A369304 - OEIS

%I #56 Dec 03 2024 12:33:33

%S 3,6,14,19,31,38,54,63,83,94,118,131,159,174,206,223,259,278,318,339,

%T 383,406,454,479,531,558,614,643,703,734,798,831,899,934,1006,1043,

%U 1119,1158,1238,1279,1363,1406,1494,1539,1631,1678,1774,1823,1923,1974,2078,2131,2239,2294,2406,2463

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

%C 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.

%C 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.

%C 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).

%C 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.

%C 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

%H Michael De Vlieger, <a href="/A369304/b369304.txt">Table of n, a(n) for n = 1..10000</a>

%H George E. Andrews, <a href="https://doi.org/10.2307/2690367">Euler's Pentagonal Number Theorem</a>, Mathematics Magazine, Vol. 56, No. 5 (Nov., 1983), pp. 279-284.

%H Vladimir Petrov Kostov, <a href="https://arxiv.org/abs/2405.18895">On universal sign patterns</a>, arXiv:2405.18895 [math.CA], 2024. See p. 5.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).

%F a(n) = ((m(n) + 3)^2 - 7)/3 where m(n) = A001651(n) is the n-th natural number not divisible by 3.

%F G.f.: (x*(1+x+x^2)*(3-x^2))/((1-x)^3*(1+x)^2). - _Joerg Arndt_, Jan 19 2024

%F 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

%F 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

%e 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.

%e 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.

%t LinearRecurrence[{1, 2, -2, -1, 1}, {3, 6, 14, 19, 31}, 56] (* _Hugo Pfoertner_, Feb 12 2024 *)

%o (PARI) isok(k) = #select(x->(x==0), Vec((x-1)^3*(x+1)^k)) > 1; \\ _Michel Marcus_, Jan 19 2024

%o (Python)

%o def A369304(n): return ((n+1<<1)-(n>>1))**2//3-2 # _Chai Wah Wu_, Mar 05 2024

%Y Cf. A001651, A022998.

%K nonn,easy

%O 1,1

%A _Vladimir Petrov Kostov_, Jan 19 2024