A049101 - OEIS

%I #37 Feb 19 2024 01:55:39

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

%T 39366,69984,139968

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

%C Next term if it exists is greater than 4*10^7. - _Michel ten Voorde_

%C 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

%C Numbers with a zero digit are not permitted. - _Harvey P. Dale_, Jul 16 2011

%C No further terms to 2.5*10^9. - _Robert G. Wilson v_, Jul 17 2011

%C Sequence is complete. - _Giovanni Resta_, Mar 20 2013

%C If product of digits is performed on nonzero digits only, then 1088 is also in the sequence. - _Giovanni Resta_, Mar 22 2013

%H Giovanni Resta, <a href="/A049101/a049101.txt">Method used to compute the full sequence</a>

%e 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

%t okQ[n_]:=Module[{idn=IntegerDigits[n]},!MemberQ[idn,0] && Divisible[ (Total[idn]*Times@@idn),n]] (* _Harvey P. Dale_, Jul 16 2011 *)

%t (* 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 *)

%Y Cf. A038369, A049102, A049105, A049106.

%K nonn,base,fini,full

%O 1,2

%A _Olivier Gérard_