A061844 Squares that remain squares if you decrease every digit by 1.

1, 36, 3136, 24336, 5973136, 71526293136, 318723477136, 264779654424693136, 24987377153764853136, 31872399155963477136, 58396845218255516736, 517177921565478376336, 252815272791521979771662766736, 518364744896318875336864648336, 554692513628187865132829886736

Offset: 1

Author

Erich Friedman, Jun 23 2001

Keywords

• base
• nonn
• nice

Comments

The terms may be calculated efficiently by solving x^2 - y^2 = 111...1; this is done by factoring 111..1 = (x + y)(x - y).

Note that some solutions will produce a square containing a zero digit so the solution is impermissible; for example, 460^2 - 317^2 = 111111 but 460^2 = 211600. - Wendy Appleby, Sep 20 2015

Except for a(1) = 1, we don't allow decreasing the digits to create a leading 0. Thus 126736 = 356^2 is not included, even though 126736 - 111111 = 15625 = 125^2. - Robert Israel, Dec 30 2015

If it exists, a(79) > 10^262. - Max Alekseyev, Sep 05 2023

From Robert Israel, Jan 04 2016: (Start)

The sequence may well be finite.

It is known that A000005(n) = O(n^epsilon) for all epsilon>0.

Therefore if 1 < c < 10/9, for d sufficiently large (10^d-1)/9 has fewer than c^d divisors, and thus fewer than c^d possible candidates for x^2 having d digits.

Heuristically, x^2 has probability ~ (9/10)^d of having no digits 0.

Thus we expect fewer than (9c/10)^d terms having d digits.

Since Sum_d (9c/10)^d converges, we expect only finitely many terms.

Of course, this is only a heuristic argument, but it seems to fit well with the data. (End)

Examples

			13225 = 115^2 and 24336 = 156^2.
		

Links

• JungHwan Min, Table of n, a(n) for n = 1..78

Crossrefs

Cf. A048379, A052382, A061843, A117755.

Programs

Maple

A:= {1}:
for d from 1 to 96 do
  r:= (10^d-1)/9;
  f:= subs(X=10,factors((X^d-1)/(X-1))[2]);
  q:= map(t -> op(map(s -> [s[1],t[2]*s[2]], ifactors(t[1])[2])),f);
  divs:= {1};
for t in q do
    divs:= map(x -> seq(x*t[1]^j,j=0..t[2]),divs)
  od;
  for t in select(s -> s^2 > r, divs) do
    x:= (t + r/t)/2;
    if ilog10(x^2) = d-1 and x^2 > 2*10^(d-1) and not has(convert(x^2,base,10),0) then
      A:= A union {x^2};
    fi
  od
od:
sort(convert(A,list)); # Robert Israel, Dec 30 2015

Mathematica

For[digits = 1, digits <= 30, digits++, n = (10^digits - 1)/9; divList = Select[Divisors[n], (#1 >= Sqrt[n])&]; For[j = 1, j <= Length[divList], j++, x = (divList[[j]] + n/divList[[j]])/2; y = (divList[[j]] - n/divList[[j]])/2; dx = IntegerDigits[x^2]; dy = IntegerDigits[y^2]; If[(Length[dx] == digits) && (Length[dy] == digits) && (Select[dx, (#1 == 0)&] == {}), Print[x^2]]]]
Flatten@Prepend[Table[Select[#[[Ceiling[(Length[#] + 1)/2] ;;]] &@(# + Reverse@#)/2 &@Divisors[(10^n - 1)/9], IntegerLength[#^2] == n && (#[[1]] != 1 && FreeQ[#, 0]&[IntegerDigits[#^2]])&]^2, {n, 30}], 1] (* JungHwan Min, Dec 29 2015 *)
Join[{1},Select[Select[Flatten[Table[#^2&/@(x/.Solve[{x^2-y^2 == FromDigits[ PadRight[{},n,1]],x>0,y>0},{x,y},Integers]),{n,2,30}]], DigitCount[ #,10,0]==0&&IntegerDigits[#][[1]]>1&]// Union,IntegerQ[ Sqrt[ FromDigits[IntegerDigits[#]-1]]]&]] (* Harvey P. Dale, Apr 16 2016 *)

Formula

a(n) = A048379(A061843(n)). - Max Alekseyev, Jul 26 2023

Extensions

More terms and program from Jonathan Cross (jcross(AT)wcox.com), Oct 08 2001