Search: id:A101702
Results 1-1 of 1 results found.

%I A101702
%S A101702 1,2,541,52100,58504,66410,430000,863180,8601400,17927300,27927300,
%T A101702 31000000,665100000,3715000000,6739630000,11000000000,21000000000,
%U A101702 53100000000,70858000000,79637300000,451000000000,1715000000000,2715000000000,
               48304000000000,340000000000000,5520000000000000
%N A101702 Numbers n such that the sum of the factorials of their digits is equal 
               to the reversal of n.
%C A101702 Next term is greater than 8*10^8. If s=sum of the factorials of digits 
               of n & reversal(n)>=s then 10^(reversal(n)-s)*n is in the sequence. 
               Example n=23; s=2!+3!; reversal(23)-s=24 & 23*10^24 is in the sequence. 
               So this sequence is infinite because there exists infinitely many 
               numbers n such that reversal(n)>s. If n is a k-digit term of this 
               sequence and the first digit of n is 1 then 10^(k-1)+n is also in 
               the sequence. Examples : n=1 so 10^(1-1)+1=2 is in the sequence, 
               n=17927300 so 10^7+17927300=27927300 is in the sequence. If n>5 then 
               10 divides a(n). If 10 doesn't divide a(n) then the reversal of n 
               is in the sequence A010480, so all terms of A010480 are : reversal(1), 
               reversal(2), reversal(541) & reversal(58504).
%e A101702 665100000 is in the sequence because reversal(665100000)=1566=
%e A101702 6!+6!+5!+1!+0!+0!+0!+0!+0!.
%t A101702 Do[h = FactorInteger[n]; l = Length[h]; If[FromDigits[Reverse[IntegerDigits[n] 
               == Sum[h[[k]]], {k, l}], Print[n]], {n, 800000000}]
%Y A101702 Cf. A014080, A049529, A101697.
%Y A101702 Sequence in context: A080778 A007513 A071613 this_sequence A119780 A120840 
               A058429
%Y A101702 Adjacent sequences: A101699 A101700 A101701 this_sequence A101703 A101704 
               A101705
%K A101702 base,nonn
%O A101702 1,2
%A A101702 Farideh Firoozbakht (mymontain(AT)yahoo.com), Dec 24 2004
%E A101702 More terms from Donovan Johnson (donovan.johnson(AT)yahoo.com), Feb 26 
               2008

Search completed in 0.001 seconds