A049101 Numbers m such that m divides (product of digits of m) * (sum of digits of m).

1, 2, 3, 4, 5, 6, 7, 8, 9, 15, 18, 24, 45, 48, 135, 144, 288, 378, 476, 756, 864, 1575, 39366, 69984, 139968

Offset: 1

Author

Olivier Gérard

Keywords

• nonn
• base
• fini
• full

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

Examples

			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
		

Links

• Giovanni Resta, Method used to compute the full sequence

Crossrefs

Cf. A038369, A049102, A049105, A049106.

Programs

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 *)