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Lexicographically earliest sequence of distinct terms > 0 such that the sum a(n) + a(n+1) is a substring of the concatenation (a(n), a(n+1)).
2

%I #45 Aug 02 2023 11:50:50

%S 1,10,99,889,8009,1101,9089,80718,100284,183899,206021,396118,215703,

%T 354632,108578,469891,229021,61195,34146,7321,13817,3536,1825,749,167,

%U 508,324,2096,4337,2958,2870,4171,12941,16470,30560,25465,21056,35296,17665,35927,23345,10106,548,279,516,1094,3228,5302

%N Lexicographically earliest sequence of distinct terms > 0 such that the sum a(n) + a(n+1) is a substring of the concatenation (a(n), a(n+1)).

%C Is this sequence a permutation of the integers > 0?

%C I conjecture that it isn't, and more specifically, that a(1) = 1 is the only single-digit term, and a(2) = 10 the only multiple of 10 below 100. See Examples for other terms of the form x*10^k, 1 <= x <= 9. - _M. F. Hasler_, Jul 03 2023

%H Eric Angelini, <a href="http://cinquantesignes.blogspot.com/2023/07/echecs-et-maths.html">Échecs et Maths</a>, Personal blog, bottom of the page, July 2023.

%H Hans Havermann, <a href="http://chesswanks.com/num/SumInConcatenation.txt">100000 terms</a>, July 2023

%e 1 + 10 = 11 and 11 is a substring of concat(1, 10) = 110.

%e 10 + 99 = 109 and 109 is a substring of concat(10, 99) = 1099.

%e 99 + 889 = 988 and 988 is a substring of concat(99, 889) = 99889.

%e 889 + 8009 = 8898 and 8898 is a substring of 8898009.

%e 8009 + 1101 = 9110 and 9110 is a substring of 80091101, etc.

%e Some examples of terms of the form x*10^k, x < 10: a(2136) = 800, a(4204) = 1000, a(6246) = 900, a(6618) = 100, a(11268) = 400, a(17446) = 10000, a(39292) = 600, a(44989) = 700, a(91359) = 300, ... - _M. F. Hasler_, Jul 03 2023

%o (PARI) A359482_first(n)={my(ok(a,k)=my(c=a*10^logint(k*10,10)+k); k=10^logint(10*a+=k,10); until(a>c\=10, c%k==a&& return(1)), U=[], a=0); vector(n,n, my(k=1); while(setsearch(U,k)|| !ok(a,k), k++); U=setunion(U,[k]); a=k)} \\ Becomes slow for n > 10. - _M. F. Hasler_, Jul 03 2023

%Y Cf. A300000.

%K base,nonn

%O 1,2

%A _Eric Angelini_ and _Hans Havermann_, Jul 03 2023