OFFSET
1,2
COMMENTS
Next term if it exists is greater than 4*10^7. - Michel ten Voorde
Sequence is finite and bounded above by 10^84, since if 10^k <= n < 10^(k+1) (product of digits of n)*(sum of digits of n) <= k*9^(k+2) which is less than 10^k for k >= 84. - Henry Bottomley, May 18 2000
Numbers with a zero digit are not permitted. - Harvey P. Dale, Jul 16 2011
No further terms to 2.5*10^9. - Robert G. Wilson v, Jul 17 2011
Sequence is complete. - Giovanni Resta, Mar 20 2013
If product of digits is performed on nonzero digits only, then 1088 is also in the sequence. - Giovanni Resta, Mar 22 2013
LINKS
EXAMPLE
139968 is in the sequence since it divides (1*3*9*9*6*8) * (1+3+9+9+6+8). - Giovanni Resta, Mar 20 2013
MATHEMATICA
okQ[n_]:=Module[{idn=IntegerDigits[n]}, !MemberQ[idn, 0] && Divisible[ (Total[idn]*Times@@idn), n]] (* Harvey P. Dale, Jul 16 2011 *)
(* full sequence *) dig[nD_] := Block[{ric, sol = {}, check}, check[mu_, minN_] := Block[{di = DigitCount@minN, k = 1, r}, While[(r = mu/k) >= minN, If[IntegerQ[r] && DigitCount[r] == di, AppendTo[sol, r]]; k++]]; ric[n_, prod_, sum_, lastd_, cnt_] := Block[{t}, If[cnt == nD, check[prod*sum, n], Do[t = nD - cnt - 1; If[n*10^(t+1) <= d*prod*9^t*(sum + d + 9*t), ric[10*n + d, d*prod, d + sum, d, cnt + 1], Break[]], {d, 9, lastd, -1}]]]; ric[0, 1, 0, 1, 0]; Print["nDig=", nD, " sol=", sol = Sort@sol]; sol]; Flatten[dig /@ Range[84]] (* Giovanni Resta, Mar 20 2013 *)
CROSSREFS
KEYWORD
nonn,base,fini,full
AUTHOR
STATUS
approved