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 A038505 Sum of every 4th entry of row n in Pascal's triangle, starting at binomial(n,2). 18
 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 of {h_i(x)} and the difference analog {H_i (n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Jun 14 2017 REFERENCES A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII. LINKS Peter Luschny, Table of n, a(n) for n = 0..1000 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 (GAP) List([0..35], n->Sum([0..n], k->Binomial(n, 2+4*k))); # Muniru A Asiru, Feb 21 2019 CROSSREFS Cf. A000749, A009116, A009545, A038503, A038504. Sequence in context: A130578 A107068 A033541 * A119971 A318290 A291986 Adjacent sequences:  A038502 A038503 A038504 * A038506 A038507 A038508 KEYWORD easy,nonn AUTHOR 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 April 17 01:26 EDT 2021. Contains 343051 sequences. (Running on oeis4.)