A116163 -id:A116163 - 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

A116112 Numbers k such that k concatenated with k-7 gives the product of two numbers which differ by 7.

Original entry on oeis.org

17, 35, 10408517, 45884051, 62918301, 1116290522645838319925, 1491109615209578451401, 2254276950187476704727, 2758431647767103545151, 3768131911733856383477, 4434103687048263321737, 5230580700713956424051

Offset: 1

Giovanni Resta, Feb 06 2006

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

			62918301//62918300 = 79321055 * 79321060, where // denotes concatenation.
62918301//62918304 = 79321056 * 79321059.
62918301//62918306 = 79321057 * 79321058.

Cf. A116107, A115426, A116113, A116119, A116243.
Cf. A116163, A115429, A116158, A116281.
Cf. A116171, A116177, A116179, A116185, A116309, A115430, A116322.

Edited by N. J. A. Sloane, Apr 15 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

A116170 Numbers k such that k concatenated with k+2 gives the product of two numbers which differ by 1.

Original entry on oeis.org

590, 738, 830, 1080, 4508, 20660, 29754, 980300, 6694218, 49826988, 117738578, 131505858, 132231404, 176445054, 177285320, 247979808, 252028388, 335180054, 336337790, 404958680, 406231130, 431477468, 499519478

Offset: 1

Giovanni Resta, Feb 06 2006

Robert Israel, Table of n, a(n) for n = 1..9000

Cf. A116163, A116171, A116099, A116301.

As:= {}:
for m from 2 to 20 do
   acands:= map(t -> rhs(op(t)), [msolve(a*(a+1)=2, 10^m+1)]);
   bcands:= map(t -> t*(t+1) mod 10^m, acands);
   good:= select(t -> bcands[t]>=10^(m-1), [$1..nops(acands)]);
   As:= As union convert(bcands[good], set);
od: map(t -> t-2, sort(convert(As, list))); # Robert Israel, Aug 20 2019

A116294 Numbers k such that k*(k+1) gives the concatenation of two numbers m and m+1.

Original entry on oeis.org

3, 7, 78, 80919, 91809, 326510, 475025, 524975, 673490, 4323777, 4767132, 5232868, 5676223, 4083911141, 4975000250, 5024999750, 5916088859, 9503960496, 9604950396, 4186904462792, 4313465946775, 5686534053225, 5813095537208, 7504715871407, 7631277355390

Offset: 1

Giovanni Resta, Feb 06 2006

			80919 * (80919 + 1) = 6547965480, the concatenation of 65479 and 65479 + 1.

Giovanni Resta, Table of n, a(n) for n = 1..2000

Cf. A116163, A116285, A116295, A116301.

Union @@ ((x /. List@ ToRules@ Reduce[x (x+1) == 10^# y +y+1 && x>0 && 10^(#-1) <= y+1 < 10^#, {x,y}, Integers]) & /@ Range[13] /. x->{}) (* Giovanni Resta, Jul 08 2018 *)

More terms from Giovanni Resta, Jul 08 2018

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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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A116112 Numbers k such that k concatenated with k-7 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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A116170 Numbers k such that k concatenated with k+2 gives the product of two numbers which differ by 1.

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A116294 Numbers k such that k*(k+1) gives the concatenation of two numbers m and m+1.

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

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