login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A265432
a(n) = smallest k with concat(1,k) and concat(n,k) both square numbers.
11
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
0,1
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
LINKS
N. J. A. Sloane, Table of n, a(n) for n = 0..20000, May 25 2016 [First 10000 terms from David W. Wilson, Dec 08 2015]
David W. Wilson, Table of n, a(n) for n = 0..10000, May 25 2016 [The graph of the first 10000 terms looks rather different from the graph of the first 20000 terms, so this file is worth preserving. - N. J. A. Sloane, May 25 2016]
EXAMPLE
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.
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 *)
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.
Sequence in context: A206649 A340501 A370611 * A247885 A345552 A345806
KEYWORD
nonn,base,look
AUTHOR
STATUS
approved