



10, 1010, 1100, 1119, 1339, 1519, 3139, 5119, 8899, 27799, 46699, 48499, 50559, 55059, 64699, 72799, 84499, 100110, 101010, 101100, 110010, 110100, 111000, 111229, 112129, 117799, 121129, 136699, 147499, 163699, 168199, 171799, 174499, 177199, 186199
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OFFSET

1,1


COMMENTS

A138147 is a subsequence. Therefore, the sequence is infinite.  David A. Corneth, Sep 17 2016
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
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, ...


LINKS

Table of n, a(n) for n=1..35.


EXAMPLE

1339 is a term because A045876(1339) = A045876(1340).
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


PROG

(PARI) A047726(n) = n=digits(n); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
A007953(n) = sumdigits(n);
lista(nn) = for(n=1, nn, if(A047726(n)*A007953(n) == A047726(n+1)*A007953(n+1), print1(n, ", ")))


CROSSREFS

Cf. A045876, A179239.
Sequence in context: A288582 A104486 A098753 * A066489 A063171 A075166
Adjacent sequences: A276755 A276756 A276757 * A276759 A276760 A276761


KEYWORD

nonn,base


AUTHOR

Altug Alkan, Sep 17 2016


STATUS

approved



