A061843 -id:A061843 - cp's OEIS frontend

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

Erich Friedman, Jun 23 2001

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)

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

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

Cf. A048379, A052382, A061843, A117755.

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

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 *)

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

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

A117755 Squares which remain squares when each digit is replaced by the next digit.

Original entry on oeis.org

0, 9, 25, 2025, 13225, 1974025, 4862025, 6943225, 60415182025, 207612366025, 916408817340025, 9960302475729225, 153668543313582025, 1978088677245614025, 13876266042653742025, 20761288044852366025, 47285734107144405625, 406066810454367265225

Offset: 1

Luc Stevens (lms022(AT)yahoo.com), Apr 14 2006

Replace 1 with 2, 2 with 3, ..., 8 with 9 and 9 with 0.

			13225 is in the sequence because (1) it is a square and (2) if we transform it we get 24336 and this is also a square.

Max Alekseyev, Table of n, a(n) for n = 1..22

Cf. A048379, A061843, A061844, A127856.

Select[Range[0, 500000]^2, IntegerQ[Sqrt[FromDigits[(1 + IntegerDigits[ # ]) /. 10-> 0]]] &] (* Harvey P. Dale, Jan 21 2007 *)

More terms from Harvey P. Dale, Jan 21 2007
3 more terms from Donovan Johnson, Apr 03 2008
a(14)-a(21) from Max Alekseyev, Oct 22 2008
a(1) = 0 prepended by Max Alekseyev, Jul 26 2023

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

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