A350869 -id:A350869 - cp's OEIS frontend

A238237 Numbers which when chopped into two parts with equal length, added and squared result in the same number.

81, 2025, 3025, 9801, 494209, 998001, 24502500, 25502500, 52881984, 60481729, 99980001, 6049417284, 6832014336, 9048004641, 9999800001, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969, 213018248521, 217930248900, 249500250000, 250500250000

Offset: 1

Arkadiusz Wesolowski, Feb 20 2014

Yet another variant of the Kaprekar numbers A006886. - N. J. A. Sloane, Aug 06 2017
From Bernard Schott, Jan 21 2022: (Start)
Three subsequences:
-> {(10^m-1)^2, m >= 1} = A059988 \ {0}; see example 9801.
-> {(10^m-1)^2 * 10^(2*m) / 4, m >= 1} = A350869 \ {0}; see example 2025.
-> {(10^m+1)^2 * 10^(2*m) / 4, m >= 1} = A038544 \ {1}, see example 3025. (End)

			2025 = (20 + 25)^2, so 2025 is in the sequence.
3025 = (30 + 25)^2, so 3025 is in the sequence.
9801 = (98 + 01)^2, so 9801 is in the sequence.

Rémy Sigrist, Table of n, a(n) for n = 1..25000

Subsequence of A102766.
Subsequence: A350870.
Cf. A006886, A038544, A059988, A350869.
For square roots see A290449.

Select[Range[600000]^2, EvenQ[len=IntegerLength[#]] && # == (Mod[#,10^(len/2)] + Floor[#/10^(len/2)])^2 &] (* Stefano Spezia, Jan 01 2025 *)

forstep(m=1, 7, 2, p=10^((m+1)/2); for(n=10^m, 10^(m+1)-1, d=lift(Mod(n, p)); if(((n-d)/p+d)^2==n, print1(n, ", "))));

a(n) = A290449(n)^2. - Bernard Schott, Jan 20 2022

a(12)-a(24) from Donovan Johnson, Feb 22 2014

A038544 a(n) = Sum_{i=0..10^n} i^3.

Original entry on oeis.org

1, 3025, 25502500, 250500250000, 2500500025000000, 25000500002500000000, 250000500000250000000000, 2500000500000025000000000000, 25000000500000002500000000000000, 250000000500000000250000000000000000, 2500000000500000000025000000000000000000

Offset: 0

Marvin Ray Burns

These terms k = x.y satisfy Diophantine equation x.y = (x+y)^2, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0. So, this is a subsequence of A350870 and A238237. - Bernard Schott, Jan 20 2022

			a(1) = Sum_{i=0..10} i^3 = (Sum_{i=0..10} i)^2 = 3025.

Cf. A037156, A238237, A350869, A350870.

sumcu(n) = for(x=0,n,y=10^x;z=y^2*(y+1)^2/4;(print1(z","))) \\ Cino Hilliard, Jun 18 2007

a(n) = (10^n+1)^2 * 10^(2*n) / 4.
From Bernard Schott, Jan 20 2022: (Start)
a(n) = A037156(n)^2.
a(n) = A350869(n) + 10^(3*n). (End)

Edited by N. J. A. Sloane, Jul 02 2008 at the suggestion of R. J. Mathar

A350870 Numbers k = x.y such that x.y = (x+y)^2, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.

Original entry on oeis.org

81, 2025, 3025, 494209, 24502500, 25502500, 52881984, 60481729, 6049417284, 6832014336, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969, 213018248521, 217930248900, 249500250000, 250500250000, 284270248900, 289940248521, 371718237969, 413908229449, 420744227904

Offset: 1

Bernard Schott, Jan 20 2022

Problem proposed on French site Diophante (see link).
We have to solve Diophantine equation (x+y)^2 = x*10^m + y where m = length(x) = length(y).
Squares of a variant of Kaprekar numbers (A045913).
Number of solutions with 2*m digits for m >= 1: 1, 2, 1, 4, 2, 21, ...
Compare with A347541 where x*y divides x.y, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.

			81 = (8+1)^2, hence 81 is a term.
3025 = (30+25)^2, hence 3025 is another term.

Diophante, A1945 - Concaténations en tous genres (in French).

Equals A238237 \ A350918.
Subsequence of A102766.
Subsequences: A038544, A350869.
Cf. A045913, A347541.

upto(n) = {i = 4; i2 = i^2; res = List(); while(i2 <= n, i++; i2 = i^2; if(#digits(i2) % 2 == 1, i = sqrtint(10^(#digits(i2))) + 1; i2 = i^2; ); if(is(i2), listput(res, i2) ); ); res }
is(n) = { my(d = digits(n), d1, d2, frd2); if(#d % 2 == 1, return(0)); d1 = vector(#d \ 2, i, d[i]); d2 = vector(#d \ 2, i, d[i + #d \ 2]); frd2 = fromdigits(d2); 10^(#d \ 2 - 1) <= frd2 && (fromdigits(d1) + frd2)^2 == n } \\ David A. Corneth, Jan 21 2022

a(n) = A045913(n+1)^2.

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A238237 Numbers which when chopped into two parts with equal length, added and squared result in the same number.

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A038544 a(n) = Sum_{i=0..10^n} i^3.

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A350870 Numbers k = x.y such that x.y = (x+y)^2, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.

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Examples

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PARI

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