terms from David Wasserman, Jul 11 2002