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A049101
Numbers m such that m divides (product of digits of m) * (sum of digits of m).
10
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
1,2
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
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
STATUS
approved