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

1,2

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)

Links

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

Formula

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

Example

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

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 *)

Crossrefs

Cf. A048379, A052382, A061843, A117755.

Sequence in context: A159728 A268365 A004706 * A036510 A232669 A338076

Adjacent sequences: A061841 A061842 A061843 * A061845 A061846 A061847

Keyword

base,nonn,nice

Author

Erich Friedman, Jun 23 2001

Extensions

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

Status

approved