A038504 Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 1".
0, 1, 2, 3, 4, 6, 12, 28, 64, 136, 272, 528, 1024, 2016, 4032, 8128, 16384, 32896, 65792, 131328, 262144, 523776, 1047552, 2096128, 4194304, 8390656, 16781312, 33558528, 67108864, 134209536, 268419072, 536854528, 1073741824
Offset: 0
Examples
a(2;1,0) = 3 since the two binary strings of trace 1, subtrace 0 and length 2 are { 10, 01 }.
References
- A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- F. Ruskey, Strings over Z_2 with given trace and subtrace
- F. Ruskey, Strings over GF(2) with 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).
Programs
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Magma
[0] cat [n le 3 select n else 4*Self(n-1) -6*Self(n-2) + 4*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 22 2012
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Mathematica
CoefficientList[Series[x(1-x)^2/((1-2x)(1-2x+2x^2)),{x,0,40}],x] (* Vincenzo Librandi, Jun 22 2012 *) LinearRecurrence[{4,-6,4},{0,1,2,3},40] (* Harvey P. Dale, Aug 23 2017 *)
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SageMath
@CachedFunction def a(n): # a = A038504 if (n<4): return n else: return 4*a(n-1) - 6*a(n-2) + 4*a(n-3) [a(n) for n in range(51)] # G. C. Greubel, Apr 20 2023
Formula
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3. - Paul Curtz, Mar 01 2008
G.f.: x*(1-x)^2/((1-2*x)*(1-2*x+2*x^2)).
From Paul Barry, Jul 25 2004: (Start)
Binomial transform of x/(1-x^4).
G.f.: x*(1-x)^2/((1-x)^4 - x^4) = x/(1-2*x) - x^3/((1-x)^4 - x^4).
a(n) = Sum_{k=0..floor(n/4)} binomial(n, 4*k+1).
a(n) = Sum_{k=0..n} binomial(n, k)*(sin(Pi*k/2)/2 + (1 - (-1)^k)/4).
a(n) = 2^(n-2) + 2^((n-2)/2)*sin(Pi*n/4) - 0^n/4. (End)
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.
(1, 2, 3, 4, 6, ...) is the binomial transform of (1, 1, 0, 0, 1, 1, ...). - Gary W. Adamson, May 15 2007
From Vladimir Shevelev, Jul 31 2017: (Start)
For n >= 1, {H_i(n)} are linearly dependent sequences: a(n) = H_2(n) = H_1(n) + H_3(n) - H_4(n);
a(n+m) = a(n)*H_1(m) + H_1(n)*a(m) + H_4(n)*H_3(m) + H_3(n)*H_4(m), where H_1 = A038503, H_3 = A038505, H_4 = A000749.
For proofs, see Shevelev's link, Theorems 2, 3. (End)
a(n) = (1/4)*(2^((n+1)/2)*ChebyshevU(n-1, 1/sqrt(2)) + 2^n - [n=0]). - G. C. Greubel, Apr 20 2023
Comments