OFFSET
1,1
COMMENTS
Also, numbers N associated with A106497.
Also, numbers k such that k concatenated with k-1 gives the product of two numbers which differ by 2. E.g., 13223140496//13223140495 = 36363636363 * 36363636365, where // denotes concatenation. - Giovanni Resta and Franklin T. Adams-Watters, Nov 13 2006
From Jianing Song, Nov 01 2024: (Start)
Numbers 10^(k-1) <= a <= 10^k - 1 such that a*(10^k + 1) is a square. Note that 10^k + 1 must be nonsquarefree, i.e., k is in A086982, otherwise a must be divisible by 10^k + 1, which is impossible.
Let v(p,m) be the p-adic valuation of m.
- If p is not in A045616, then v(p,10^k+1) = r > 0 if and only if v(p,gcd(n,10^k+1)) = r-1.
- If p is in A045616, let e be the multiplicative order of 10 modulo p, then v(p,10^k+1) > 0 if and only if e is even and k is an odd multiple of e/2, in which case v(p,10^k+1) = v(p,10^e-1) + v(p,k) = v(p,10^e-1) + v(p,gcd(k,10^k+1)).
This helps to find the terms. (End)
REFERENCES
Andrew Bridy, Robert J. Lemke Oliver, Arlo Shallit, and Jeffrey Shallit, The Generalized Nagell-Ljunggren Problem: Powers with Repetitive Representations, Experimental Math, 28 (2019), 428-439.
R. Ondrejka, Problem 1130: Biperiod Squares, Journal of Recreational Mathematics, Vol. 14:4 (1981-82), 299. Solution by F. H. Kierstead, Jr., JRM, Vol. 15:4 (1982-83), 311-312.
LINKS
David W. Wilson, Table of n, a(n) for n = 1..1098
Dr Barker, Can Numbers Like These Be Square?, YouTube video, 2023.
Andrew Bridy, Robert J. Lemke Oliver, Arlo Shallit, and Jeffrey Shallit, The Generalized Nagell-Ljunggren Problem: Powers with Repetitive Representations, preprint arXiv:1707.03894 [math.NT], July 14 2017.
EXAMPLE
13223140496 concatenated with 13223140496 is 1322314049613223140496 = 36363636364^2.
40495867769 is in the sequence because writing it twice gives the square number 4049586776940495867769 = 63636363637^2.
MAPLE
with(numtheory): Digits:=50:for d from 1 to 35 do tendp1:=10^d+1: tendp1fact:=ifactors(tendp1)[2]: n:=mul(piecewise(tendp1fact[i][2] mod 2=1, tendp1fact[i][1], 1), i=1..nops(tendp1fact)):for i from ceil(sqrt((10^(d-1))/n)) to floor(sqrt((10^d-1)/n)) do printf("%d, ", n*i^2) od: od:
MATHEMATICA
A102567L[n_] := Catenate@Table[Module[{fac = FactorInteger[10^k + 1], min}, If[Max@fac[[All, -1]] == 1, {}, min = Times @@ Cases[fac, {a_, _?OddQ} :> a]; Table[min s^2, {s, Ceiling@Sqrt[10^(k - 1)/min], Floor@Sqrt[(10^k - 1)/min]}]]], {k, n}]; A102567L[30] (* JungHwan Min, Dec 11 2016 *)
A102567Q = IntegerQ@Sqrt@FromDigits[Join[#, #] &@IntegerDigits[#]] & (* JungHwan Min, Dec 11 2016 *)
PROG
(Python)
from itertools import count, islice
from sympy import sqrt_mod
def A102567_gen(): # generator of terms
for j in count(0):
b = 10**j
a = b*10+1
for k in sorted(sqrt_mod(0, a, all_roots=True)):
if a*b <= k**2 < a*(a-1):
yield k**2//a
(PARI) p = [3, 487, 56598313]; \\ A045616
b(n) = my(d = gcd(n, lift(Mod(10, n)^n)+1), s = 1); for(j=1, #p, my(e = znorder(Mod(10, p[j]))); if((e % 2 == 0) && (n % (e/2) == 0) && (n/(e/2) % 2 == 1), my(v = valuation(d, p[j])); d /= p[j]^v; s *= p[j]^((v+valuation(10^e-1, p[j]))\2))); my(f = factor(d)); for(i=1, #f~, s *= f[i, 1]^((f[i, 2]+1)\2)); s; \\ giving s such that 10^n + 1 = s^2*t where t is squarefree, considering only the three already-known terms of A045616
A102567_length_n(n) = my(t = (10^n+1)/b(n)^2, lowlim = 1+sqrtint(10^(n-1)\t), uplim = sqrtint((10^n-1)\t)); vector(uplim-lowlim+1, i, (lowlim-1+i)^2 * t) \\ terms of the form a^2*t such that 10^(n-1) <= a^2*t <= 10^n - 1
\\ Jianing Song, Nov 01 2024
CROSSREFS
KEYWORD
easy,nonn,base
AUTHOR
C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 15 2005
EXTENSIONS
Entry revised by N. J. A. Sloane, Nov 14 2006 and also Nov 27 2006
Definition edited and reference added by William Rex Marshall, Nov 12 2010
STATUS
approved