cp's OEIS Frontend

A102567 Numbers k such that the concatenation of k with itself is a biperiod square.

%I A102567 #42 Nov 01 2024 23:49:27
%S A102567 13223140496,20661157025,29752066116,40495867769,52892561984,
%T A102567 66942148761,82644628100,183673469387755102041,326530612244897959184,
%U A102567 510204081632653061225,734693877551020408164
%N A102567 Numbers k such that the concatenation of k with itself is a biperiod square.
%C A102567 Also, numbers N associated with A106497.
%C A102567 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
%C A102567 From _Jianing Song_, Nov 01 2024: (Start)
%C A102567 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.
%C A102567 Let v(p,m) be the p-adic valuation of m.
%C A102567  - 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.
%C A102567  - 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)).
%C A102567 This helps to find the terms. (End)
%D A102567 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.
%D A102567 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.
%H A102567 David W. Wilson, <a href="/A102567/b102567.txt">Table of n, a(n) for n = 1..1098</a>
%H A102567 Dr Barker, <a href="https://www.youtube.com/watch?v=c1peEN5Q-0c">Can Numbers Like These Be Square?</a>, YouTube video, 2023.
%H A102567 Andrew Bridy, Robert J. Lemke Oliver, Arlo Shallit, and Jeffrey Shallit, <a href="https://arxiv.org/abs/1707.03894">The Generalized Nagell-Ljunggren Problem: Powers with Repetitive Representations</a>, preprint arXiv:1707.03894 [math.NT], July 14 2017.
%e A102567 13223140496 concatenated with 13223140496 is 1322314049613223140496 = 36363636364^2.
%e A102567 40495867769 is in the sequence because writing it twice gives the square number 4049586776940495867769 = 63636363637^2.
%p A102567 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:
%t A102567 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 *)
%t A102567 A102567Q = IntegerQ@Sqrt@FromDigits[Join[#, #] &@IntegerDigits[#]] & (* _JungHwan Min_, Dec 11 2016 *)
%o A102567 (Python)
%o A102567 from itertools import count, islice
%o A102567 from sympy import sqrt_mod
%o A102567 def A102567_gen(): # generator of terms
%o A102567     for j in count(0):
%o A102567         b = 10**j
%o A102567         a = b*10+1
%o A102567         for k in sorted(sqrt_mod(0,a,all_roots=True)):
%o A102567             if a*b <= k**2 < a*(a-1):
%o A102567                 yield k**2//a
%o A102567 A102567_list = list(islice(A102567_gen(),10)) # _Chai Wah Wu_, Feb 19 2024
%o A102567 (PARI) p = [3, 487, 56598313]; \\ A045616
%o A102567 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
%o A102567 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
%o A102567 \\ _Jianing Song_, Nov 01 2024
%Y A102567 Cf. A092118, A054214, A116163, A116136, A116279.
%K A102567 easy,nonn,base
%O A102567 1,1
%A A102567 C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 15 2005
%E A102567 Entry revised by _N. J. A. Sloane_, Nov 14 2006 and also Nov 27 2006
%E A102567 Definition edited and reference added by _William Rex Marshall_, Nov 12 2010