A140414 Triangle T(p,s) showing the coefficients of sequences which are half their p-th differences.
3, 2, 1, 3, -3, 3, 4, -6, 4, 1, 5, -10, 10, -5, 3, 6, -15, 20, -15, 6, 1, 7, -21, 35, -35, 21, -7, 3, 8, -28, 56, -70, 56, -28, 8, 1, 9, -36, 84, -126, 126, -84, 36, -9, 3, 10, -45, 120, -210, 252, -210, 120, -45, 10, 1
Offset: 1
Examples
The triangle starts in row p=0 as: 3; (p=1, example A000244, a(n+1)=3*a(n) ) 2, 1; (p=2 example A000244 or A000129, a(n+2) = 2*a(n+1)+a(n) ) 3, -3, 3; (p=3 example A052103 or A136297, a(n+3) = 3*a(n+2)-3*a(n+1)+3*a(n) ) 4, -6, 4, 1; 5,-10, 10, -5, 3; 6,-15, 20, -15, 6, 1; 7,-21, 35, -35, 21, -7, 3; 8,-28, 56, -70, 56, -28, 8, 1; 9,-36, 84,-126,126, -84, 36, -9, 3; 10,-45,120,-210,252,-210,120,-45,10,1;
Crossrefs
Cf. A135356.
A141775 Binomial transform of (1, 2, 0, 1, 2, 0, 1, 2, 0, ...).
1, 3, 5, 8, 15, 31, 64, 129, 257, 512, 1023, 2047, 4096, 8193, 16385, 32768, 65535, 131071, 262144, 524289, 1048577, 2097152, 4194303, 8388607, 16777216, 33554433, 67108865, 134217728, 268435455, 536870911, 1073741824, 2147483649, 4294967297, 8589934592, 17179869183
Offset: 0
Comments
From Paul Curtz, Jun 15 2011: (Start)
A square array of a(n) and its higher order differences is defined by T(0,k) = a(k) and T(n,k) = T(n-1,k+1)-T(n-1,k):
1, 3, 5, 8, 15, 31,
2, 2, 3, 7, 16, 33,
0, 1, 4, 9, 17, 32, see A130785(n).
1, 3, 5, 8, 15, 31,
2, 2, 3, 7, 16, 33,
a(n) is identical to its third differences: T(n+3,k) = T(n,k).
The main diagonal is T(n,n) = 2^n. Subdiagonals are T(n,n-1) = A014551(n) and T(n,n-2) = A062510(n).
(End)
Examples
a(4) = 8 = (1, 2, 0, 1) dot (1, 3, 3, 1) = (1 + 6 + 0 + 1).
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,-3,2).
Programs
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Magma
I:=[1,3,5]; [n le 3 select I[n] else 3*Self(n-1) - 3*Self(n-2) + 2*Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 15 2018
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Mathematica
LinearRecurrence[{3,-3,2},{1,3,5},40] (* Harvey P. Dale, May 29 2012 *) -
PARI
x='x+O('x^30); Vec((x-1)*(1+x)/((2*x-1)*(x^2-x+1))) \\ G. C. Greubel, Jan 15 2018
Formula
From Paul Curtz, Jun 15 2011: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3).
a(n) = 2^n - A128834(n).
a(n) - 2a(n-1)= A057079(n+1).
a(n) + a(n+3) = 9*2^n.
a(n+6) - a(n) = 63*2^n.
G.f.: (x-1)*(1+x) / ( (2*x-1)*(x^2-x+1) ). - R. J. Mathar, Jun 22 2011
a(n) = 2^n + (2*sin((Pi*n)/3))/sqrt(3). - Colin Barker, Feb 10 2017
A330863 Decimal expansion of Product_{k>=1} (1 + 1/(-2)^k).
5, 6, 8, 6, 9, 8, 9, 4, 6, 2, 6, 5, 4, 2, 8, 5, 0, 5, 9, 5, 4, 9, 7, 6, 7, 3, 7, 0, 7, 4, 4, 4, 4, 6, 5, 4, 2, 9, 0, 8, 5, 2, 4, 5, 1, 3, 8, 9, 3, 5, 9, 0, 2, 9, 3, 1, 9, 3, 4, 4, 0, 4, 6, 0, 1, 8, 3, 5, 3, 5, 6, 3, 2, 3, 0, 9, 1, 2, 6, 4, 0, 9, 6, 1, 4, 6, 4, 4, 1, 1, 7, 3, 0, 6, 1, 4, 8, 6, 0, 4, 8, 0, 2, 7, 2, 6, 9, 4, 1, 8
Offset: 0
Examples
(1 - 1/2) * (1 + 1/2^2) * (1 - 1/2^3) * (1 + 1/2^4) * (1 - 1/2^5) * ... = 0.568698946265428505954976737...
Crossrefs
Programs
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Mathematica
RealDigits[QPochhammer[-1, -1/2]/2, 10, 110] [[1]] N[3/QPochhammer[-2, 1/4], 120] (* Vaclav Kotesovec, Apr 28 2020 *)
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PARI
prodinf(k=1, 1 + 1/(-2)^k) \\ Michel Marcus, Apr 28 2020
Formula
Equals Product_{k>=1} 1/(1 + 1/2^(2*k-1)).
Equals exp(Sum_{k>=1} A000593(k)/(k*(-2)^k)).
From Peter Bala, Dec 15 2020: (Start)
Constant C = (2/3) - (1/3)*Sum_{n >= 0} (-1)^n * 2^(n^2)/( Product_{k = 1..n+1} 4^k - 1 ).
C = Sum_{n >= 0} 1/( Product_{k = 1..n} (-2)^k - 1 ) = 1 - 1/3 - 1/9 + 1/81 + 1/1215 - - + + ... = Sum_{n >= 0} 1/A216206(n).
C = 1 + Sum_{n >= 0} (-1/2)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
3*C = 2 - Sum_{n >= 0} (1/4)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
9*C = 5 - Sum_{n >= 0} (-1/8)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
81*C = 46 + Sum_{n >= 0} (1/16)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
1215*C = 691 + Sum_{n >= 0} (-1/32)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
The sequence [1, 2, 5, 46, 691, ...] is the sequence of numerators of the partial sums of the series Sum_{n >= 0} 1/A216206(n). (End)
A323210 a(n) = 9*J(n)^2 where J(n) are the Jacobsthal numbers A001045 with J(0) = 1.
1, 9, 9, 81, 225, 1089, 3969, 16641, 65025, 263169, 1046529, 4198401, 16769025, 67125249, 268402689, 1073807361, 4294836225, 17180131329, 68718952449, 274878955521, 1099509530625, 4398050705409, 17592177655809, 70368760954881, 281474943156225, 1125899973951489
Offset: 0
Comments
Colin Barker conjectures that A208556 is a shifted version of this sequence.
Links
- Index entries for linear recurrences with constant coefficients, signature (3,6,-8).
Programs
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Maple
gf := (8*x^3 - 24*x^2 + 6*x + 1)/((4*x - 1)*(2*x + 1)*(x - 1)): ser := series(gf,x,32): seq(coeff(ser,x,n), n=0..25);
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Mathematica
LinearRecurrence[{3, 6, -8}, {1, 9, 9, 81}, 25] -
Sage
# Demonstrates the product formula. CC = ComplexField(200) def t(n,k): return CC(3)*cos(CC(pi*k/n)) - CC(i)*sin(CC(pi*k/n)) def T(n,k): return t(n,k)*(t(n,k).conjugate()) def a(n): return prod(T(n,k) for k in (1..n)) print([a(n).real().round() for n in (0..29)])
Formula
a(n) = Product_{k=1..n} T(n, k) where T(n, k) = t(n,k)*conjugate(t(n,k)) and t(n,k) = 3*cos(Pi*k/n) - i*sin(Pi*k/n), i is the imaginary unit.
a(n) = [x^n] (8*x^3 - 24*x^2 + 6*x + 1)/((4*x - 1)*(2*x + 1)*(x - 1)).
a(n) = n! [x^n] (1 + exp(x) - 2*exp(-2*x) + exp(4*x)).
a(n) = 3*a(n-1) + 6*a(n-2) - 8*a(n-3) for n >= 4.
A062510(n) = sqrt(a(n)) for n > 0.
a(n) = 4^n-2*(-2)^n+1, n>0. - R. J. Mathar, Mar 06 2022
Comments