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%I #30 May 05 2023 01:35:28
%S 1,3,4,7,8,10,11,18,19,21,22,25,26,28,29,47,48,50,51,54,55,57,58,65,
%T 66,68,69,72,73,75,76,123,124,126,127,130,131,133,134,141,142,144,145,
%U 148,149,151,152,170,171,173,174
%N Phi-antipalindromic numbers.
%C We call m a phi-antipalindromic number if for the vector (a,...,b) (a<...<b) of exponents of its base-phi expansion, we have (a,...,b)=(-b,...,-a). For n>=2, either a(n)+1 or a(n)-1 is in the sequence; also either a(n)+3 or a(n)-3 is in the sequence.
%C Conjecture: this is the sequence of numbers k for which f(k) is an integer, where f(x) is the change-of-base function defined at A214969 using b=phi and c=b^2. - _Clark Kimberling_, Oct 17 2012
%C There is a 21-state automaton accepting the Zeckendorf representations of those n in this sequence. - _Jeffrey Shallit_, May 03 2023
%H Amiram Eldar, <a href="/A178482/b178482.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..3071 from R. J. Mathar)
%H Jeffrey Shallit, <a href="https://arxiv.org/abs/2305.02672">Proving Properties of phi-Representations with the Walnut Theorem-Prover</a>, arXiv:2305.02672 [math.NT], 2023.
%F For k>=1, a(2^k)=A005248(k); if 2^k<n<2^(k+1), then a(n)=a(2^k)+a(n-2^k).
%e The vectors of exponents of 4 and 5 are (-2,0,2) and (-4,-1,3) correspondingly (cf.A104605). Therefore by definition 4 is a phi-antipalindromic number, while 5 is not. Let n=38. Then k=5. Thus a(38)=A005248(5)+a(6)=123+10=133. The vector of exponents of phi in the base-phi expansion of 133 is (-10,-4,-2,2,4,10).
%t phiAPQ[1] = True; phiAPQ[n_] := Module[{d = RealDigits[n, GoldenRatio, 2*Ceiling[Log[GoldenRatio, n]]]}, e = d[[2]] - Flatten @ Position[d[[1]], 1]; Reverse[e] == -e]; Select[Range[200], phiAPQ] (* _Amiram Eldar_, Apr 23 2020 *)
%Y Cf. A005248, A055778, A104605, A104626, A104627, A104628.
%Y For bisections see A171070, A171071.
%K nonn,base
%O 1,2
%A _Vladimir Shevelev_, May 28 2010
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