A369304 Numbers k for which the polynomial (x-1)^3*(x+1)^k has more than one zero coefficient.

3, 6, 14, 19, 31, 38, 54, 63, 83, 94, 118, 131, 159, 174, 206, 223, 259, 278, 318, 339, 383, 406, 454, 479, 531, 558, 614, 643, 703, 734, 798, 831, 899, 934, 1006, 1043, 1119, 1158, 1238, 1279, 1363, 1406, 1494, 1539, 1631, 1678, 1774, 1823, 1923, 1974, 2078, 2131, 2239, 2294, 2406, 2463

Offset: 1

Author

Vladimir Petrov Kostov, Jan 19 2024

Keywords

• nonn
• easy

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).

a(n) = A001082(n+2)-2. Indeed, this formula is valid for n=1,...,20 and the even and odd terms of both sequences A001082 and A369304 are the values of quadratic polynomials in n.

The sequence terms are the exponents in the expansion of Sum_{n >= 1} (-1)^(n+1) * x^(3*n) * Product_{k = 2..n} (1 + x^(2*k-1)) = x^3 - x^6 + x^14 - x^19 + x^31 - x^38 + x^54 - x^63 + x^83 - x^94 + ... (set x = -q and replace q with q^2 in Andrews, equation 8). - Peter Bala, Nov 19 2024

Examples

			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.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• George E. Andrews, Euler's Pentagonal Number Theorem, Mathematics Magazine, Vol. 56, No. 5 (Nov., 1983), pp. 279-284.
• 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).

Crossrefs

Cf. A001651, A022998.

Programs

Mathematica

LinearRecurrence[{1, 2, -2, -1, 1}, {3, 6, 14, 19, 31}, 56] (* Hugo Pfoertner, Feb 12 2024 *)

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

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