A048460 Total of odd numbers in the generations from 2 onwards.

2, 3, 3, 3, 4, 6, 5, 3, 4, 6, 6, 6, 8, 12, 9, 3, 4, 6, 6, 6, 8, 12, 10, 6, 8, 12, 12, 12, 16, 24, 17, 3, 4, 6, 6, 6, 8, 12, 10, 6, 8, 12, 12, 12, 16, 24, 18, 6, 8, 12, 12, 12, 16, 24, 20, 12, 16, 24, 24, 24, 32, 48, 33, 3, 4, 6, 6, 6, 8, 12, 10, 6, 8, 12, 12, 12, 16, 24, 18, 6, 8, 12, 12, 12, 16

Offset: 2

Author

Patrick De Geest, May 15 1999

Keywords

• nonn

Examples

			a(7)=6 because in generation 7 there are six odd numbers: 127,237,403,729,879,1109.
		

Crossrefs

For "Generations" see A048448-A048455. See also A047844.

Cf. A220466.

Programs

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 *)

Python

def A048460(n): return (1<>1 # Chai Wah Wu, Jul 30 2025

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

Extensions

More terms from R. J. Mathar, Aug 07 2010