A203568 a(n) = A026837(n) - A026838(n).

0, 1, -1, 0, 0, 1, 0, -1, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0

Offset

0,1

Links

Robert Israel, Table of n, a(n) for n = 0..10000

George E. Andrews, Euler's Pentagonal Number Theorem, Mathematics Magazine, Vol. 56, No. 5 (Nov., 1983), pp. 279-284.

Formula

G.f.: Sum_{k in Z} sign(k) * x^(k * (3*k - 1) / 2).

G.f.: Sum_{k>0} x^(k * (3*k - 1) / 2) * (1 - x^k). - Michael Somos, Jul 12 2015

G.f.: x - x^2 * (1 + x) + x^3 * (1 + x) * (1 + x^2) - x^4 * (1 + x) * (1 + x^2) * (1 + x^3) + .... - Michael Somos, Jul 12 2015

G.f.: x / (1 + x) - x^3 / ((1 + x) * (1 + x^2)) + x^6 / ((1 + x) * (1 + x^2) * (1 + x^3)) - .... - Michael Somos, Jul 12 2015

G.f.: x / (1 + x^2) - x^2 / ((1 + x^2) * (1 + x^4)) + x^3 / ((1 + x^2 ) * (1 + x^4) * (1 + x^6)) - .... - Michael Somos, Jul 12 2015

a(n) = - A143062(n) unless n=0. - Michael Somos, Jul 12 2015

For k >= 1, a((3*k^2 - k)/2) = 1, a((3*k^2 + k)/2) = -1. a(n) = 0 otherwise. - Robert Israel, Nov 24 2015

From Peter Bala, Feb 11 2021: (Start)

G.f.: A(x) = Sum_{n >= 1} x^(n*(2*n-1))/Product_{k = 1..2*n} 1 + x^k = x - x^2 + x^5 - x^7 + x^12 - x^15 + - ..., follows by adding terms in pairs in the above g.f. Sum_{n >= 1} (-1)^(n+1)*x^(n*(n+1)/2)/Product_{k = 1..n} 1 + x^k of Somos, dated Jul 12 2015.

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

A(x) = Sum_{n >= 0} (-1)^n * x^(n+1)*Product_{k = 1..n} 1 + x^k. (Set x = -1 in Andrews, equation 8. For similar results see the Examples below.)

Conjectural g.f: A(x) = Sum_{n >= 1} (-1)^(n+1) * x^(2*n-1)/Product_{k = 1..n} 1 + x^(2*k-1) = x - x^2 + x^5 - x^7 + x^12 - x^15 + - ....

More generally, for positive integer N, we appear to have the identity

A(x) = Product_{j = 1..N-1} 1/(1 + x^(2*j)) * ( P(N,x) + Sum_{n >= 1} (-1)^(n+N) * x^(2*N*n-N)/Product_{k = 1..n} 1 + x^(2*k-1) ), where P(N,x) is a polynomial in x of degree N^2 - N - 1 for N > 1, with the first few values given empirically by P(1,x) = 0, P(2,x) = x, P(3,x) = x - x^2 + x^5 and P(4,x) = x - x^2 + x^3 + x^5 + x^7 - x^8 + x^11. Cf. A186424. (End)

Example

G.f. = x - x^2 + x^5 - x^7 + x^12 - x^15 + x^22 - x^26 + x^35 - x^40 + x^51 - ...
G.f. = q^25 - q^49 + q^121 - q^169 + q^289 - q^361 + q^529 - q^625 + ..
From Peter Bala, Feb 13 2020: (Start)
G.f.s for the tails of A(x):
Sum_{n >= 1} (-1)^(n+1) * x^(2*n+3)*Product_{k = 2..n} 1 + x^k = x^5 - x^7 + x^12 - x^15 + x^22 - ....
Sum_{n >= 2} (-1)^n * x^(3*n+6)*Product_{k = 3..n} 1 + x^k = x^12 - x^15 + x^22 - x^26 + x^35 - ....
Sum_{n >= 3} (-1)^(n+1) * x^(4*n+10)*Product_{k = 4..n} 1 + x^k =
x^22 - x^26 + x^35 - x^40 + x^51 - .... (End)

Maple

N:= 1000: # to get a(0) to a(N)
V:= Array(0..N):
for k from 1 to floor((sqrt(1+24*N)-1)/6) do V[(3*k^2-k)/2]:= 1 od:
for k from 1 to floor((sqrt(1+24*N)+1)/6) do V[(3*k^2+k)/2]:= -1 od:
convert(V, list); # Robert Israel, Nov 24 2015

Mathematica

a[ n_] := Which[ n < 1, 0, SquaresR[ 1, 24 n + 1] == 2, -(-1)^Quotient[ Sqrt[24 n + 1], 3], True, 0]; (* Michael Somos, Jul 12 2015 *)

Prog

(PARI) {a(n) = if( n<1, 0, if( issquare( 24*n + 1, &n), - kronecker( -12, n)))};

Crossrefs

Cf. A003406, A010815, A026837, A026838, A143062.

Sequence in context: A189706 A188321 A257628 * A327974 A286049 A287657

Adjacent sequences: A203565 A203566 A203567 * A203569 A203570 A203571

Keyword

sign

Author

Michael Somos, Jan 03 2012

Status

approved