cp's OEIS Frontend

A083178 Numbers with a digit sum of n and a maximum product of digits. In case of two identical products choose the largest number.

%I A083178 #18 Dec 12 2015 08:00:44
%S A083178 1,2,3,22,32,33,322,332,333,3322,3332,3333,33322,33332,33333,333322,
%T A083178 333332,333333,3333322,3333332,3333333,33333322,33333332,33333333,
%U A083178 333333322,333333332,333333333,3333333322,3333333332,3333333333
%N A083178 Numbers with a digit sum of n and a maximum product of digits. In case of two identical products choose the largest number.
%C A083178 The corresponding maximum product is A000792(n). - _Lekraj Beedassy_, Nov 13 2009
%C A083178 Except for the first term, terms in the sequence are exactly those numbers formed by sequence of digits 3 followed by either zero, one or two digits 2. - _Chai Wah Wu_, Dec 11 2015
%F A083178 Conjecture: a(n) = 10*a(n-3)+a(n-6)-10*a(n-9) for n>10. - _Colin Barker_, Oct 14 2014
%F A083178 Empirical g.f.: x*(90*x^6+10*x^4+11*x^3+3*x^2+2*x+1) / ((x-1)*(x^2+x+1)*(10*x^3-1)). - _Colin Barker_, Oct 14 2014
%F A083178 For n > 7, a(n) = 11*a(n-3)-10*a(n-6). For n > 4, a(n-3) + 3*10^(floor((n-1)/3)). For n > 1, (2*10^(floor((n+2)/3))+(63*m^2-129*m-2))/6, where m is the least nonnegative residue of n mod 3. - _Chai Wah Wu_, Dec 11 2015
%o A083178 (Python)
%o A083178 from __future__ import division
%o A083178 def A083178(n):
%o A083178     return 1 if n == 1 else (2*10**((n+2)//3)+(63*(n%3)**2-129*(n%3)-2))//6 # _Chai Wah Wu_, Dec 11 2015
%Y A083178 Cf. A000792.
%K A083178 base,nonn,easy
%O A083178 1,2
%A A083178 _Amarnath Murthy_, Apr 26 2003
%E A083178 Corrected and extended by _David Wasserman_, Oct 25 2004