OFFSET
1,1
COMMENTS
In this sequence, pairs of consecutive even numbers (excluding the leading term) alternate with pairs of consecutive odd numbers. When in the sequence a(n) is even (resp. when a(n) is odd), the polynomial (x-1)^3*(x+1)^a(n) has two (resp. three) vanishing coefficients.
These are the coefficients of x^j with j = (m(n) +- 2)*(m(n) +- 1)/6, where m(n) = (6*n - 3 - (-1)^n)/4, and for odd a(n), also with j = (a(n) + 3)/2.
The first differences are a(n) - a(n-1) = n+1 if n even, or 2*(n+1) if n odd, for n >= 2 (A022998).
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000
Vladimir Petrov Kostov, On universal sign patterns, arXiv:2405.18895 [math.CA], 2024. See p. 5.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).
FORMULA
a(n) = ((m(n) + 3)^2 - 7)/3 where m(n) = A001651(n) is the n-th natural number not divisible by 3.
G.f.: (x*(1+x+x^2)*(3-x^2))/((1-x)^3*(1+x)^2). - Joerg Arndt, Jan 19 2024
E.g.f.: (4 + (3*x^2 + 13*x - 4)*cosh(x) + (3*x^2 + 11*x - 1)*sinh(x))/4. - Stefano Spezia, Feb 13 2024
Sum_{n>=1} 1/a(n) = 3/2 + (tan((1+2*sqrt(7))*Pi/6) - cot((1+sqrt(7))*Pi/3)) * Pi/(2*sqrt(7)). - Amiram Eldar, Mar 07 2024
EXAMPLE
For n=1, a(1)=3 and the polynomial (x-1)^3*(x+1)^3 = x^6 - 3*x^4 + 3*x^2 - 1 has three vanishing coefficients, those of x^5, x^3 and x.
For n=2, a(2)=6 and the polynomial (x-1)^3*(x+1)^6 = x^9 + 3*x^8 - 8*x^6 - 6*x^5 + 6*x^4 + 8*x^3 - 3*x - 1 has two vanishing coefficients, those of x^7 and x^2.
MATHEMATICA
LinearRecurrence[{1, 2, -2, -1, 1}, {3, 6, 14, 19, 31}, 56] (* Hugo Pfoertner, Feb 12 2024 *)
PROG
(PARI) isok(k) = #select(x->(x==0), Vec((x-1)^3*(x+1)^k)) > 1; \\ Michel Marcus, Jan 19 2024
(Python)
def A369304(n): return ((n+1<<1)-(n>>1))**2//3-2 # Chai Wah Wu, Mar 05 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Petrov Kostov, Jan 19 2024
STATUS
approved