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 1-5 of 5 results.

A068536 Numbers m such that m^2 + (reversal of m)^2 is a square. (Leading 0's are ignored.)

Original entry on oeis.org

88209, 90288, 125928, 196020, 368280, 829521, 1978020, 2328480, 5513508, 8053155, 19798020, 86531940, 197998020, 554344560, 556326540, 1960396020, 1979998020, 5543944560, 5925169800, 8820988209, 9028890288, 12592925928, 14011538112, 19602196020, 19799998020
Offset: 1

Views

Author

Joseph L. Pe, Mar 22 2002

Keywords

Comments

The sequence is infinite, even if it is restricted to terms that end with a nonzero digit, as any term generates an infinite number of other terms by the following scheme: If m is a term of the sequence and d(m) denotes the number of digits of m, then set m' = m*10^d'+m with d' >= d(m). For d' >= d(m) we have reverse(m') = reverse(m)*10^d' + reverse(m) and thus (m')^2 + reverse(m')^2 = (m*10^d'+m)^2 + (reverse(m)*10^d'+reverse(m))^2 = (m^2+reverse(m)^2)*(10^d'+1)^2. As m^2+reverse(m)^2 is a perfect square by assumption, the product on the right-hand side of the equation is also a perfect square and m' is part of the sequence. The calculation works also with m' = m*(10^(k*d')+...+10^d'+1). As an example take a(1)=88209. All numbers 8820988209, 882098820988209, 88209882098820988209, ... and 88209088209, 882090088209, 8820900088209, ... are also terms of the sequence. - Matthias Baur, May 01 2020

Examples

			88209^2 + 90288^2 = 126225^2, so 88209 belongs to the sequence.
		

Crossrefs

Cf. A004086 (digit reversal), A202386, A256515.

Programs

  • Mathematica
    Do[If[IntegerQ[Sqrt[n^2 + FromDigits[Reverse[IntegerDigits[n]]]^2]], Print[n]], {n, 1, 10^6}]

Extensions

a(7)-a(15) from Donovan Johnson, Apr 09 2010
a(16)-a(25) from Donovan Johnson, Jul 15 2011

A256515 Nonpalindromic positive integers k such that the absolute value of k^2 - reverse(k)^2 is a square.

Original entry on oeis.org

56, 65, 5265, 5625, 5656, 6565, 12705, 44370, 50721, 51557, 55517, 56056, 59248, 65065, 71555, 75515, 84295, 139755, 273728, 360145, 481610, 523908, 541063, 557931, 560056, 560439, 565656, 606056, 621770, 650065, 650606, 656565, 697996, 699796, 809325, 827372
Offset: 1

Views

Author

Bui Quang Tuan, Apr 01 2015

Keywords

Examples

			The nonpalindromic number 5265 is a term because abs(5265^2 - 5625^2) = 1980^2.
		

Crossrefs

Cf. A004086 (digit reversal), A202386, A068536.

Programs

  • Magma
    [n: n in [0..10^6] | Intseq(n) ne Reverse(Intseq(n)) and IsSquare(s) where s is Abs(n^2-Seqint(Reverse(Intseq(n)))^2)]; // Bruno Berselli, Apr 01 2015
    
  • Mathematica
    Select[Range[200000], ! PalindromeQ@ # && IntegerQ@ Sqrt@ Abs[#^2 - IntegerReverse[#]^2] &] (* Michael De Vlieger, Mar 02 2022 *)
  • Python
    from sympy.ntheory.primetest import is_square
    def R(n): return int(str(n)[::-1])
    def ok(n): Rn = R(n); return n != Rn and is_square(abs(n**2 - Rn**2))
    print([k for k in range(10**6) if ok(k)]) # Michael S. Branicky, Mar 02 2022

A113798 Numbers k such that k^2 plus the reverse of k is a square.

Original entry on oeis.org

117, 119817, 13101687, 119819817, 13101801687, 119819819817, 11983101689817, 13101801801687, 119819819819817
Offset: 1

Views

Author

Giovanni Resta, Jan 21 2006

Keywords

Comments

a(10) > 1.36*10^14. The sequence contains two infinite sets, 117*1000^k + 2817*(1000^k - 1)/999 and 13101687*1000^k + 114687*(1000^k - 1)/999, for k >= 0. Terms of both sets satisfy the equation rev(x) = 6*x + 9, and thus x^2 + rev(x) = (x+3)^2. - Giovanni Resta, Aug 26 2019

Examples

			117^2 + 711 = 120^2, thus 117 is a term.
		

Crossrefs

Extensions

a(5)-a(6) from Giovanni Resta, May 10 2017
a(7)-a(9) from Giovanni Resta, Aug 26 2019

A289140 Positive numbers k such that rev(k)^2 + rev(k^2) is a square, where rev(n) = A004086(n) is the digital reverse of n.

Original entry on oeis.org

998586, 3632658, 9985860, 36326580, 74471091, 99664458, 99858600, 363265800, 634826115, 743193501, 744710910, 756335085, 759317343, 996644580, 998586000, 3632658000, 6348261150, 7177621788, 7431935010, 7447109100, 7563350850, 7593173430, 9966445800
Offset: 1

Views

Author

Giovanni Resta, Jun 26 2017

Keywords

Comments

Every term must be a multiple of 3.

Examples

			998586 is a term since rev(998586^2) + 685899^2 = 1079100^2.
		

Crossrefs

Programs

  • Mathematica
    rev[n_] := FromDigits@ Reverse@ IntegerDigits@ n; Parallelize@ Select[3 Range[4 10^6], IntegerQ@ Sqrt[rev[#^2] + rev[#]^2] &]
  • PARI
    isok(n) = issquare(fromdigits(Vecrev(digits(n)))^2 + fromdigits(Vecrev(digits(n^2)))); \\ Michel Marcus, Jun 29 2017

A332850 Numbers k = a^2 + b^2 such that reversal(k) = a^2 - b^2 for a > b > 0, where reversal is A004086.

Original entry on oeis.org

699796, 4854634, 6752626, 84036010, 931910661, 21584860960, 52554850525, 467170024564, 637843128736, 638730439636, 638734039636, 638943127636, 727830438745, 727834038745, 746710459825, 746754019825, 748943127625, 9894192267061, 401309596403104, 844181015028970
Offset: 1

Views

Author

Metin Sariyar, Feb 26 2020

Keywords

Comments

When b=0, the palindromic numbers m = a^2 + b^2 such that reversal(m) = a^2 - b^2, are A002779 (palindromic squares).
a(19) > 3*10^14, if it exists. - Giovanni Resta, Feb 27 2020

Examples

			699796 = 836^2 + 30^2 and 697996 = 836^2 - 30^2.
		

Crossrefs

Programs

  • Mathematica
    Do[If[IntegerReverse[a^2+b^2]==a^2-b^2,Print[{a^2+b^2,a,b}]],{a,1,50000},{b,1,a-1}]
  • PARI
    isok(k) = {my(r = fromdigits(Vecrev(digits(k))), s = r+k, d = k-r); d && !(s % 2) && issquare(s/2) && !(d % 2) && issquare(d/2); } \\ Michel Marcus, Feb 27 2020

Extensions

a(6)-a(18) from Giovanni Resta, Feb 27 2020
a(19)-a(20) from Jinyuan Wang, Apr 10 2025
Showing 1-5 of 5 results.