A238208 The total number of 1's in all partitions of n into an odd number of distinct parts.

0, 1, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 7, 8, 10, 12, 14, 17, 20, 24, 28, 33, 38, 45, 52, 60, 69, 80, 91, 105, 120, 137, 156, 178, 202, 230, 261, 295, 334, 378, 426, 481, 542, 609, 685, 769, 862, 966, 1082, 1209, 1351, 1508, 1681, 1873, 2086, 2319, 2578

Offset: 0

Author

Mircea Merca, Feb 20 2014

Keywords

• nonn
• easy

Comments

The g.f. for "number of k's" is (1/2)*(x^k/(1+x^k))*(Product_{n>=1} 1 + x^n) + (1/2)*(x^k/(1-x^k))*(Product_{n>=1} 1 - x^n).

Or: the number of partitions of n-1 into an even number of distinct parts >=2. - R. J. Mathar, May 11 2016

Examples

			a(10) = 3 because the partitions in question are: 7+2+1, 6+3+1, 5+4+1.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from Andrew Howroyd)

Crossrefs

Column k=1 of A238450.

Cf. A067659, A067661, A133280.

Programs

Maple

A238208 := proc(n)
    local a,L,Lset;
    a := 0 ;
    L := combinat[firstpart](n) ;
    while true do
        # check that parts are distinct
        Lset := convert(L,set) ;
        if nops(L) = nops(Lset) then
            # check that number is odd
            if type(nops(L),'odd') then
                if 1 in Lset then
                    a := a+1 ;
                end if;
            end if;
        end if;
        L := combinat[nextpart](L) ;
        if L = FAIL then
            return a;
        end if;
    end do:
    a ;
end proc: # R. J. Mathar, May 11 2016
# second Maple program:
b:= proc(n, i, t) option remember; `if`(n=0, t,
     `if`(i>n, 0, b(n, i+1, t)+b(n-i, i+1, 1-t)))
    end:
a:= n-> b(n-1, 2, 1):
seq(a(n), n=0..100);  # Alois P. Heinz, May 01 2020

Mathematica

b[n_, i_, t_] := b[n, i, t] = If[n == 0, t, If[i > n, 0, b[n, i+1, t] + b[n-i, i+1, 1-t]]];
a[n_] := b[n-1, 2, 1];
a /@ Range[0, 100] (* Jean-François Alcover, May 17 2020, after Alois P. Heinz *)

PARI

seq(n)={my(A=O(x^n)); Vec(x*(eta(x^2 + A)/(eta(x + A)*(1+x)) + eta(x + A)/(1-x))/2, -(n+1))} \\ Andrew Howroyd, May 01 2020

Formula

a(n) = Sum_{j=1..round(n/2)} A067661(n-(2*j-1)) - Sum_{j=1..floor(n/2)} A067659(n-2*j).

G.f.: (1/2)*(x/(1+x))*(Product_{n>=1} 1 + x^n) + (1/2)*(x/(1-x))*(Product_{n>=1} 1 - x^n).

a(n) ~ exp(Pi*sqrt(n/3)) / (16 * 3^(1/4) * n^(3/4)). - Vaclav Kotesovec, May 17 2020

From Peter Bala, Feb 02 2021: (Start)

a(n+1) = d(n) - ( d(n-1) + d(n-3) ) + ( d(n-4) + d(n-6) + d(n-8) ) - ( d(n-9) + d(n-11) + d(n-13) + d(n-15) ) + ( d(n-16) + d(n-18) + d(n-20) + d(n-22) + d(n-24) ) - ( d(n-25) + d(n-27) + d(n-29) + d(n-31) + d(n-33) + d(n-35) ) + ..., where d(n) = A000009(n) is the number of partitions of n into distinct parts, with the convention that d(n) = 0 for n < 0.

G.f.: x/(1 - x^2)*Sum_{n >= 0} (-1)^n*x^((n^2+n+1-(-1)^n)/2)/Product_{k = 1..n} 1 - x^k.

Alternative g.f.: ( Product_{k >= 1} 1 + x^k ) * x*Sum_{n >= 0} (-1)^n*x^(n^2)*(1 - x^(2*n+2))/(1 - x^2).

Faster converging g.f. (conjecture): Sum_{n >= 0} x^((n+1)*(2*n+1))/ Product_{k = 1..2*n} 1 - x^k. (End)

Extensions

a(51)-a(60) from R. J. Mathar, May 11 2016