A275986 Positive integers of the form x*10^k + y which also equal x^2 + y^2 (x, y and k being positive integers).

101, 1233, 8833, 10001, 10100, 990100, 1000001, 5882353, 94122353, 99009901, 100000001, 100010000, 1765038125, 2584043776, 7416043776, 8235038125, 9901009901, 10000000001, 48600220401, 116788321168, 123288328768, 601300773101, 876712328768, 883212321168, 990100990100, 999900010000, 1000000000001, 1000001000000

Offset: 1

Author

Douglas E. Iannucci, Aug 15 2016

Keywords

• nonn

Comments

The condition x^2 + y^2 = x*10^k + y is equivalent to (2x-10^k)^2 + (2y-1)^2 = 10^2k + 1, so to find these sequence elements it is necessary to write 10^2k + 1 as the sum of two squares.

The number of elements in this sequence corresponding to a fixed k is tau(10^2k + 1) - 1, where tau counts the (positive) divisors of a natural number. For all k, 10^2k + 1 is itself a member of the sequence corresponding to k, and is the only one such if it is prime. The elements themselves are arranged according to magnitude, indexed here by n. There is some disruption of the order of the terms versus the corresponding exponent k. For example, the twelfth member of the sequence, 100010000, corresponds to k=6, yet the thirteenth, 1765038125, corresponds to the smaller k=5.

Contains 10^(2*i) + 10^(4*i) and 10^(6*i) - 10^(4*i) + 10^(2*i) for each i >= 1 (corresponding to k = 3*i). - Robert Israel, Mar 30 2017

Examples

			a(1) = 101 corresponds to k = 1, x = 10, and y = 1.
a(2) = 1233 corresponds to k = 2, x = 12, y = 33.
		

Links

• Steven Charlton, Square sum concatenation - Number theory challenge
• A. van der Poorten, K. Thomsen, and M. Wiebe, A curious cubic identity and self-similar sums of squares, The Mathematical Intelligencer, v.29(2), pp. 69-73, June 2007.