A348832 Positive numbers whose square starts and ends with exactly 444.

666462, 666538, 666962, 667038, 2107462, 2107538, 2107962, 2108038, 2108462, 2108538, 2108962, 2109038, 2109462, 6663462, 6663538, 6663962, 6664038, 6664462, 6664538, 6664962, 6665038, 6665462, 6665538, 6665962, 6666038, 6667462, 6667538, 6667962, 6668038, 6668462, 6668538, 6668962

Offset

1,1

Comments

The 1st problem of British Mathematical Olympiad (BMO) in 1995 (see link) asked to find all positive integers whose squares end in three 4’s (A039685); this sequence is the subsequence of these integers whose squares also start in precisely three 4's (no four or more 4's). Two such infinite subsequences are proposed below.

When a square starts and ends with digits ddd, then ddd is necessarily 444.

The first 3 digits of terms are either 210, 666 or 667, while the last 3 digits are either 038, 462, 538 or 962 (see examples).

From Marius A. Burtea, Nov 09 2021 : (Start)

The sequence is infinite because the numbers 667038, 6670038, 66700038, 667000038, ..., 667*10^k + 38, k >= 3, are terms because are square 444939693444, 44489406921444, 4448895069201444, 444889050692001444, 44488900506920001444, ...

Also, 6663462, 66633462, 666333462, 6663333462, ..., (1999*10^k + 386) / 3, k >= 4, are terms and have no digits 0, because their squares are 44401725825444, 4440018258105444, 444000282580905444, 44400012825808905444,

4440001128258088905444, ... (End)

References

A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 1 pp. 55 and 95-96 (1995)

Links

Table of n, a(n) for n=1..32.

British Mathematical Olympiad 1975, Problem 1.

Example

666462 is a term since 666462^2 = 444171597444.
21038 is not a term since 21038^2 = 442597444.

Mathematica

Select[Range[100, 7*10^6], (d = IntegerDigits[#^2])[[1 ;; 3]] == d[[-3 ;; -1]] == {4, 4, 4} && d[[-4]] != 4 && d[[4]] != 4 &] (* Amiram Eldar, Nov 09 2021 *)

Prog

(Python)
from itertools import count, takewhile
def ok(n):
  s = str(n*n); return len(s.rstrip("4")) == len(s.lstrip("4")) == len(s)-3
def aupto(N):
  ends = [38, 462, 538, 962]
  r = takewhile(lambda x: x<=N, (1000*i+d for i in count(0) for d in ends))
  return [k for k in r if ok(k)]
print(aupto(6668962)) # Michael S. Branicky, Nov 09 2021
(Magma) fd:=func<n|Seqint(Intseq(n*n)) mod 1000 eq 444 and Seqint(Intseq(n*n)) mod 10000 ne 4444>; fs:=func<n|Seqint(Reverse(Intseq(n*n))) mod 1000 eq 444 and Seqint(Reverse(Intseq(n*n))) mod 10000 ne 4444>; [n:n in [1..6700000]|fd(n) and fs(n)]; // Marius A. Burtea, Nov 09 2021

Crossrefs

Cf. A017317, A328886.

Subsequence of A039685, A045858, A273375, A305719, A346892.

Similar to: A348488 (d=4), A348831 (dd=44), this sequence (ddd=444).

Sequence in context: A144132 A069373 A270801 * A162840 A233817 A204887

Adjacent sequences: A348829 A348830 A348831 * A348833 A348834 A348835

Keyword

nonn,base

Author

Bernard Schott, Nov 09 2021

Status

approved