cp's OEIS Frontend

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

%I A369304 #56 Dec 03 2024 12:33:33
%S A369304 3,6,14,19,31,38,54,63,83,94,118,131,159,174,206,223,259,278,318,339,
%T A369304 383,406,454,479,531,558,614,643,703,734,798,831,899,934,1006,1043,
%U A369304 1119,1158,1238,1279,1363,1406,1494,1539,1631,1678,1774,1823,1923,1974,2078,2131,2239,2294,2406,2463
%N A369304 Numbers k for which the polynomial (x-1)^3*(x+1)^k has more than one zero coefficient.
%C A369304 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 A369304 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 A369304 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 A369304 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 A369304 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 A369304 Michael De Vlieger, <a href="/A369304/b369304.txt">Table of n, a(n) for n = 1..10000</a>
%H A369304 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 A369304 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 A369304 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F A369304 a(n) = ((m(n) + 3)^2 - 7)/3 where m(n) = A001651(n) is the n-th natural number not divisible by 3.
%F A369304 G.f.: (x*(1+x+x^2)*(3-x^2))/((1-x)^3*(1+x)^2). - _Joerg Arndt_, Jan 19 2024
%F A369304 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 A369304 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 A369304 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 A369304 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 A369304 LinearRecurrence[{1, 2, -2, -1, 1}, {3, 6, 14, 19, 31}, 56] (* _Hugo Pfoertner_, Feb 12 2024 *)
%o A369304 (PARI) isok(k) = #select(x->(x==0), Vec((x-1)^3*(x+1)^k)) > 1; \\ _Michel Marcus_, Jan 19 2024
%o A369304 (Python)
%o A369304 def A369304(n): return ((n+1<<1)-(n>>1))**2//3-2 # _Chai Wah Wu_, Mar 05 2024
%Y A369304 Cf. A001651, A022998.
%K A369304 nonn,easy
%O A369304 1,1
%A A369304 _Vladimir Petrov Kostov_, Jan 19 2024