A276758 Numbers n such that A045876(n) = A045876(n+1).

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

Offset: 1

Author

Altug Alkan, Sep 17 2016

Keywords

• nonn
• base

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, ...

Examples

			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
		

Crossrefs

Cf. A045876, A179239.

Programs

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, ", ")))