OFFSET
1,3
COMMENTS
Let A be the sequence A114729 (1, 1, 1, 2, 3, 2, 2, 1, 1, 1, ...), B be the sequence A114730 (1, 1, 2, 2, 1, 1, 1, 2, 3, 4, ...) and C be the sequence A114731 (1, 2, 1, 1, 1, 2, 3, 3, 2, 2, ...). Let D be the sequence A114732 (1, 2, 3, 1, 1, 2, 3, 4, 5, 6, ...), E be the sequence A114733 (1, 2, 1, 2, 3, 4, 5, 3, 1, 1, ...) and F be the sequence A114734 (1, 1, 2, 3, 4, 2, 1, 2, 3, 4, ...). Then:
- A upper trims to B
- B upper trims to C
- C upper trims to A
- A lower trims to C
- B lower trims to A
- C lower trims to B
- D gives the number of times each element of A occurs
- E gives the number of times each element of B occurs
- F gives the number of times each element of C occurs
- A gives the number of times each element of D occurs
- B gives the number of times each element of E occurs
- C gives the number of times each element of F occurs
- D lower trims to E
- E lower trims to F
- F lower trims to D
- D upper trims to F
- E upper trims to D
- F upper trims to E
EXAMPLE
C(10)=2 and it is the fourth 2 in the sequence, so F(10)=4.
MATHEMATICA
c[n_] := Flatten[ Table[{Range[3 Floor[(k - 1)/2] + 2],
Table[{i, i}, {i, Floor[k/2] + k, 1, -1}]}, {k, n}]];
uppertrim[list_] := Fold[DeleteCases[#1, #2, 1, 1] &, list, Range[Max[list]]];
lowertrim[list_] := DeleteCases[list - 1, 0];
numbertimes[list_] := Table[Length@Position[Take[list, k], list[[k]]], {k, Length[list]}];
a[n_] := uppertrim[c[n]];
b[n_] := uppertrim[a[n]];
d[n_] := numbertimes[a[n]];
e[n_] := numbertimes[b[n]];
f[n_] := numbertimes[c[n]];
f[6] (* Birkas Gyorgy, Apr 21 2011 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Kerry Mitchell, Dec 28 2005
STATUS
approved