A055618 -id:A055618 - cp's OEIS frontend

A055616 Numbers, with an even number of digits, that are the sum of the squares of their two halves (leading zeros allowed only for the second half).

1233, 8833, 990100, 94122353, 1765038125, 2584043776, 7416043776, 8235038125, 9901009901, 116788321168, 123288328768, 876712328768, 883212321168, 999900010000, 13793103448276, 15348303604525, 84651703604525, 86206903448276, 91103202846976, 92318202663025

Offset: 1

Ulrich Schimke (ulrschimke(AT)aol.com)

The sequence is infinite since it contains several infinite subsequences (see A055617, etc.).
If x = A*10^n+B is an element not beginning with 9, then (10^n-A)*10^n+B is another (e.g. 1233 <-> 8833).
Numbers that can be written as n = A*10^d + B with 10^(d-1) <= A < 10^d, 0 <= B < 10^d, and A^2 + B^2 = n. - Robert Israel, May 10 2015

			8833 is ok, since 8833 = 88^2 + 33^2.

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A064942 for the number of solutions, where leading zeros are allowed.

Cf. A055617, A055618, A055619.

dmax:= 8: # to get all entries with at most 2*dmax digits
Res:= NULL:
for d from 2 to dmax  do
     cands:= map(t -> subs(t,[x,y]), [isolve(x^2 + y^2 = 10^(2*d)+1)]);
     cands:= select(t -> t[1]::even and t[1]>=0 and t[2]>0, cands);
     cands:= map(t -> ([(10^d + t[1])/2, (t[2]+1)/2], [(10^d-t[1])/2, (t[2]+1)/2]), cands);
     cands:= select(t -> (t[1]>= 10^(d-1) and t[1] < 10^d and t[2] <= 10^d), cands);
     Res:= Res, op(map(t -> 10^d*t[1]+t[2], cands));
od:
sort([Res]); # Robert Israel, May 10 2015

fQ[n_] := Block[{d = IntegerDigits@ n}, If[OddQ[Length@ d], False, Plus[FromDigits[Take[d, Length[d]/2]]^2, FromDigits[Take[d, -Length[d]/2]]^2]] == n]; Select[Range@ 1000000, fQ] (* Michael De Vlieger, May 09 2015 *)

select( {is_A055616(n, L=logint(n,10))=L%2 && n==norml2(divrem(n,10^(L\/2)))}, [1..10^5]) \\ M. F. Hasler, Dec 20 2024
for(L=1,oo, for(n=10^L,10^L++, is_A055616(n)&& print1(n", "))) \\ slow beyond 10^6

def a():
  n = 1
  while n < 10**6:
    st = str(n)
    if len(st) % 2 == 0:
      s1 = st[:int(len(st)/2)]
      s2 = st[int(len(st)/2):int(len(st))]
      if int(s1)**2+int(s2)**2 == int(st):
        print(n,end=', ')
        n += 1
      else:
        n += 1
    else:
      n = 10*n
a()
# Derek Orr, Jul 08 2014

Definition corrected by Derek Orr, Jul 09 2014

A064942 Decimal numbers n such that after possibly prefixing leading 0's to n, the resulting number n' can be broken into 2 numbers of equal length, n' = xy, such that x^2+y^2 = n (y may also have leading zeros).

Original entry on oeis.org

1, 1233, 8833, 10100, 990100, 5882353, 94122353, 99009901, 100010000, 1765038125, 2584043776, 7416043776, 8235038125, 9901009901, 48600220401, 116788321168, 123288328768, 601300773101, 876712328768, 883212321168, 990100990100

Offset: 1

Ulrich Schimke (ulrschimke(AT)aol.com)

If A*10^m+B is an element, then so is (10^m-A)*10^m+B (e.g. 1233 <-> 8833 or 010100 <-> 990100)

Since A^2+B^2 = A*10^m+B can be written as 10^(2*m)+1 = (2*A-10^m)^2 + (2*B-1)^2 the number of solutions with 2*m digits (necessary leading zeros count) can be reduced to finding the ways 10^(2*m)+1 can be written as sum of 2 squares. For the following results, see A002654. Since 10^(2*m)+1 is odd and has no prime factors of the form 4*r+3 the number of ways writing 10^(2*m)+1 as sum of 2 squares is just tau(10^(2*m)+1) (order matters). Since changing the order does not lead to a new solution (2*A-10^m is always the even square and 2*B-1 is always the odd square) and since the trivial 10^(2*m)+1 = (10^m)^2 + 1^2 leads to the ambiguous A = 0 and B = 1 there are only tau(10^(2*m)+1)/2-1 relevant ways. Because of the transformation from A to (10^m-A) every of these possibilities leads to a pair of solutions. So the number of solutions with 2*m digits is tau(10^(2*m)+1)-2, see A064943

			8833 = 88^2 + 33^2, 5882353 = 0588^2 + 2353^2.

IBM Corp., IBM March 2000 Challenge solution
C. Rivera, Problem 104 of the prime puzzle pages with other approaches

Cf. A064943 for the number of solutions, A055616 for the solutions where leading zeros are not allowed, A055617, A055618, A055619 for some infinite subsequences and A002654 for finding the number of ways writing an integer as sum of two squares.

A101311 is another version.

with (numtheory): for m from 1 to 10 do: for i in sum2sqr(10^(2*m)+1) do: if i[1] > 1 and i[1] < 10^m then if type(i[1],odd) then a := (i[2]+10^m)/2: b := (i[1]+1)/2: else a := (i[1]+10^m)/2: b := (i[2]+1)/2: fi: print("Length =", 2*m, "Solution =", (10^m-a)*10^m+b): print(Length = 2*m, Solution = a*10^m+b): fi: od: od:

Edited by N. J. A. Sloane, Jul 31 2007

cp's OEIS Frontend

A055616 Numbers, with an even number of digits, that are the sum of the squares of their two halves (leading zeros allowed only for the second half).

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