A030466 Squares that are concatenations of two consecutive nonzero numbers.

183184, 328329, 528529, 715716, 60996100, 1322413225, 4049540496, 106755106756, 453288453289, 20661152066116, 29752082975209, 2214532822145329, 2802768328027684, 110213248110213249, 110667555110667556, 147928995147928996, 178838403178838404, 226123528226123529

Offset

1,1

References

British Mathematical Olympiad, 1993, Round 1, Question 1: "Find, showing your method, a six-digit integer n with the following properties: (i) n is a perfect square, (ii) the number formed by the last three digits of n is exactly one greater than the number formed by the first three digits of n. (Thus n might look like 123124, although this is not a square.)"

Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 1 of the British Mathematical Olympiad 1993, page 164.

Links

Iain Fox, Table of n, a(n) for n = 1..10471.

British Mathematical Olympiad 1993, Round 1, Problem 1.

Michael Penn, British Mathematics Olympiad 1993 Round 1 Question 1, YouTube, Apr 24, 2020.

Pante Stanica, Squares as concatenation of consecutive integers, Slides, West Coast Number Theory, Dec 17 2017.

Index to sequences related to Olympiads.

Formula

a(n) = A030465(n)*(10^A055642(A030465(n))+1)+1. - Iain Fox, Oct 16 2021

Mathematica

fQ[n_] := IntegerQ[Sqrt[n*10^Floor[1 + Log10[n + 1]] + n + 1]]; (* Robert G. Wilson v, Dec 27 2017 *)

Prog

(PARI) lista(nn) = forstep(n=183, nn, [3, 5, 7, 5, 3, 1, 4, 7, 5, 3, 5, 7, 5, 3, 5, 7, 5, 3, 5, 7, 4, 1], my(s = eval(concat(Str(n), Str(n+1)))); if(issquare(s), print1(s, ", "))) \\ Iain Fox, Dec 27 2017
(PARI) eea(x, y) = my(a=max(x, y), b=min(x, y), s=0, so=1, st, r=b, ro=a, rt, q, t); while(r, q=ro\r; rt=r; r=ro-q*r; ro=rt; st=s; s=so-q*s; so=st); t=(ro-so*a)\b; if(x>y, [so, t], [t, so]) \\ Extended Euclidean Algorithm
lista(nn) = my(res=Set(), b, f2, c, s); for(d=3, nn, b=10^d+1; fordiv(b, f, if(f!=1 && f!=b, f2=b/f; if(gcd(f, f2)==1, c=eea(f, f2); if(c[1]<0, s=f*(f2+2*c[1])*f2*(f-2*c[2])+1, s=f*(2*c[1])*f2*(-2*c[2])+1); if(#digits(s)==d*2, res=setunion(res, Set(s))))))); Vec(res) \\ (Will find all values of length nn*2 or shorter) Iain Fox, Oct 16 2021

Crossrefs

Cf. A000993, A030465, A030467, A054214, A054215, A054216, A020339, A020340.

Sequence in context: A250013 A258842 A368939 * A233956 A249232 A086478

Adjacent sequences: A030463 A030464 A030465 * A030467 A030468 A030469

Keyword

nonn,base

Author

Patrick De Geest

Extensions

a(15)-a(17) from Arkadiusz Wesolowski, Apr 02 2014

a(18) from Iain Fox, Dec 27 2017

Status

approved