cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A220672 Coefficients of powers of x^2 of polynomials, called h(2,n,x^2), appearing in a conjecture on alternating sums of fifth powers of odd-indexed Chebyshev S polynomials stated in A220671.

Original entry on oeis.org

-14, 6, 5, -12, 3, 46, -95, 16, 75, -69, 24, -3, 106, -520, 928, -607, -351, 894, -651, 234, -42, 3, 186, -1600, 5840, -11355, 11005, -1110, -9615, 11580, -6906, 2433, -513, 60, -3, 286, -3775, 22360, -75595, 153515, -177565, 77115, 84495, -171324, 145302, -75831, 26235, -6057, 900, -78, 3
Offset: 0

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Author

Wolfdieter Lang, Jan 14 2013

Keywords

Comments

The row lengths sequence for this irregular triangle is 3*n+1 = A016777(n).
A generalized Melham conjecture involving fifth powers (m=2) of odd-indexed Chebyshev S polynomials (see A049310) is H(2,n,x^2):= (x^2-3)*(x^4-5*x^2+5)*sum(((-1)^k)*(S(2*k-1,x)/x)^(2*m+1), k=0..n)/((1 - (-1)^n*S(2*n,x))/x^2)^2 = h(2,n,x^2) - 3*z(n) + 8*z(n)^2 + 4*z(n)^3, with z(n):= ((-1)^n)*S(2*n,x), and h an integer polynomial of degree 3*n.
The present array a(n,p) appears as h(2,n,x^2) = sum(a(n,p)*x^(2*p),p=0..3*n), n >= 1. The entry a(0,0) := -14 has been used because, in accordance with the original Melham conjecture (see a comment on A220671), h(2,n,i^2), with the imaginary unit i, is conjectured to be -14, for all n >= 1.
[-14, -3, 8, 4] is row m=2 of A217475.

Examples

			The array a(n,p) begins:
n\p   0     1    2     3     4    5     6    7    8  9
0:  -14
1:    6     5  -12     3
2:   46   -95   16    75   -69   24    -3
3:  106  -520  928  -607  -351  894  -651  234  -42  3
...
Row n=4: [186, -1600, 5840, -11355, 11005, -1110, -9615, 11580, -6906, 2433, -513, 60, -3];
Row n=5: [286, -3775, 22360, -75595, 153515, -177565, 77115, 84495, -171324, 145302, -75831, 26235, -6057, 900, -78, 3].
Thus the conjecture is true at least for n=1..5.
		

Crossrefs

Formula

a(n,p) = [x^(2p)] h(0,2,n,x^2), with the polynomial h defined above in a comment. The conjecture is that h is an integer polynomial of degree 3n in x^2.

A220670 Coefficient triangle for powers of x^2 of polynomials appearing in a generalized Melham conjecture on alternating sums of third powers of Chebyshev's S polynomials with odd indices. Coefficients in powers of x^2 of 2 + (-1)^n*S(2*n,x).

Original entry on oeis.org

3, 3, -1, 3, -3, 1, 3, -6, 5, -1, 3, -10, 15, -7, 1, 3, -15, 35, -28, 9, -1, 3, -21, 70, -84, 45, -11, 1, 3, -28, 126, -210, 165, -66, 13, -1, 3, -36, 210, -462, 495, -286, 91, -15, 1, 3, -45, 330, -924, 1287, -1001, 455, -120, 17, -1, 3, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1
Offset: 0

Views

Author

Wolfdieter Lang, Jan 07 2013

Keywords

Comments

For the original Melham conjecture on sums of odd powers of even-indexed Fibonacci numbers see the references given in A217475. See especially the Wang and Zhang reference given there.
An analog conjecture stated for Chebyshev's S polynomials (see A049310) is product(tau(j,x),j=0..m)*sum(((-1)^k)*(S(2*k-1,x)/x)^(2*m+1),k=0..n)/(P(n,x^2)^2) = H(m,n,x^2), with P(n,x^2) := (1 - (-1)^n*S(2*n,x))/x^2 and certain integer polynomials H with degree (2*m-1)*n + binomial(m-1,2) in x^2. The coefficients of powers of x^2 of the monic integer polynomials tau(n,x):= 2*T(2*n+1,x/2)/x, with Chebyshev's T polynomials, are given by the signed A111125 triangle (see a comment there from Oct 23 2012). The coefficients of the powers of x^(2*j) of the polynomials P(n,x^2) are found in (-1)^(n-1)*A109954(n-1,j).
Here the conjecture is considered for m=1 (third powers): H(1,n,x^2) = sum(a(n,p)*x^(2*p),p=0..n), n >= 1. It is conjectured that in fact H(1,n,x^2) = 2 + (-1)^n*S(2*n,x). This has been checked by Maple for n=1..100. Therefore we have added a(0,0) = 3 (in the conjecture above this would be the undetermined 0/0).
The original Melham conjecture for m=1 (third powers), appears by putting x = i (the imaginary unit): 1*4*sum(F(2*k)^3)/(1-F(2*n+1))^2 = sum(a(n,p)*(-1)^p) = 2 + F(2*n+1) (the unsigned row sums of the present triangle). This m=1 identity is, of course, proved.
The row sums of this triangle are given by 2 + (-1)^n*S(2*n,1) = 2 + (-1)^n*((2/sqrt(3))*sin((2*n+1)*Pi/3)) producing the period 6 sequence periodic (3, 2, 1, 1, 2, 3).

Examples

			The triangle a(n,p) begins:
n\p 0    1    2     3    4     5    6    7   8   9 10 ...
0:  3
1:  3   -1
2:  3   -3    1
3:  3   -6    5    -1
4:  3  -10   15    -7    1
5:  3  -15   35   -28    9    -1
6:  3  -21   70   -84   45   -11    1
7:  3  -28  126  -210  165   -66   13  -1
8:  3  -36  210  -462  495  -286   91  -15   1
9:  3  -45  330  -924 1287 -1001  455 -120  17  -1
10: 3  -55  495 -1716 3003 -3003 1820 -680 153 -19  1
...
Row n=2: H(1,2,x^2) := (-3+x^2)*(0 - (S(1,x)/x)^3 + (S(3,x)/x)^3)/((1 - S(4,x))/x^2)^2 = 3 - 3*x^2 + x^4 =
  2 + S(4,x).
Row n=3:  H(1,3,x^2) := (-3+x^2)*(0 - (S(1,x)/x)^3 + (S(3,x)/x)^3 - (S(5,x)/x)^3 )/((1 + S(6,x))/x^2)^2 =  3-6*x^2+5*x^4-x^6 = 2 - S(6,x).
		

Crossrefs

Cf, A049310, A111125 (signed), A109954 (signed), A217475, A220671 (fifth powers).

Formula

a(n,p) = [x^(2*p)] H(1,n,x^2), with H(1,n,x^2) := (-3+x^2)*sum(((-1)^k)*(S(2*k-1,x)/x)^3,k=0..n)/((1 - (-1)^n*S(2*n,x))/x^2)^2, n >= 1, p = 0..n, and a(0,0):=3.
a(n,p) = [x^(2*p)] (2 + (-1)^n*S(2*n,x)), n >= 0, p = 0..n.
Showing 1-2 of 2 results.