A020339 a(n)^2 is the least square base-n doublet (base-n representation is the concatenation of 2 identical strings).
6, 2, 615, 84, 119973, 4, 3, 23620, 36363636364, 6, 24766945690, 17928148, 915, 4, 86808207405692007605, 6, 130, 10, 2667, 95530227420606, 10623969116570, 12, 5, 343872950627253606, 9, 14, 59239353339085, 8130
Offset: 2
Examples
The first few squares in binary are 1, 100, 1001, 10000, 11001, 100100. Thus we see that 100100, which is 36 in decimal, the square of 6, is the first square which is the concatenation of two identical bit patterns, and therefore a(2) = 6.
References
- 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.
Links
- Robert Israel, Table of n, a(n) for n = 2..99
- 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
Programs
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Maple
f:= proc(b) local d,F,x,t,j; for d from 1 do F:= select(t -> t[2]::odd, ifactors(1+b^d)[2]); x:= mul(t[1],t=F); if x >= b^d then next fi; j:= ceil(sqrt(b^(d-1)/x)); if j^2*x < b^d then return j*sqrt(x*(1+b^d)) fi od end proc: map(f, [$2..40]); # Robert Israel, May 19 2024
Formula
a(j*k^2-1) = j if k >= 2 and j is squarefree. - Robert Israel, May 19 2024
Extensions
Name slightly adjusted by Alonso del Arte, Jun 20 2018
Comments