A054214 -id:A054214 - cp's OEIS frontend

A116123 Duplicate of A054214.

82, 8242, 9802, 538277, 998002, 77837026, 99980002, 7922547265, 8643251345, 9223797610, 9999800002, 106710893290, 453378226757, 491023832065, 945958034530, 999998000002, 11916002265170, 15790977390245

Offset: 1

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A116142 Duplicate of A054214.

Original entry on oeis.org

82, 8242, 9802, 538277, 998002, 77837026, 99980002, 7922547265, 8643251345, 9223797610, 9999800002, 106710893290, 453378226757, 491023832065, 945958034530, 999998000002, 11916002265170, 15790977390245

Offset: 1

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

Original entry on oeis.org

13223140496, 20661157025, 29752066116, 40495867769, 52892561984, 66942148761, 82644628100, 183673469387755102041, 326530612244897959184, 510204081632653061225, 734693877551020408164

Offset: 1

C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 15 2005

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)

			13223140496 concatenated with 13223140496 is 1322314049613223140496 = 36363636364^2.
40495867769 is in the sequence because writing it twice gives the square number 4049586776940495867769 = 63636363637^2.

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.

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.

Cf. A092118, A054214, A116163, A116136, A116279.

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:

A102567L[n_] := Catenate@Table[Module[{fac = FactorInteger[10^k + 1], min}, If[Max@fac[[All, -1]] == 1, {}, min = Times @@ Cases[fac, {a_, A102567L%5B30%5D%20(*%20_JungHwan%20Min">?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 *)

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

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
A102567_list = list(islice(A102567_gen(),10)) # Chai Wah Wu, Feb 19 2024

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

A030467 Numbers k such that k^2 is a concatenation of two successive numbers.

Original entry on oeis.org

428, 573, 727, 846, 7810, 36365, 63636, 326734, 673267, 4545454, 5454547, 47058823, 52941178, 331983807, 332667334, 384615386, 422892898, 475524477, 524475524, 577107103, 615384615, 667332667, 668016194, 719964246, 758241758, 804511280, 810873337, 857142859

Offset: 1

Patrick De Geest

			428^2 = 183184, the concatenation of 183 and 184.

Chai Wah Wu, Table of n, a(n) for n = 1..10471 (terms n = 1..53 from Donovan Johnson, terms n = 54..1000 from Giovanni Resta)
Pante Stanica, Squares as concatenation of consecutive integers, Slides, West Coast Number Theory, Dec 17 2017.

Cf. A030465, A030466, A054214, A054215, A054216, A020339, A020340.

t={}; Do[If[EvenQ[y=Length[x=IntegerDigits[n^2]]] && Differences[FromDigits/@Partition[x,y/2]]=={1},AppendTo[t,n]],{n, 5.5*10^6}]; t (* Jayanta Basu, May 25 2013 *)
Sqrt[#]&/@(Select[FromDigits[Flatten[IntegerDigits/@#]]&/@ (Partition[ Range[735*10^6],2,1]),IntegerQ[Sqrt[#]]&]) (* The program takes a long time to run. *) (* Harvey P. Dale, Oct 10 2017 *)

for(n=1, 10^9, t=eval(concat(Str(n),Str(n+1))); if(issquare(t,&s), print1(s,", "))); /* Antonio Roldán and Joerg Arndt, Dec 31 2012 */

a(17) corrected by Donovan Johnson, Jan 03 2013

A030465 Numbers k such that k concatenated with k+1 is a square.

Original entry on oeis.org

183, 328, 528, 715, 6099, 13224, 40495, 106755, 453288, 2066115, 2975208, 22145328, 28027683, 110213248, 110667555, 147928995, 178838403, 226123528, 275074575, 333052608, 378698224, 445332888, 446245635

Offset: 1

Patrick De Geest

Also called Sastry numbers. - Lekraj Beedassy, Jul 18 2008

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 183, p. 56, Ellipses, Paris 2008.

Giovanni Resta, Table of n, a(n) for n = 1..198 (terms < 10^20)

Cf. A030466, A030467, A054214, A054215, A054216, A020339, A020340.

Select[{#,FromDigits[Join[IntegerDigits[#], IntegerDigits[1 + #]]]} & /@
  Flatten[Table[10*n + {0, 3, 4, 5, 8, 9}, {n, 10^5}]], IntegerQ[Sqrt[#[[2]]]] &] (* Hans Rudolf Widmer, Jun 30 2021 *)

isok(k) = issquare(eval(concat(Str(k), Str(k+1)))); \\ Michel Marcus, Jun 30 2021

A030466 Squares that are concatenations of two consecutive nonzero numbers.

Original entry on oeis.org

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

Offset: 1

Patrick De Geest

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.

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.

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

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

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

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

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

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

a(18) from Iain Fox, Dec 27 2017

A054216 Numbers m such that m^2 is a concatenation of two consecutive decreasing numbers.

Original entry on oeis.org

91, 9079, 9901, 733674, 999001, 88225295, 99990001, 8900869208, 9296908812, 9604060397, 9999900001, 326666333267, 673333666734, 700730927008, 972603739727, 999999000001, 34519562953737, 39737862788838, 49917309624956

Offset: 1

Patrick De Geest, Feb 15 2000

Obviously b(n) = 100^n - 10^n + 1 = (91, 9901, 999001, 99990001, ...) is a subsequence. Are { b(2), b(4), b(6), b(8) } the only terms of this sequence that are prime? - M. F. Hasler, Mar 30 2008. Answer: The smallest prime in this sequence that is not of the form b(n) is A054216(155) = 811451682377384625400019885321 [Max Alekseyev, Oct 08 2008]. See A145381 for further prime terms.
Other subsequences are c(n) = ( 10^(6n) - 2*10^(5n) - 10^(3n) - 2*10^n + 1 )/3 (n>=2), d(n) = (33/101)*(100^(404n+71)+1)+10^(404n+71) (n>=0) and e(n) = (33/101)*(100^(404n-71)+1)+10^(404n-71) (n>=1). Primes among these include c(10), c(14) and d(0). - M. F. Hasler, Oct 09 2008
A positive integer m is in this sequence if and only if m^2 == -1 (mod 10^k + 1) where k is the number of decimal digits in m. Note that k cannot be odd, since in this case 11 divides 10^k + 1 while -1 is not a square modulo 11. - Max Alekseyev, Oct 09 2008

			'8242' + '8242-1' gives 82428241 which is 9079^2.
Leading zeros are not allowed, which is why c(1)=266327 is not in this sequence although c(1)^2 = 070930 070929.

Luca, Florian, and Pantelimon Stănică. "Perfect Squares as Concatenation of Consecutive Integers." The American Mathematical Monthly 126.8 (2019): 728-734.

Giovanni Resta, Table of n, a(n) for n = 1..1000

Cf. A020339, A020340, A030465, A030466, A030467, A054214, A054215, A145381.

isA054216(n)={ 1==[1,-1]*divrem(n^2,10^(#Str(n^2)\2)) & #Str(n^2)%2==0 }

a(n) = sqrt(A054215(n)). - Max Alekseyev, May 14 2007

More terms from Max Alekseyev, May 14 2007

Several corrections and additions from M. F. Hasler, Oct 09 2008

A054215 Squares that are concatenations of two consecutive decreasing numbers.

Original entry on oeis.org

8281, 82428241, 98029801, 538277538276, 998002998001, 7783702677837025, 9998000299980001, 79225472657922547264, 86432513458643251344, 92237976109223797609, 99998000029999800001, 106710893290106710893289

Offset: 1

Patrick De Geest, Feb 15 2000

Infinitely many terms of this sequence are provided by A168624(k)^2 for k>0. - Bruno Berselli, Mar 13 2018

			E.g. '8242' + '8242-1' gives 82428241 which is 9079^2.

Luca, Florian, and Pantelimon Stănică. "Perfect Squares as Concatenation of Consecutive Integers." The American Mathematical Monthly 126.8 (2019): 728-734.

Cf. A054214, A054216, A030465, A030466, A030467, A020339, A020340.

a(n) = concatenation of A054214(n) and A054214(n)-1. - Max Alekseyev, May 14 2007

More terms from Max Alekseyev, May 14 2007

82848241 corrected to 82428241 by Dominick Cancilla, Jul 21 2010

A020339 a(n)^2 is the least square base-n doublet (base-n representation is the concatenation of 2 identical strings).

Original entry on oeis.org

6, 2, 615, 84, 119973, 4, 3, 23620, 36363636364, 6, 24766945690, 17928148, 915, 4, 86808207405692007605, 6, 130, 10, 2667, 95530227420606, 10623969116570, 12, 5, 343872950627253606, 9, 14, 59239353339085, 8130

Offset: 2

David W. Wilson

The identical strings must contain at least one nonzero digit, so that a(n) > 0. - Alonso del Arte, Jun 20 2018

In Bridy et al. it is shown how to construct an example (although not necessarily the least example) for each integer base n >= 2. - Jeffrey Shallit, Jun 14 2021

			The first few squares in binary are 1, 100, 1001, 10000, 11001, 100100. Thus we see that 100100, which is 36 in decimal, the square of 6, is the first square which is the concatenation of two identical bit patterns, and therefore a(2) = 6.

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.
David Wells, "The Penguin Dictionary of Curious and Interesting Numbers", Revised Edition 1997, p. 189.

Robert Israel, Table of n, a(n) for n = 2..99
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.
A. Ottens, The arithmetic-digits-squares-three.digits problem

Cf. A020340, A054214, A054215, A054216, A030465, A030466, A030467.

f:= proc(b)
  local d,F,x,t,j;
  for d from 1 do
    F:= select(t -> t[2]::odd, ifactors(1+b^d)[2]);
    x:= mul(t[1],t=F);
    if x >= b^d then next fi;
    j:= ceil(sqrt(b^(d-1)/x));
    if j^2*x < b^d then return j*sqrt(x*(1+b^d)) fi
  od
end proc:
map(f, [$2..40]); # Robert Israel, May 19 2024

a(j*k^2-1) = j if k >= 2 and j is squarefree. - Robert Israel, May 19 2024

Name slightly adjusted by Alonso del Arte, Jun 20 2018

A020340 Least square base n doublet (written in base 10).

Original entry on oeis.org

36, 4, 378225, 7056, 14393520729, 16, 9, 557904400, 1322314049613223140496, 36, 613401598811409576100, 321418490709904, 837225, 16, 7535664872989640713426833504575377836025, 36, 16900, 100, 7112889

Offset: 2

David W. Wilson

In Bridy et al. it is shown how to construct infinitely many examples for any given base n >= 2. - Jeffrey Shallit, Jun 14 2021

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.
David Wells, "The Penguin Dictionary of Curious and Interesting Numbers", Revised Edition 1997, p. 189.

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.
A. Ottens, The arithmetic-digits-squares-three.digits problem

Cf. A020339, A054214, A054215, A054216, A030465, A030466, A030467.

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A116123 Duplicate of A054214.

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A116142 Duplicate of A054214.

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

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A030467 Numbers k such that k^2 is a concatenation of two successive numbers.

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A030465 Numbers k such that k concatenated with k+1 is a square.

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A030466 Squares that are concatenations of two consecutive nonzero numbers.

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A054216 Numbers m such that m^2 is a concatenation of two consecutive decreasing numbers.

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A054215 Squares that are concatenations of two consecutive decreasing numbers.

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