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A038505
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Sum of every 4th entry of row n in Pascal's triangle, starting at binomial(n,2).
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18
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0, 0, 1, 3, 6, 10, 16, 28, 56, 120, 256, 528, 1056, 2080, 4096, 8128, 16256, 32640, 65536, 131328, 262656, 524800, 1048576, 2096128, 4192256, 8386560, 16777216, 33558528, 67117056, 134225920, 268435456, 536854528, 1073709056
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OFFSET
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0,4
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COMMENTS
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Number of strings over Z_2 of length n with trace 0 and subtrace 1.
Same as number of strings over GF(2) of length n with trace 0 and subtrace 1.
Binomial transform of (0,1,1,0,0,1,1,0,...) gives a(n) for n >= 1. - Paul Barry, Jul 07 2003
M^n * [1,0,0,0] = [A038503(n), A000749(n), a(n), A038504(n)]; where M = a 4 X 4 matrix [1,1,0,0; 0,1,1,0; 0,0,1,1; 1,0,0,1]. Sum of terms = 2^n.
Example: M^6 * [1,0,0,0] [16, 20, 16, 12]; sum = 2^6 = 64. (End)
{A038503, A038504, A038505, A000749} is the difference analog of the hyperbolic functions of order 4, {h_1(x), h_2(x), h_3(x), h_4(x)}. For a definition of {h_i(x)} and the difference analog {H_i (n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Jun 14 2017
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REFERENCES
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A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
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LINKS
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FORMULA
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a(n; t, s) = a(n-1; t, s) + a(n-1; t+1, s+t+1) where t is the trace and s is the subtrace.
a(n) = Sum_{k=0..n} binomial(n, 2 + 4*k), n >= 0.
a(n) = Sum_{k=0..n} (1/2)*C(n, k)*(-1)^C(k+3, 3) for n >= 1. - Paul Barry, Jul 07 2003
G.f.: x^2*(1-x)/((1-x)^4-x^4) = x^2*(1-x)/((1-2*x)*(1-2*x+2*x^2));
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*(1-(-1)^k)/2. (End)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3; sequence is identical to its fourth differences. - Paul Curtz, Dec 21 2007
a(n+m) = a(n)*H_1(m) + H_2(n)*H_2(m) + H_1(n)*a(m) + H_4(n)*H_4(m),
a(n) = n! [x^n] (1 + exp(2*x) - 2*exp(x)*cos(x))/4.
a(n) = (2^n - (1-i)^n - (1+i)^n) / 4 for n >= 1. Compare V. Shevelevs' formula (1) in A000749. (End)
Proof of the conjecture by Creighton Dement (May 22 2005): using the first formula of Theorem 1 in [Shevelev link] for n=4, omega=i=sqrt(-1), i:=1,2,3,4, m:=n>=1, we have
a(n) = (1/2)*(2^(n-1)-2^(n/2)*cos(Pi*n/4)), A038504(n) = (1/2)*(2^(n-1)+2^(n/2)* sin(Pi*n/4)), A000749(n) = (1/2)*(2^(n-1)-2^(n/2)*sin(Pi*n/4)). Finally we use the formula by Paul Barry: A009545(n) = 2^(n/2)*sin(Pi*n/4) = 2^(n/2)*(-cos(Pi*(n+2)/4)). Now it is easy to obtain the hypothetical formula. (End)
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EXAMPLE
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a(3; 0, 1) = 3 since the three binary strings of trace 0, subtrace 1 and length 3 are { 011, 101, 110 }.
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MAPLE
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s := sqrt(2): h := n -> [-2, -s, 0, s, 2, s, 0, -s][1 + (n mod 8)]:
a := n -> `if`(n=0, 0, (2^n + 2^(n/2)*h(n))/4): seq(a(n), n=0..32);
# Alternatively:
egf := (1 + exp(2*x) - 2*exp(x)*cos(x))/4:
series(egf, x, 33): seq(n!*coeff(%, x, n), n=0..32); # (End)
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MATHEMATICA
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Table[If[n==0, 0, 2^(n-2) - 2^(n/2-1) Cos[Pi*n/4]], {n, 0, 32}] (* Vladimir Reshetnikov, Sep 16 2016 *)
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PROG
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(Magma) I:=[0, 0, 1, 3]; [n le 3 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 22 2012
(Haskell)
a038505 n = a038505_list !! n
a038505_list = tail $ zipWith (-) (tail a000749_list) a000749_list
(Sage)
A = lambda n: (2^n - (1-I)^n - (1+I)^n) / 4 if n != 0 else 0
(GAP) List([0..35], n->Sum([0..n], k->Binomial(n, 2+4*k))); # Muniru A Asiru, Feb 21 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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