cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Original entry on oeis.org

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

Views

Author

Arkadiusz Wesolowski, Feb 20 2014

Keywords

Comments

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)

Examples

			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.
		

Crossrefs

Subsequence of A102766.
Subsequence: A350870.
For square roots see A290449.

Programs

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

Formula

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

Extensions

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

Views

Author

Keywords

Comments

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

Examples

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

Crossrefs

Programs

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

Formula

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)

Extensions

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

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

Views

Author

Bernard Schott, Jan 20 2022

Keywords

Comments

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.

Examples

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

Crossrefs

Programs

  • Mathematica
    a[n_] := (10^n*(10^n - 1)/2)^2; Array[a, 11, 0] (* Amiram Eldar, Jan 20 2022 *)
  • PARI
    a(n) = my(x=10^n-1); (x*(x+1)/2)^2; \\ Michel Marcus, Jan 22 2022

Formula

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

Views

Author

Bernard Schott, Jan 22 2022

Keywords

Comments

Problem proposed on French site Diophante (see link).

Examples

			(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.
		

Crossrefs

Equals A238237 \ A350870.
A059988 \ {0, 81} is a subsequence.
Showing 1-4 of 4 results.