cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A272685 a(n) = smallest m such that A265432(m) = A272684(n), or -1 if no such m exists.

Original entry on oeis.org

1, 15, 14, 13, 12, 217, 0, 215, 45, 213, 44, 43, 209, 42, 207, 2, 573, 1327, 572, 130, 185, 570, 1492, 569, 78, 568, 128, 567, 1318, 1498, 565, 188, 564, 10, 563, 1312, 562, 1504, 1309, 560, 1507, 693, 74, 558, 1510, 557, 192, 2360, 6085, 350, 1480, 708, 6083, 3643
Offset: 1

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Keywords

Comments

It is conjectured that all terms of A272684 eventually appear in A265432.
See A273369 for a version that ignores the fact that terms in A272671 ending in 0 cannot appear in A265432.

Examples

			A272684(1) = 6 first appears in A265432 at index 1, so a(1) = 1.
A272684(7) = 225 first appears in A265432 at index 0, so a(7) = 0.
		

Crossrefs

See A273369 for another version.

A272671 Numbers k such that the decimal number 1k is a square.

Original entry on oeis.org

6, 21, 44, 69, 96, 156, 225, 296, 369, 444, 521, 600, 681, 764, 849, 936, 1025, 1236, 1449, 1664, 1881, 2100, 2321, 2544, 2769, 2996, 3225, 3456, 3689, 3924, 4161, 4400, 4641, 4884, 5129, 5376, 5625, 5876, 6129, 6384, 6641, 6900, 7161, 7424, 7689, 7956, 8225, 8496, 8769, 9044, 9321, 9600
Offset: 1

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Examples

			44 is a member because 144 = 12^2 is a square.
0 is not a member because 10 is not a square.
		

Crossrefs

Cf. A265432, A272672, A045855 (squares beginning with 1), A272684, A272685.

Programs

  • Magma
    [n: n in [1..10000 ] | IsSquare(Seqint(Intseq(n) cat Intseq(1)))]; // Marius A. Burtea, Mar 21 2019
    
  • Maple
    t1:=[];
    for k from 1 to 20000 do
    if issqr(k+10^length(k)) then t1:=[op(t1),k]; fi;
    od;
    t1;
  • Mathematica
    Flatten[n /. Solve[10^# + n == a^2 && 10^(# - 1) <= n < 10^# && a > 0, {n, a}, Integers] & /@ Range[3]] (* Davin Park, Feb 05 2017 *)
    Select[Range[10000],IntegerQ[Sqrt[10^IntegerLength[#]+#]]&] (* Harvey P. Dale, Jul 20 2025 *)
  • PARI
    isok(n) = issquare(eval(concat(1, Str(n)))); \\ Michel Marcus, Mar 21 2019
    
  • Python
    from sympy.ntheory.primetest import is_square
    def ok(n): return is_square(int('1'+str(n)))
    print(list(filter(ok, range(9601)))) # Michael S. Branicky, Jun 21 2021

Extensions

Extended by Davin Park, Feb 05 2017

A265432 a(n) = smallest k with concat(1,k) and concat(n,k) both square numbers.

Original entry on oeis.org

225, 6, 1025, 6, 225, 9937257544619140625, 80625, 225, 19025, 14797831640625, 5625, 89450791534674072265625, 96, 69, 44, 21, 1993672119140625, 2002541101386962890625, 225, 6, 8734765625, 99758030615478515625, 5625, 863225, 80625, 6, 40625, 225, 890625, 158764150390625
Offset: 0

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Comments

k must be a positive integer (and of course cannot begin with 0). - N. J. A. Sloane, May 19 2016
Every term is a member of A272671, by definition. Certainly every term of A272671 which is a power of 100 times an earlier term of A272671 (such as 600, 2100, 4400) will not appear, by the "smallest k" condition. Does every other term of A272671 (that is, the terms of A272684) eventually appear? See A272685 and A273369 for the first appearance of these terms. - Nathan Fox, Brooke Logan, and N. J. A. Sloane, May 23 2016

Examples

			a(0) = 225 because 1225 is a square as is (0)225. (In other words, 225 is the first term in A272672 that is itself a square). - _N. J. A. Sloane_, May 21 2016
a(2) = 1025 because concat(1,1025) = 11025 = 105^2 and concat(2,1025) = 21025 = 145^2.
		

Crossrefs

Cf. A045855, A272671, A018796, A272684, A272685 and A273369 (smallest inverse).
For records see A272674, A272675.
For square roots referred to in definition see A272682, A272683.
A018851 is a simpler sequence in the same spirit.

Programs

  • Mathematica
    << Combinatorica`
    A265432[0] = 225;
    A265432[1] = 6;
    A265432[n_] := Block[{x = {-1, 1, 0, 1}[[Mod[n, 4, 1]]], d = Infinity, l, i}, While[d > Sqrt[10.0^(x - 1)] (Sqrt[10.0 n + 1] - Sqrt[11.0]), x++; d = Infinity; l = Divisors[((n - 1) 10^x)/4]; i = BinarySearch[l, 0.5 Sqrt[(n + 1) 10.0^x - 1] - 0.5 Sqrt[2*10.0^x - 1]]; If[i <= Length@l, d = 2*l[[i + 1/2]]]]; (((n - 1) 10^x - d^2)/(2 d))^2 - 10^x] (* Davin Park, Apr 11 2017 *)
Showing 1-3 of 3 results.