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Numbers k whose last digit is the k-th digit of sqrt(2).
3

%I #21 Jan 29 2023 16:59:47

%S 15,18,20,21,39,45,47,49,67,87,92,117,122,126,147,149,162,165,170,179,

%T 192,197,205,235,237,264,272,274,275,279,282,286,289,292,297,305,313,

%U 318,335,340,344,345,356,381,384,394,403,409,418,433,446,449,471,476

%N Numbers k whose last digit is the k-th digit of sqrt(2).

%C The k-th digit of sqrt(2) is calculated starting after the decimal point. - _Harvey P. Dale_, Jan 29 2023

%H Chai Wah Wu, <a href="/A067757/b067757.txt">Table of n, a(n) for n = 1..10000</a>

%e Let d(n) be the n-th digit of sqrt(2)=1.4142... so d(1)=4, d(2)=1 etc. d(122)=2 and 2 is the last digit of 122 hence 122 is in the sequence.

%t Module[{nn=500,sq2},sq2=RealDigits[Sqrt[2],10,nn][[1]];Select[Range[nn-1],Mod[ #,10] == sq2[[#+1]]&]] (* _Harvey P. Dale_, Jan 29 2023 *)

%o (Python 3.8+)

%o from math import isqrt

%o A067757_list = [i for i,d in enumerate(str(isqrt(2*100**100000))) if i % 10 == int(d)] # _Chai Wah Wu_, Jul 08 2021

%Y Cf. A002193, A067750, A067789.

%K easy,nonn,base

%O 1,1

%A _Benoit Cloitre_, Mar 07 2002