OFFSET
0,2
COMMENTS
Compare to the curious identities:
(1) Sum_{n=-oo..+oo} x^n * (1 - x^n)^n = 0.
(2) Sum_{n=-oo..+oo} (-x)^n * (1 + x^n)^n = 0.
Name changed for clarity by Paul D. Hanna, Dec 10 2024; prior name was "G.f.: (1/2) * Sum_{n=-oo..+oo} x^n * (1 + x^n)^n, an even function."
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..8200
FORMULA
The g.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) A(x^2) = (1/2) * Sum_{n=-oo..+oo} x^n * (1 + x^n)^n.
(2) A(x^2) = (1/2) * Sum_{n=-oo..+oo} (-x)^n * (1 - x^n)^n.
(3) A(x^2) = (1/2) * Sum_{n=-oo..+oo} x^(n^2-n) / (1 + x^n)^n.
(4) A(x^2) = (1/2) * Sum_{n=-oo..+oo} x^(n^2-n) / (1 - x^n)^n.
(5) A(x) = Sum_{n=-oo..+oo} x^n * (1 + x^n)^(2*n).
(6) A(x) = Sum_{n=-oo..+oo} x^n * (1 - x^n)^(2*n).
(7) A(x) = Sum_{n=-oo..+oo} x^(2*n^2-n) / (1 - x^n)^(2*n).
(8) A(x) = Sum_{n=-oo..+oo} x^(2*n^2-n) / (1 + x^n)^(2*n).
a(2^n) = 1 for n > 0 (conjecture).
a(p) = p+1 for primes p > 3 (conjecture).
From Peter Bala, Jan 23 2021: (Start)
The following are conjectural:
A(x^2) = Sum_{n = -oo..+oo} x^(2*n+1)*(1 + x^(2*n+1) )^(2*n+1).
Equivalently: A(x^2) = Sum_{n = -oo..+oo} x^(4*n^2 + 2*n)/(1 + x^(2*n+1))^(2*n+1).
a(2*n+1) = [x^(2*n+1)] Sum_{n = -oo..+oo} x^(2*n+1)*(1 + x^(2*n+1))^(4*n+2)
More generally, for k = 1,2,3,..., a((2^k)*(2*n + 1)) = [x^(2*n+1)] Sum_{n = -oo..+oo} x^(2*n+1)*(1 + x^(2*n+1))^(2^(k+1)*(2*n+1)).
a(2*n+1) = [x^(2*n+1)] Sum_{n = -oo..+oo} (-1)^(n+1)*x^n*(1 + x^n)^(2*n) = [x^(2*n+1)] Sum_{n = -oo..+oo} (-1)^(n+1)*x^n*(1 - x^n)^(2*n).
More generally, for k = 1,2,3,...,
a((2^k)*(2*n+1)) = [x^(2*n+1)] Sum_{n = -oo..+oo} (-1)^(n+1)*x^n*(1 + x^n)^(2^(k+1)*n) = [x^(2*n+1)] Sum_{n = -oo..+oo} (-1)^(n+1)*x^n*(1 - x^n)^(2^(k+1)*n).
a(4*n+2) = [x^(4*n+2)] Sum_{n = -oo..+oo} (-1)^n*x^n*(1 + x^n)^(2*n) = [x^(4*n+2)] Sum_{n = -oo..+oo} (-1)^n*x^n*(1 - x^n)^(2*n).
a(n) = [x^(2*n)] Sum_{n = -oo..+oo} (-1)^n*x^(2*n+1)*(1 + (-1)^n* x^(2*n+1) )^(2*n+1).
For k = 1,2,3,...,
a((2^k)*(2*n+1)) = [x^(2*n+1)] Sum_{n = -oo..+oo} x^(2*n+1)*(1 + (-1)^n* x^(2*n+1) )^(2^(k+1)*(2*n+1)).
(End)
From Peter Bala, Mar 02 2025: (Start)
a(n) = Sum_{d divides n} (binomial(2*n/d, d-1) + (-1)^(n/d+1) * binomial(n/d, 2*d-1)) for n >= 1.
Hence, a(p) = p + 1 for primes p > 3 and a(2^n) = 1 for n > 0 as conjectured above. (End)
EXAMPLE
G.f.: A(x) = 1 + 2*x^2 + x^4 + 5*x^6 + x^8 + 6*x^10 + 8*x^12 + 8*x^14 + x^16 + 25*x^18 + 12*x^20 + ...
where 2*A(x) = 1 + P(x) + N(x) with
P(x) = x*(1+x) + x^2*(1+x^2)^2 + x^3*(1+x^3)^3 + x^4*(1+x^4)^4 + x^5*(1+x^5)^5 + ...
N(x) = 1/(1+x) + x^2/(1+x^2)^2 + x^6/(1+x^3)^3 + x^12/(1+x^4)^4 + x^20/(1+x^5)^5 + ...
Explicitly,
P(x) = x + 2*x^2 + x^3 + 3*x^4 + x^5 + 5*x^6 + x^7 + 5*x^8 + 4*x^9 + 6*x^10 + x^11 + 14*x^12 + x^13 + 8*x^14 + 11*x^15 + 13*x^16 + x^17 + 25*x^18 + x^19 + 22*x^20 + 22*x^21 + 12*x^22 + x^23 + 61*x^24 + 6*x^25 +...+ A217668(n)*x^n + ...
N(x) = 1 - x + 2*x^2 - x^3 - x^4 - x^5 + 5*x^6 - x^7 - 3*x^8 - 4*x^9 + 6*x^10 - x^11 + 2*x^12 - x^13 + 8*x^14 - 11*x^15 - 11*x^16 - x^17 + 25*x^18 - x^19 + 2*x^20 - 22*x^21 + 12*x^22 - x^23 - 3*x^24 - 6*x^25 +...+ A260148(n)*x^n + ...
From Paul D. Hanna, Dec 10 2024: (Start)
SPECIFIC VALUES.
A(z) = 0 at z = -0.404783857785183643579648014798209689698619095608142590080356...
where 0 = Sum_{n=-oo..+oo} z^n * (1 + z^n)^(2*n).
A(t) = 8 at t = 0.66184860446935243758952792459096102121713616089603...
A(t) = 7 at t = 0.64280265347584821638335226655422639958638446962646...
A(t) = 6 at t = 0.61846293982236470622283664293769398297407552626520...
A(t) = 5 at t = 0.58591538561726828976301449562779896617926938759041...
A(t) = 4 at t = 0.53948212974289878102531393938569583066950526874204...
A(t) = 3 at t = 0.46633361832235508894561538442655261465230172977527...
A(t) = 2 at t = 0.33014122063168294490173944063355594394361494532642...
where 2 = Sum_{n=-oo..+oo} t^n * (1 + t^n)^(2*n).
A(t) = -1 at t = -0.57221202613754835881500708971837082259712665852148...
A(t) = -2 at t = -0.66124771863833308133360587362156745037996654826889...
A(t) = -3 at t = -0.72841228559829175547612598129696947453305714538354...
A(t) = -4 at t = -0.90975449896027994776675798799643226140294233213401...
A(4/5) = 39.597156112579883800797829785472315940190856875500...
A(3/4) = 18.522637966827153559321082877260756270457362912092...
A(2/3) = 8.2917909754417331599016245586686519315443444070756...
A(3/5) = 5.3942577326786364433206097043093210828422082884565...
A(1/2) = 3.3971121875472777749836900920631175982646917998641...
where A(1/2) = Sum_{n=-oo..+oo} (2^n + 1)^(2*n) / 2^(2*n^2+n).
A(2/5) = 2.4226617866265771206729430879848898772232404418272...
A(1/3) = 2.0164022766484546805373278337731916678136050742206...
where A(1/3) = Sum_{n=-oo..+oo} (3^n + 1)^(2*n) / 3^(2*n^2+n).
A(1/4) = 1.6529591092151291503041860933179009814428123139546...
where A(1/4) = Sum_{n=-oo..+oo} (4^n + 1)^(2*n) / 4^(2*n^2+n).
A(1/5) = 1.4841513733060571811336245213703004776194631749017...
where A(1/5) = Sum_{n=-oo..+oo} (5^n + 1)^(2*n) / 5^(2*n^2+n).
(End)
MAPLE
with(numtheory):
seq(add(binomial(2*n/d, d-1) + (-1)^(n/d+1) * binomial(n/d, 2*d-1), d in divisors(n)), n = 1..70); # Peter Bala, Mar 02 2025
MATHEMATICA
terms = 100; max = 2 terms; 1/2 Sum[x^n*(1 + x^n)^n, {n, -max, max}] + O[x]^max // CoefficientList[#, x^2]& (* Jean-François Alcover, May 16 2017 *)
PROG
(PARI) {a(n) = my(A=1); A = sum(k=-2*n-2, 2*n+2, x^k*(1+x^k)^k/2 + O(x^(2*n+2)) ); polcoef(A, 2*n)}
for(n=0, 60, print1(a(n), ", "))
(PARI) {a(n) = my(A=1); A = sum(k=-2*n-2, 2*n+2, x^(k^2-k) / (1 + x^k)^k /2 + O(x^(2*n+2)) ); polcoef(A, 2*n)}
for(n=0, 60, print1(a(n), ", "))
(PARI) {a(n) = my(A=1); A = sum(k=-sqrtint(n)-1, n+1, x^k*((1+x^k)^(2*k) + (1-x^k)^(2*k))/2 + O(x^(n+1)) ); polcoef(A, n)}
for(n=0, 60, print1(a(n), ", "))
(PARI) {a(n) = my(A=1); A = sum(k=-n-1, n+1, x^k*(1+x^k)^(2*k) + O(x^(n+1)) ); polcoef(A, n)}
for(n=0, 60, print1(a(n), ", "))
(PARI) {a(n) = my(A=1); A = sum(k=-n-1, n+1, x^(2*k^2-k)/(1-x^k + O(x^(n+1)))^(2*k) ); polcoef(A, n)}
for(n=0, 60, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Jul 17 2015
STATUS
approved
