OFFSET
2,1
FORMULA
It appears that a(n) = A105321(n)/2. - Omar E. Pol, May 29 2010. Proof from Nathaniel Johnston, Nov 07 2010: If you remove every 2nd row from Pascal's triangle then the rule for constructing the parity of the next row from the current row is the same as the rule for constructing generation n+1 of the primes from generation n: add up the previous and next term in the current row.
a((2*n-3)*2^p) = (2^(p-1)+1)*A001316(n-2), p >= 0 and n >= 2. - Johannes W. Meijer, Jan 22 2013
EXAMPLE
a(7)=6 because in generation 7 there are six odd numbers: 127,237,403,729,879,1109.
MAPLE
A048460 := proc(nmax) local par, c, r, prevc, prevl, cpar; par := [[], [1, 1]] ; for c from 3 to nmax do prevc := op(-1, par) ; prevl := nops(prevc) ; if nops(prevc) < 2 then cpar := [0] ; else cpar := [op(2, prevc)] ; end if; for r from 2 to prevl-1 do cpar := [op(cpar), ( op(r-1, prevc) + op(r+1, prevc)) mod 2] ; end do: cpar := [op(cpar), op(prevl-1, prevc), 1] ; par := [op(par), cpar] ; end do: cpar := [] ; for c from 2 to nops(par) do add(r, r=op(c, par)) ; cpar := [op(cpar), %] ; end do: cpar ; end proc: A048460(120) ; # R. J. Mathar, Aug 07 2010
nmax := 86: A001316 := n -> if n <=- 1 then 0 else 2^add(i, i=convert(n, base, 2)) fi: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 2 to nmax/(p+2) do a((2*n-3)*2^p) := (2^(p-1)+1)*A001316(n-2) od: od: seq(a(n), n=2..nmax); # Johannes W. Meijer, Jan 22 2013
MATHEMATICA
A105321[n_] := Sum[Binomial[1, n-k] Mod[Binomial[k, j], 2], {k, 0, n}, {j, 0, k}];
a[n_] := A105321[n]/2;
Table[a[n], {n, 2, 86}] (* Jean-François Alcover, Oct 25 2023 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Patrick De Geest, May 15 1999
EXTENSIONS
More terms from R. J. Mathar, Aug 07 2010
STATUS
approved