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A100258
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Triangle of coefficients of normalized Legendre polynomials, with increasing exponents.
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15
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1, 0, 1, -1, 0, 3, 0, -3, 0, 5, 3, 0, -30, 0, 35, 0, 15, 0, -70, 0, 63, -5, 0, 105, 0, -315, 0, 231, 0, -35, 0, 315, 0, -693, 0, 429, 35, 0, -1260, 0, 6930, 0, -12012, 0, 6435, 0, 315, 0, -4620, 0, 18018, 0, -25740, 0, 12155, -63, 0, 3465, 0, -30030, 0, 90090, 0, -109395, 0, 46189
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OFFSET
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0,6
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COMMENTS
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For a relation to Jacobi quartic elliptic curves, see the MathOverflow link. For a self-convolution of the polynomials relating them to the Chebyshev and Fibonacci polynomials, see A049310 and A053117. For congruences and connections to other polynomials (Jacobi, Gegenbauer, and Chebyshev) see the Allouche et al. link. For relations to elliptic cohomology and modular forms, see references in Copeland link.- Tom Copeland, Feb 04 2016
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 798.
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
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FORMULA
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The n-th normalized Legendre polynomial is generated by 2^(-n-a(n)) (d/dx)^n (x^2-1)^n / n! with a(n) = A005187(n/2) for n even and a(n) = A005187((n-1)/2) for n odd. The non-normalized polynomials have the o.g.f. 1 / sqrt(1 - 2xz + z^2). - Tom Copeland, Feb 07 2016
The consecutive nonzero entries in the m-th row are, in order, (c+b)!/(c!(m-b)!(2b-m)!*A048896(m-1)) with sign (-1)^b where c = m/2-1, m/2, m/2+1, ..., (m-1) and b = c+1 if m is even and sign (-1)^c with c = (m-1)/2, (m-1)/2+1, (m-1)/2+2, ..., (m-1) with b = c+1 if m is odd. For the 9th row the 5 consecutive nonzero entries are 315, -4620, 18018, -25740, 12155 given by c = 4,5,6,7,8 and b = 5,6,7,8,9. - Richard Turk, Aug 22 2017
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EXAMPLE
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Triangle begins:
1;
0, 1;
-1, 0, 3;
0, -3, 0, 5;
3, 0, -30, 0, 35;
0, 15, 0, -70, 0, 63;
-5, 0, 105, 0, -315, 0, 231;
0, -35, 0, 315, 0, -693, 0, 429;
35, 0, -1260, 0, 6930, 0, -12012, 0, 6435;
...
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MATHEMATICA
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row[n_] := CoefficientList[ LegendreP[n, x], x]*2^IntegerExponent[n!, 2]; Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Jan 15 2015 *)
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PROG
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(PARI) a(k, n)=polcoeff(pollegendre(k, x), n)*2^valuation(k!, 2)
(Python)
from mpmath import *
mp.dps=20
def a007814(n):
return 1 + bin(n - 1)[2:].count('1') - bin(n)[2:].count('1')
for n in range(11):
y=2**sum(a007814(i) for i in range(2, n+1))
l=chop(taylor(lambda x: legendre(n, x), 0, n))
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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