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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.

A256133 Numbers that have unique expansion with minimal digit sum in terms of Fibonacci numbers F_k (k > 1).

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%I A256133 #47 Jan 05 2025 19:51:40
%S A256133 1,2,3,5,7,8,9,11,13,14,15,18,20,21,22,23,24,29,30,32,34,35,36,37,39,
%T A256133 41,47,48,49,52,54,55,56,57,58,60,62,63,64,66,76,77,78,79,84,85,87,89,
%U A256133 90,91,92,94,96,97,98,100,102,103,104,107,109,123,124,125,126
%N A256133 Numbers that have unique expansion with minimal digit sum in terms of Fibonacci numbers F_k (k > 1).
%C A256133 This sequence shows that the "proper digital expansion" mentioned in the introduction to the paper by Drmota and Gajdosik (see Links) is not unique.
%C A256133 This sequence consists of all positive integers that have Zeckendorf expansions not containing any ...1001... and not ending in ...101. Example: 20 is in and it has 20 = 13 + 5 + 2 with Zeckendorf expansion 101010, while 19 is not in and has 19 = 13 + 5 + 1 with Zeckendorf expansion 101001. - _Thomas Bier_, Oct 09 2015
%H A256133 Patrick Okolo Edeogu, <a href="/A256133/b256133.txt">Table of n, a(n) for n = 1..141</a>
%H A256133 M. Drmota and M. Gajdosik, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/36-1/drmota.pdf">The parity of the sum of digits function of generalized Zeckendorf expansions</a>, The Fibonacci Quarterly, 36:1 (1988), pp. 3-19.
%e A256133 7 = 5 + 2 is unique with respect to its minimal digit sum 1 + 1 = 2.
%e A256133 But 10 = 8 + 2 = 5 + 5 is not unique with respect to its minimal digit sum 1 + 1 = 2.
%p A256133 x0:=0: x1:=1: ML:=[]: L:=[]: mes:=0:  for r from 2 to 14 do: z:=x1+x0: x0:=x1: x1:=z:  rj:=12: L:=[op(L),z]: ML:=[z,op(ML)]: od: XL:=[]: for m from 1 to 400 do:  NL:=[]: n:=m: for j from 12 to 1 by -1 do: if L[j+1]>n and L[j]-1 < n then NL:=[op(NL),j]: n:=n-L[j]: fi: od: eps:=0: nx:=nops(NL):  for j from 1 to nx-1 do: if NL[j]-NL[j+1]=3 then eps:=1: fi: if NL[nx-1]-NL[nx]=2 and NL[nx]=1 then eps:=1: fi:od: if eps=0 then XL:=[op(XL),m]: fi: od: print(XL);
%Y A256133 Cf. A000045.
%K A256133 nonn
%O A256133 1,2
%A A256133 _Patrick Okolo Edeogu_, Jul 10 2015