|
| |
|
|
A178482
|
|
Phi-antipalindromic numbers.
|
|
8
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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
|
|
|
|
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
|
|
|
|
STATUS
|
approved
|
| |
|
|
