A343740 - OEIS

%I #16 Aug 06 2021 05:53:31

%S -1,19,23,-1,37,39,45,36,-1,27,17,25,15,36,19,-1,20,36,25,37,28,13,27,

%T 52,-1,39,17,38,27,26,17,23,24,37,19,-1,25,26,26,41,58,57,25,12,25,22,

%U 24,19,-1,33,48,23,41,49,23,32,32,23,30,19,17,31,27,-1,24

%N a(n) is the digit position of the first appearance of the last digit to appear in sqrt(n) (or -1 if n is a square).

%C A343739(n) is the last digit to appear in the decimal expansion of sqrt(n) (or -1 if n is a square), so a(n) is the digit position of the first appearance of the digit A343739(n) in sqrt(n).

%C (The first digit of sqrt(n) is counted as digit position 1; the decimal point is disregarded.)

%F a(100^q*n) = a(n), q > 0. - _Bernard Schott_, Jul 29 2021

%e a(2)=19 because A343739(2)=8 and the first appearance of an 8 in sqrt(2) = 1.414213562373095048... is at the 19th digit;

%e a(24)=52 because A343739(24)=0 and the first appearance of a 0 in sqrt(24) = 4.898979485566356196394568149411782783931894961313340... is at the 52nd digit.

%t Table[If[IntegerQ@ Sqrt@ n, -1, Function[s, Max@ Array[FirstPosition[s, #][[1]] &, 10, 0]]@ RealDigits[Sqrt[n], 10, 120][[1]]], {n, 65}] (* _Michael De Vlieger_, Jul 06 2021 *)

%o (Python 3.8+)

%o from math import isqrt

%o def A343740(n):

%o     i = isqrt(n)

%o     if i**2 == n:

%o         return -1

%o     m, dset, pos = n, set(), 0

%o     for d in (int(s) for s in str(i)):

%o         dset.add(d)

%o         pos += 1

%o         if len(dset) == 10:

%o             return pos

%o     while len(dset) < 10:

%o         m *= 100

%o         d = isqrt(m) % 10

%o         dset.add(d)

%o         pos += 1

%o     return pos # _Chai Wah Wu_, Jul 07 2021

%Y Cf. A023961, A343739, A343741, A343742.

%K sign,base

%O 1,2

%A _Jon E. Schoenfield_, Jul 05 2021