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Least k>0 such that the decimal expansion of k^2 contains k+n as a substring.
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%I #16 Mar 02 2024 14:07:21

%S 1,11,2,13,104,14,3,15,108,16,11,17,4,39,18,77,760,19,52,117,5,118,34,

%T 21,120,121,22,41,123,23,6,125,12,24,42,128,504,25,352,130,16,26,7,

%U 133,377,27,322,135,136,44,26,393,24,747,139,29,8,141,108,142,30,143,22,144,380,31,606,146,1064,147,32

%N Least k>0 such that the decimal expansion of k^2 contains k+n as a substring.

%C This sequence is defined for all n. Proof: Given n, consider k = 10^x + n where 10^x > n^2. Since k^2 = (k+n) * 10^x + n^2, k^2 contains k+n as a substring. Furthermore, x = ceiling(log_10(1+n^2)) satisfies the inequality, therefore a(n) <= 10^ceiling(log_10(1+n^2)) + n. - _Jason Yuen_, Feb 26 2024

%H Michael S. Branicky, <a href="/A370004/b370004.txt">Table of n, a(n) for n = 0..10000</a>

%e a(3) = 13 because 13 is the least positive integer such that 13^2 = 169 contains 13 + 3 = 16 as a substring.

%e a(4) = 104 because 104 is the least positive integer such that 104^2 = 10816 contains 104 + 4 = 108 as a substring.

%t Table[k=1;While[!StringContainsQ[ToString[k^2],ToString[k+n]],k++];k,{n,0,70}]

%o (PARI) a(n) = my(k=1); while (#strsplit(Str(k^2), Str(k+n))<2, k++); k; \\ _Michel Marcus_, Feb 07 2024

%o (Python)

%o from itertools import count

%o def a(n): return next(k for k in count(1) if str(k+n) in str(k*k))

%o print([a(n) for n in range(71)]) # _Michael S. Branicky_, Feb 07 2024

%Y Cf. A000290, A018834.

%K nonn,base

%O 0,2

%A _Giorgos Kalogeropoulos_, Feb 07 2024