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%I #26 Jun 08 2017 16:33:47
%S 10,1010,1100,1119,1339,1519,3139,5119,8899,27799,46699,48499,50559,
%T 55059,64699,72799,84499,100110,101010,101100,110010,110100,111000,
%U 111229,112129,117799,121129,136699,147499,163699,168199,171799,174499,177199,186199
%N Numbers n such that A045876(n) = A045876(n+1).
%C A138147 is a subsequence. Therefore, the sequence is infinite. - _David A. Corneth_, Sep 17 2016
%C Suppose a term is of the form SDN, where S is a sequence of digits without leading zeros, D is a digit less than 9 and N is a sequence of digits 9 (possibly 0 nines; terms from A002283) and SDN is a concatenation of S, D and N. Let S' be a permutation of digits of S without leading zeros. Then S'DN is also in the sequence. To search terms one may choose S from A179239. - _David A. Corneth_, Sep 18 2016
%C Since (n + 8*k) = (n - k + 1)*(n - k) has solutions that are n = k + 3*sqrt(k) and n = k - 3*sqrt(k), for square values of k there are infinitely many terms such that: 1119, 1111119999, 111111111999999999, ...
%e 1339 is a term because A045876(1339) = A045876(1340).
%e See 2nd comment. As 27799 is in the sequence, we can see S = 27, D = 7 and N = 99. Now all permutations S' (distinct) of S without leading zeros give terms. They are 72, giving term 72799. - _David A. Corneth_, Sep 18 2016
%o (PARI) A047726(n) = n=digits(n); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
%o A007953(n) = sumdigits(n);
%o lista(nn) = for(n=1, nn, if(A047726(n)*A007953(n) == A047726(n+1)*A007953(n+1), print1(n, ", ")))
%Y Cf. A045876, A179239.
%K nonn,base
%O 1,1
%A _Altug Alkan_, Sep 17 2016
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