A020339 -id:A020339 - cp's OEIS frontend

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

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

A054214 Numbers n such that n concatenated with n-1 is a square.

Original entry on oeis.org

82, 8242, 9802, 538277, 998002, 77837026, 99980002, 7922547265, 8643251345, 9223797610, 9999800002, 106710893290, 453378226757, 491023832065, 945958034530, 999998000002, 11916002265170, 15790977390245, 24917378001937, 25082758752026, 36315251812570

Offset: 1

Patrick De Geest, Feb 15 2000

Also, numbers k such that k concatenated with k-2 gives the product of two numbers which differ by 2.
Also, numbers n such that n concatenated with n-5 gives the product of two numbers which differ by 4.
Every term contains an even number of digits. - Max Alekseyev, May 14 2007
If n=(10^m-1)^2+1 where m is a positive integer then n is in the sequence. Because then n has 2m digits and n concatenated with n-1 is n*10^(2m)+(n-1) = (10^(2m)-10^m+1)^2. for example, taking m=1 we get 82, the first term of the sequence. - Farideh Firoozbakht, Aug 22 2013
As pointed out by Georg Fischer, it is very plausible that all of A054214, A116123, and A116142 contain exactly the same terms. If that is indeed true, then A054214 should be edited to mention the alternative constructions, and the other two sequences declared "dead". However, this needs careful analysis to deal with the possibilities that n, n-2, and n-5 may not all have the same number of digits. - N. J. A. Sloane, Oct 30 2018. Nov 05 2018: Thanks to Giovanni Resta_ (see below), this has now been done. - N. J. A. Sloane, Nov 05 2018
From Giovanni Resta, Nov 05 2018: (Start)
For n and n-5 to have a different digit length, we must have n = 10^k+h with 0<=h<=4.
We want to prove that in this case the concatenation of n and n-5 cannot be of the form m(m+4). The numbers m(m+4) modulo 9 can only be equal to 0, 3, 5, or 6, but it is easy to see that the concatenation of 10^k+h and 10^k+h-5 can be equal to one of these values modulo 9 only if h=0.
Now, the concatenation of 10^k and 10^k-5 is equal to 3 modulo 4 for every k>1, but m(m+4) modulo 4 can only be equal to 0 or 1, so A116123 is indeed equal to this sequence.
Using an identical argument (with mods 9 and 4) we can prove that the concatenation of n and n-2, when n and n-2 have a different number of digits, cannot be equal to m(m+2) and so A116142 is equal to this sequence. (End)

			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.

Giovanni Resta, Table of n, a(n) for n = 1..4751

Cf. A054215, A054216, A030465, A030466, A030467, A020339, A020340.

More terms from Max Alekseyev, May 14 2007
a(20)-a(21) from Giovanni Resta, Nov 05 2018
Edited by N. J. A. Sloane, Nov 05 2018

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

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

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A054214 Numbers n such that n concatenated with n-1 is a square.

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

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