A116136 -id:A116136 - cp's OEIS frontend

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

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

A115426 Numbers k such that the concatenation of k with k+2 gives a square.

Original entry on oeis.org

7874, 8119, 69476962, 98010199, 108746354942, 449212110367, 544978035127, 870501316279, 998001001999, 1428394731903223, 1499870932756487, 1806498025502498, 1830668275445687, 1911470478658759, 2255786189655202

Offset: 1

Giovanni Resta, Jan 24 2006

Numbers k such that k concatenated with k+1 gives the product of two numbers which differ by 2.
Numbers k such that k concatenated with k-2 gives the product of two numbers which differ by 4.
Numbers k such that k concatenated with k-7 gives the product of two numbers which differ by 6.

			8119//8121 = 9011^2, where // denotes concatenation.
98010199//98010200 = 99000100 * 99000102.
98010199//98010197 = 99000099 * 99000103.

Chai Wah Wu, Table of n, a(n) for n = 1..1000

Cf. A030465, A102567, A115427, A115428, A115429, A115430, A115431, A115432, A115433, A115434, A115435, A115436, A115437.
Cf. A116106, A116110, A116112, A116125, A116242.
Cf. A116136, A116143, A116107, A116275, A116163, A116099, A116177, A116295.

from itertools import count, islice
from sympy import sqrt_mod
def A115426_gen(): # generator of terms
    for j in count(0):
        b = 10**j
        a = b*10+1
        for k in sorted(sqrt_mod(2,a,all_roots=True)):
            if a*(b-2) <= k**2-2 < a*(a-3):
                yield (k**2-2)//a
A115426_list = list(islice(A115426_gen(),40)) # Chai Wah Wu, Feb 20 2024

Edited by N. J. A. Sloane, Apr 13 2007

A116099 Numbers k such that k concatenated with k-9 gives the product of two numbers which differ by 7.

Original entry on oeis.org

69, 59898667, 79493157, 13412927190959690154913903, 14163000698458955079906403, 38895475965785687555173929, 40165600438484442828161229, 74294440818366638194239027

Offset: 1

Giovanni Resta, Feb 06 2006

Also numbers k such that k concatenated with k-3 gives the product of two numbers which differ by 5.
Also numbers k such that k concatenated with k+1 gives the product of two numbers which differ by 3.
Also numbers k such that k concatenated with k+3 gives the product of two numbers which differ by 1.

			79493157//79493154 = 89158933 * 89158938, where // denotes concatenation.
79493157//79493158 = 89158934 * 89158937.
79493157//79493160 = 89158935 * 89158936.
79493157//79493148 = 89158932 * 89158939.

Cf. A116098, A116193, A116107, A116230, A116130, A116136, A116205, A116107, A116268.

Cf. A115426, A115428, A116171, A116296, A116170, A116177, A116307.

Edited by N. J. A. Sloane, Apr 12 2007

A116130 Numbers k such that k concatenated with k-4 gives the product of two numbers which differ by 5.

Original entry on oeis.org

8, 98, 590, 738, 830, 998, 1080, 4508, 9998, 20660, 29754, 99998, 980300, 999998, 6694218, 9999998, 49826988, 99999998, 117738578, 131505858, 132231404, 176445054, 177285320, 247979808, 252028388, 335180054, 336337790

Offset: 1

Giovanni Resta, Feb 06 2006

Also numbers k such that k concatenated with itself gives the product of two numbers which differ by 3.

			7531357568//7531357564 = 8678339452 * 8678339457, where // denotes concatenation.

Cf. A116124, A116129, A116131, A116099, A116261, A116163, A116136, A116125, A116287.

Edited by N. J. A. Sloane, Apr 13 2007

A116098 Numbers k such that k concatenated with k-9 gives the product of two numbers which differ by 6.

Original entry on oeis.org

11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 13223140496, 20661157025, 29752066116, 40495867769, 52892561984, 66942148761, 82644628100, 100000000001, 1000000000001

Offset: 1

Giovanni Resta, Feb 06 2006

From Robert Israel, Aug 13 2018: (Start)
Contained in, and apparently identical, to A116129.
Numbers k such that k*(10^d+1) is a square, where k-9 has d decimal digits.
(End)

			100000001//99999992 = 99999998 * 100000004, where // denotes
concatenation.

Robert Israel, Table of n, a(n) for n = 1..134

Cf. A116097, A116099, A116106, A054214, A116128, A116129, A116130, A116136, A116229, A116260.

g:= proc(d) local r,c,a,b;
   r:= mul(t[1],t=select(s -> s[2]::odd, ifactors(10^d+1)[2]))
   c:= ceil((10^(d-1)+9)/r);
   a:= isqrt(c);
   if a^2 < c then a:= a+1 fi;
   c:= floor((10^d+8)/r);
   b:= isqrt(c);
   if b^2 > c then b:= b-1 fi;
   seq(r*y^2, y = a..b)
end proc:
seq(g(d),d=1..60); # Robert Israel, Aug 13 2018

A116267 Numbers k such that k*(k+4) gives the concatenation of two numbers m and m-3.

Original entry on oeis.org

8, 98, 426, 571, 725, 844, 998, 7808, 9998, 36363, 63634, 99998, 326732, 673265, 999998, 4545452, 5454545, 9999998, 47058821, 52941176, 99999998, 331983805, 332667332, 384615384, 422892896, 475524475, 524475522, 577107101, 615384613, 667332665, 668016192, 719964244

Offset: 1

Giovanni Resta, Feb 06 2006

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A116136, A116260, A116266, A116268, A116275.

from itertools import count, islice
from sympy import sqrt_mod
def A116267_gen(): # generator of terms
    for j in count(0):
        b = 10**j
        a = b*10+1
        for k in sorted(sqrt_mod(1,a,all_roots=True)):
            if a*(b+3) <= k**2-1 < a*(a+2):
                yield k-2
A116267_list = list(islice(A116267_gen(),20)) # Chai Wah Wu, Feb 19 2024

a(29)-a(32) from Chai Wah Wu, Feb 19 2024

A116286 Numbers k such that k*(k+2) gives the concatenation of a number m with itself.

Original entry on oeis.org

9, 99, 427, 572, 726, 845, 999, 7809, 9999, 36364, 63635, 99999, 326733, 673266, 999999, 4545453, 5454546, 9999999, 47058822, 52941177, 99999999, 331983806, 332667333, 384615385, 422892897, 475524476, 524475523, 577107102, 615384614, 667332666, 668016193, 719964245, 758241757, 804511279

Offset: 1

Giovanni Resta, Feb 06 2006

From Robert Israel, Apr 08 2025: (Start)
Numbers k such that k * (k + 2) = (10^d + 1) * m for some d and m where m has d digits.
Contains 10^d - 1 for all d >= 1. (End)

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A116136, A116279, A116285, A116287, A116295.

q:= proc(d,m) local R,t,a,b,x,q;
   t:= 10^d+1;
   R:= NULL;
   for a in numtheory:-divisors(t) do
     b:= t/a;
     if igcd(a,b) > 1 then next fi;
     for x from chrem([0,-m],[a,b]) by t do
       q:= x*(x+m)/t;
       if q >= 10^d then break fi;
       if q >= 10^(d-1) then R:= R, x fi;
   od od;
   sort(convert({R},list));
end proc:
A:=[seq(op(q(d,2)),d=1..10)]; # Robert Israel, Apr 08 2025

More terms from Robert Israel, Apr 08 2025

A116129 Numbers k such that k concatenated with k-4 gives the product of two numbers which differ by 4.

Original entry on oeis.org

11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 13223140496, 20661157025, 29752066116, 40495867769, 52892561984, 66942148761, 82644628100, 100000000001, 1000000000001

Offset: 1

Giovanni Resta, Feb 06 2006

From Robert Israel, Aug 13 2018: (Start)
Contains, and appears to be identical to, A116098.
Numbers k such that (10^d+1)*k is a square, where k-4 has d digits. (End)

			100000001//99999997 = 99999999 * 100000003, where // denotes concatenation.

Cf. A116098, A054214, A116128, A116130, A116136, A116260.

g:= proc(d) local r,c,a,b;
   r:= mul(t[1],t=select(s -> s[2]::odd, ifactors(10^d+1)[2]));
   c:= ceil((10^(d-1)+4)/r);
   a:= isqrt(c);
   if a^2 < c then a:= a+1 fi;
   c:= floor((10^d+3)/r);
   b:= isqrt(c);
   if b^2 > c then b:= b-1 fi;
   seq(r*y^2, y = a..b)
end proc:
map(g, [$1..60]); # Robert Israel, Aug 13 2018

A116154 Numbers k such that k concatenated with itself gives the product of two numbers which differ by 1.

Original entry on oeis.org

132, 406, 510, 852, 7930, 66942, 113322, 440056, 5289256, 58477510, 111333222, 164378892, 183673470, 200444410, 206611570, 224376732, 277008310, 297520662, 305024040, 326530612, 353505556, 444000556, 479289942

Offset: 1

Giovanni Resta, Feb 06 2006

Cf. A116141, A116136, A116163, A116285.

A116135 Duplicate of A115431.

Original entry on oeis.org

6, 5346, 8083, 10578, 45531, 58626, 2392902, 2609443, 7272838, 51248898, 98009803, 159728062051, 360408196038, 523637103531, 770378933826, 998000998003, 1214959556998, 1434212848998, 3860012299771, 4243705560771

Offset: 1

dead

Cf. A115431, A116117, A116121, A116134, A116136, A054214, A116266.

cp's OEIS Frontend

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

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A115426 Numbers k such that the concatenation of k with k+2 gives a square.

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A116099 Numbers k such that k concatenated with k-9 gives the product of two numbers which differ by 7.

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A116130 Numbers k such that k concatenated with k-4 gives the product of two numbers which differ by 5.

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A116098 Numbers k such that k concatenated with k-9 gives the product of two numbers which differ by 6.

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A116267 Numbers k such that k*(k+4) gives the concatenation of two numbers m and m-3.

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A116286 Numbers k such that k*(k+2) gives the concatenation of a number m with itself.

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A116129 Numbers k such that k concatenated with k-4 gives the product of two numbers which differ by 4.

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