A238237 -id:A238237 - cp's OEIS frontend

A290449 Square roots of terms in A238237.

9, 45, 55, 99, 703, 999, 4950, 5050, 7272, 7777, 9999, 77778, 82656, 95121, 99999, 318682, 329967, 351352, 356643, 390313, 461539, 466830, 499500, 500500, 533170, 538461, 609687, 643357, 648648, 670033, 681318, 791505, 812890, 818181, 851851, 857143, 961038

Offset: 1

N. J. A. Sloane, Aug 06 2017, following a suggestion from Max Alekseyev

Yet another variant of the Kaprekar numbers A006886.

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

Cf. A238237, A006886.

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

A059988 a(n) = (10^n - 1)^2.

Original entry on oeis.org

0, 81, 9801, 998001, 99980001, 9999800001, 999998000001, 99999980000001, 9999999800000001, 999999998000000001, 99999999980000000001, 9999999999800000000001, 999999999998000000000001, 99999999999980000000000001, 9999999999999800000000000001, 999999999999998000000000000001

Offset: 0

Henry Bottomley, Mar 07 2001

From James D. Klein, Feb 05 2012: (Start)
The periods of the reciprocals of a(n) are the consecutive integers from 0 to 10^n-1, omitting the one integer 10^n-2, right-justified in field widths of size n.
E.g.:
  1/81 = 0.012345679...
  1/9801 = 0.000102030405060708091011...9799000102...
  1/998001 = 0.000001002003004005...997999000001002... (End)
Sum of first 10^n - 1 odd numbers. - Arkadiusz Wesolowski, Jun 12 2013

			From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 81 = 9^2;
n=2: ................... 9801 = 99^2;
n=3: ................. 998001 = 999^2;
n=4: ............... 99980001 = 9999^2;
n=5: ............. 9999800001 = 99999^2;
n=6: ........... 999998000001 = 999999^2;
n=7: ......... 99999980000001 = 9999999^2;
n=8: ....... 9999999800000001 = 99999999^2;
n=9: ..... 999999998000000001 = 999999999^2. (End)

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 32 at p. 61.
Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 34.

Harry J. Smith, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417, A178630, A178631, A178632, A178633, A178634, A178635, A272066, A272067, A272068.
Subsequence of A238237.
Cf. A002275, A002283, A002477.

A059988:=n->(10^n-1)^2; seq(A059988(n), n=0..20); # Wesley Ivan Hurt, Nov 21 2013

Table[(10^n-1)^2, {n,0,20}] (* Wesley Ivan Hurt, Nov 21 2013 *)

a(n) = { (10^n - 1)^2 } \\ Harry J. Smith, Jul 01 2009

def a(n): return (10**n - 1)**2  # Martin Gergov, Sep 10 2022

a(n) = 81*A002477(n) = A002283(n)^2 = (9*A002275(n))^2.
a(n) = {999... (n times)}^2 = {999... (n times), 000... (n times)} - {999... (n times)}. For example, 999^2 = 999000 - 999 = 998001. - Kyle D. Balliet, Mar 07 2009
a(n) = (A002283(n-1)*10 + 8) * 10^(n-1) + 1, for n>0. - Reinhard Zumkeller, May 31 2010
From Ilya Gutkovskiy, Apr 19 2016: (Start)
O.g.f.: 81*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
E.g.f.: (1 - 2*exp(9*x) + exp(99*x))*exp(x). (End)
Sum_{n>=1} 1/a(n) = (log(10)*(QPolyGamma(0, 1, 1/10) - log(10/9)) + QPolyGamma(1, 1, 1/10))/log(10)^2 = 0.012448721523422795191... . - Stefano Spezia, Jul 31 2024
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3). - Elmo R. Oliveira, Aug 02 2025

A102766 Numbers n that can be chopped into two parts, which when added and squared result in n.

Original entry on oeis.org

1, 81, 100, 2025, 3025, 9801, 10000, 88209, 494209, 998001, 1000000, 4941729, 7441984, 23804641, 24502500, 25502500, 28005264, 52881984, 60481729, 99980001, 100000000, 300814336, 493817284, 1518037444, 6049417284, 6832014336, 9048004641, 9999800001, 10000000000

Offset: 1

Bodo Zinser, Feb 10 2005

			a(7) = 88209 is a term as (88+209)^2 = 297^2 = 88209.

Michael S. Branicky, Table of n, a(n) for n = 1..131
Henry Ernest Dudeney, Amusements in Mathematics, 1917. See problem 113, "The torn number".

Supersequence of A238237.

from math import isqrt
from itertools import count, islice
def ok(n):
    if n == 1: return True
    r = isqrt(n)
    if r**2 != n: return False
    s = str(n)
    return any(int(s[:i])+int(s[i:])== r for i in range(1, len(s)))
def agen(): yield from (k**2 for k in count(1) if ok(k**2))
print(list(islice(agen(), 29))) # Michael S. Branicky, Dec 29 2024

a(n) = A248353(n)^2.

a(1)=1 prepended by Max Alekseyev, Aug 04 2017

a(27) and beyond from Michael S. Branicky, Dec 29 2024

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.

A350869 a(n) = Sum_{i=0..10^n-1} i^3.

Original entry on oeis.org

0, 2025, 24502500, 249500250000, 2499500025000000, 24999500002500000000, 249999500000250000000000, 2499999500000025000000000000, 24999999500000002500000000000000, 249999999500000000250000000000000000, 2499999999500000000025000000000000000000

Offset: 0

Bernard Schott, Jan 20 2022

These terms k = x.y satisfy 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.

			a(1) = Sum_{i=0..9} i^3 = (Sum_{i=0..9} i)^2 = 2025.

Index entries for linear recurrences with constant coefficients, signature (11100,-11100000,1000000000).

Cf. A000217, A037156, A037182, A038544, A238237.

a[n_] := (10^n*(10^n - 1)/2)^2; Array[a, 11, 0] (* Amiram Eldar, Jan 20 2022 *)

a(n) = my(x=10^n-1); (x*(x+1)/2)^2; \\ Michel Marcus, Jan 22 2022

a(n) = 10^(2n) * (10^n-1)^2 / 4 = A037182(n)^2.
a(n) = A000217(10^n-1)^2.
a(n) = A038544(n) - 10^(3*n).

A350918 Numbers k = x.y which when split into two parts x and y of equal length, added and squared result in the same number k, '.' means concatenation, and the second part y starts with 0.

Original entry on oeis.org

9801, 998001, 99980001, 9048004641, 9999800001, 923594037444, 989444005264, 999998000001, 7901234409876544, 8434234407495744, 8934133805179209, 9999999800000001, 999999998000000001, 79012345680987654321, 82644628100826446281, 83407877440792003584, 87138706300620940900

Offset: 1

Bernard Schott, Jan 22 2022

Problem proposed on French site Diophante (see link).

			(998+001)^2 = 999^2 = 998001, as x = 998 and y = 001 starts with 0, 998001 is a term.
(30+25)^2 = 55^2 = 3025, here x = 30 but y = 25 does not start with 0, hence 3025 is not a term.

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

Equals A238237 \ A350870.

A059988 \ {0, 81} is a subsequence.

cp's OEIS Frontend

A290449 Square roots of terms in A238237.

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A059988 a(n) = (10^n - 1)^2.

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A102766 Numbers n that can be chopped into two parts, which when added and squared result in n.

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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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A350869 a(n) = Sum_{i=0..10^n-1} i^3.

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A350918 Numbers k = x.y which when split into two parts x and y of equal length, added and squared result in the same number k, '.' means concatenation, and the second part y starts with 0.

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