OFFSET
1,2
COMMENTS
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.
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
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.
FORMULA
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).
EXAMPLE
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).
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 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Vladimir Shevelev, May 28 2010
STATUS
approved