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

Author

Patrick De Geest

Keywords

• nonn
• base

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.

Crossrefs

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

Programs

Mathematica

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

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

Formula

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

Extensions

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

a(18) from Iain Fox, Dec 27 2017