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A038505 Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 2". 17
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

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

From Gary W. Adamson, Mar 13 2009: (Start)

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 see the reference "Higher Trancendental Functions" and the Shevelev link. - Vladimir Shevelev, Jun 14 2017

REFERENCES

Higher Trancendental Functions, Bateman Manuscript Project, Vol. 3, ed. A. Erdelyi, 1983 (chapter XVIII).

LINKS

Peter Luschny, Table of n, a(n) for n = 0..1000

F. Ruskey, Strings over Z_2 of given Trace and Subtrace

F. Ruskey, Strings over GF(2) of given Trace and Subtrace

Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4).

FORMULA

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

From Paul Barry, Nov 29 2004: (Start)

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)

Conjecture: 2*a(n+2) = A038504(n+2) + A000749(n+2) + 2*A009545(n) - Creighton Dement, May 22 2005

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) = A000749(n+1) - A000749(n). - Reinhard Zumkeller, Jul 15 2013

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),

where H_1=A038503, H_2=A038504, H_4=A000749. - Vladimir Shevelev, Jun 14 2017

From Peter Luschny, Jun 15 2017: (Start)

a(n) = n! [x^n] (1 + exp(2*x) - 2*exp(x)*cos(x))/4.

a(n) = A038503(n+2) - 2*A038503(n+1) + A038503(n).

a(n) = 2^(n-2) - A046980(n)*2^(A004525(n-3)) for n>=1.

a(n) = (2^n - (1-i)^n - (1+i)^n) / 4 for n>=1. Compare V. Shevelevs' formula (1) in A000749. (End)

From Vladimir Shevelev, Jun 16 2017: (Start)

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)

EXAMPLE

a(3; 0, 1) = 3 since the three binary strings of trace 0, subtrace 1 and length 3 are { 011, 101, 110 }.

MAPLE

# From Peter Luschny, Jun 15 2017: (Start)

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)

MATHEMATICA

LinearRecurrence[{4, -6, 4}, {0, 0, 1, 3}, 40] (* Vincenzo Librandi, Jun 22 2012 *)

Table[If[n==0, 0, 2^(n-2) - 2^(n/2-1) Cos[Pi*n/4]], {n, 0, 32}] (* Vladimir Reshetnikov, Sep 16 2016 *)

PROG

(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

-- Reinhard Zumkeller, Jul 15 2013

(Sage)

A = lambda n: (2^n - (1-I)^n - (1+I)^n) / 4 if n <> 0 else 0

print [A(n) for n in (0..32)] # Peter Luschny, Jun 16 2017

CROSSREFS

Cf. A000749, A009116, A009545, A038503, A038504.

Sequence in context: A130578 A107068 A033541 * A119971 A094272 A236326

Adjacent sequences:  A038502 A038503 A038504 * A038506 A038507 A038508

KEYWORD

easy,nonn,changed

AUTHOR

Frank Ruskey

EXTENSIONS

Missing 0 prepended by Vladimir Shevelev, Jun 14 2017

Edited by Peter Luschny, Jun 16 2017

STATUS

approved

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Last modified June 25 22:15 EDT 2017. Contains 288730 sequences.