A178482 Phi-antipalindromic numbers.

1, 3, 4, 7, 8, 10, 11, 18, 19, 21, 22, 25, 26, 28, 29, 47, 48, 50, 51, 54, 55, 57, 58, 65, 66, 68, 69, 72, 73, 75, 76, 123, 124, 126, 127, 130, 131, 133, 134, 141, 142, 144, 145, 148, 149, 151, 152, 170, 171, 173, 174

Offset: 1

Author

Vladimir Shevelev, May 28 2010

Keywords

• nonn
• base
• changed

Comments

We call m a phi-antipalindromic number if for the vector (a,...,b) (a<...=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.

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

There is a 21-state automaton accepting the Zeckendorf representations of those n in this sequence. - Jeffrey Shallit, May 03 2023

Kimberling's conjecture has been proven by Ingrid Vukusic and myself. Along the way we prove an alternate characterization of the sequence: they are the positive integers whose base-phi expansion consists only of even exponents of phi. - Jeffrey Shallit, Aug 28 2025

Alternatively, this sequence consists of those numbers k such that either k or k-1 can be written as the (possibly empty) sum of distinct Lucas numbers L_i where i>=2 and i is even. - Jeffrey Shallit, Aug 28 2025

Examples

			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).
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..3071 from R. J. Mathar)
• Jeffrey Shallit, Proving Properties of phi-Representations with the Walnut Theorem-Prover, arXiv:2305.02672 [math.NT], 2023.

Crossrefs

Cf. A005248, A055778, A104605, A104626, A104627, A104628.

For bisections see A171070, A171071.

Programs

Mathematica

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 *)

Formula

For k>=1, a(2^k)=A005248(k); if 2^k

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