A064943 -id:A064943 - cp's OEIS frontend

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).

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

A258482 Positive numbers n with concatenations n=x//y such that n=x^2-y^2.

Original entry on oeis.org

100, 147, 10000, 13467, 1000000, 1010100, 1016127, 1034187, 1140399, 1190475, 1216512, 1300624, 1334667, 1416767, 1484847, 1530900, 100000000, 102341547, 102661652, 116604399, 133346667, 159809775, 10000000000, 10101010100, 13333466667, 14848484847

Offset: 1

Pieter Post, May 31 2015

The terms in this sequence have only an odd number of digits. If they would have an even number of digits both parts would have the same length. The maximum difference x^2 - y^2 would be (10^m-1)^2 - 1^2, which is (10^m-2)*10^m. But this is always less than (10^m-1)^2 + 1, so m never equals x^2 - y^2.
For example m=3: 999^2 - 1^2 < 999001.
The terms in this sequence all start with the digit '1'. Suppose they would start with the digit '2' (or more) the smallest possiblity of x^2 - y^2 would be (2*10^m)^2 - (10^m-1)^2 = 3*10^2*m + 2*10^m-1, but this is always more than 2*10^2*m + 10^3-1, so m never equals x^2 - y^2.
For example m=3: 2000^2 - 999^2 > 2000999.
This sequence has an infinite subsequence, since (10^m+(10^m+2)/3)*10^m+(2*10^m+1)/3 equals (10^m+(10^m+2)/3)^2 - ((2*10^m+1)/3)^2 for every positive m.
For example m=3: 1334667 = 1334^2 - 667^2.
This set is a subset of A113797.

			147 is a member, since 147 = 14^2 - 7^2.
1484847 is a member, since 1484847 = 1484^2- 847^2.
48 is a member of A113797 since 48 = |4^2 - 8^2|, but 48 is not equal to 4^2 - 8^2, so 48 is not a member of this sequence.

Giovanni Resta, Table of n, a(n) for n = 1..4397 (terms < 10^60)

Cf. A002654, A055616, A064942, A064943, A113797, A101311, A162700, A257796.

isok(n) = {d = digits(n); if (#d > 1, for (k=1, #d-1, vba = Vecrev(vector(k, i, d[i])); vbb = Vecrev(vector(#d-k, i, d[k+i])); da = sum(i=1, #vba, vba[i]*10^(i-1)); db = sum(i=1, #vbb, vbb[i]*10^(i-1)); if (da^2 - db^2 == n, return(1));););} \\ Michel Marcus, Jun 14 2015

for p in range(1, 7):
    for i in range(10**p, 10**(p + 1)):
        c = 10**(int((p - 1) / 2) + 1)
        a, b = i // c, i % c
        if i == a**2 - b**2:
            print(i, end=",")

n=x*10^d+y, where 10^(d-1)<=x<10^d and 0<=y<10^d and n=x^2-y^2.

More terms from Giovanni Resta, Jun 14 2015

cp's OEIS Frontend

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).

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