|
|
A053381
|
|
Maximal number of linearly independent smooth nowhere-zero vector fields on a (2n+1)-sphere.
|
|
6
|
|
|
1, 3, 1, 7, 1, 3, 1, 8, 1, 3, 1, 7, 1, 3, 1, 9, 1, 3, 1, 7, 1, 3, 1, 8, 1, 3, 1, 7, 1, 3, 1, 11, 1, 3, 1, 7, 1, 3, 1, 8, 1, 3, 1, 7, 1, 3, 1, 9, 1, 3, 1, 7, 1, 3, 1, 8, 1, 3, 1, 7, 1, 3, 1, 15, 1, 3, 1, 7, 1, 3, 1, 8, 1, 3, 1, 7, 1, 3, 1, 9, 1, 3, 1, 7, 1, 3, 1, 8, 1, 3, 1, 7, 1, 3, 1, 11, 1, 3, 1, 7, 1, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
The corresponding terms for a 2n-sphere are all 0 ("you can't comb the hair on a billiard ball"). The "3" and "7" come from the quaternions and octonions.
b(n) = a(n-1): b(2^e) = ((e+1) idiv 4) + 2^((e+1) mod 4) - 1, b(p^e) = 1, p>2. - Christian G. Bower, May 18 2005
|
|
LINKS
|
|
|
FORMULA
|
Let f(n) be the number of linearly independent smooth nowhere-zero vector fields on an n-sphere. Then f(n) = 2^c + 8d - 1 where n+1 = (2a+1) 2^b and b = c+4d and 0 <= c <= 3. f(n) = 0 if n is even.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 10/3. - Amiram Eldar, Nov 29 2022
|
|
MAPLE
|
with(numtheory): for n from 1 to 601 by 2 do c := irem(ifactors(n+1)[2, 1, 2], 4): d := iquo(ifactors(n+1)[2, 1, 2], 4): printf(`%d, `, 2^c+8*d-1) od:
nmax:=101: A047530 := proc(n): ceil(n/4) + 2*ceil((n-1)/4) + 4*ceil((n-2)/4) + ceil((n-3)/4) end: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 0 to ceil(nmax/(p+2))+1 do A053381((2*n+1)*2^p-1) := A047530(p+1): od: od: seq(A053381(n), n=0..nmax); # Johannes W. Meijer, Jun 07 2011, revised Jan 29 2013
|
|
MATHEMATICA
|
a[n_] := Module[{b, c, d, rho, n0}, n0 = 2*n; b = 0; While[BitAnd[n0, 1] == 0, n0 /= 2; b++]; c = BitAnd[b, 3]; d = (b - c)/4; rho = 2^c + 8*d; Return[rho - 1]]; Table[a[n], {n, 1, 102}] (* Jean-François Alcover, May 16 2013, translated from C *)
|
|
PROG
|
(C) int MaxLinInd(int n){ /* Returns max # linearly indep smooth nowhere zero * vector fields on S^{n-1}, n=1, 2, ... */ int b, c, d, rho; b = 0; while((n & 1)==0){ n /= 2; b++; } c = b & 3; d = (b - c)/4; rho = (1 << c) + 8*d; return( rho - 1); }
(C, gcc/clang) int MaxLinInd(int n) { int b = __builtin_ctz(n); return (1<<b%4) + b/4*8 - 1; } /* Jeremy Tan, Apr 09 2021 */
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,nice,easy,mult
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|