A350870 - OEIS

%I #29 Jul 27 2024 23:53:35

%S 81,2025,3025,494209,24502500,25502500,52881984,60481729,6049417284,

%T 6832014336,101558217124,108878221089,123448227904,127194229449,

%U 152344237969,213018248521,217930248900,249500250000,250500250000,284270248900,289940248521,371718237969,413908229449,420744227904

%N Numbers k = x.y such that x.y = (x+y)^2, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.

%C Problem proposed on French site Diophante (see link).

%C We have to solve Diophantine equation (x+y)^2 = x*10^m + y where m = length(x) = length(y).

%C Squares of a variant of Kaprekar numbers (A045913).

%C Number of solutions with 2*m digits for m >= 1: 1, 2, 1, 4, 2, 21, ...

%C Compare with A347541 where x*y divides x.y, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.

%H Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a1-pot-pourri/1024-a1945-concatenations-en-tous-genres">A1945 - Concaténations en tous genres</a> (in French).

%F a(n) = A045913(n+1)^2.

%e 81 = (8+1)^2, hence 81 is a term.

%e 3025 = (30+25)^2, hence 3025 is another term.

%o (PARI) upto(n) = {i = 4; i2 = i^2; res = List(); while(i2 <= n, i++; i2 = i^2; if(#digits(i2) % 2 == 1, i = sqrtint(10^(#digits(i2))) + 1; i2 = i^2; ); if(is(i2), listput(res, i2) ); ); res }

%o is(n) = { my(d = digits(n), d1, d2, frd2); if(#d % 2 == 1, return(0)); d1 = vector(#d \ 2, i, d[i]); d2 = vector(#d \ 2, i, d[i + #d \ 2]); frd2 = fromdigits(d2); 10^(#d \ 2 - 1) <= frd2 && (fromdigits(d1) + frd2)^2 == n } \\ _David A. Corneth_, Jan 21 2022

%Y Equals A238237 \ A350918.

%Y Subsequence of A102766.

%Y Subsequences: A038544, A350869.

%Y Cf. A045913, A347541.

%K nonn,base

%O 1,1

%A _Bernard Schott_, Jan 20 2022