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A038503
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Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 0".
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26
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1, 1, 1, 1, 2, 6, 16, 36, 72, 136, 256, 496, 992, 2016, 4096, 8256, 16512, 32896, 65536, 130816, 261632, 523776, 1048576, 2098176, 4196352, 8390656, 16777216, 33550336, 67100672, 134209536, 268435456, 536887296, 1073774592, 2147516416, 4294967296, 8589869056, 17179738112
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OFFSET
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0,5
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COMMENTS
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Number of strings over Z_2 of length n with trace 0 and subtrace 0.
Same as number of strings over GF(2) of length n with trace 0 and subtrace 0.
M^n = [1,0,0,0] = [a(n), A000749(n), A038505(n), A038504(n)]; where M = the 4 X 4 matrix [1,1,0,0; 0,1,1,0; 0,0,1,1; 1,0,0,1]. Sum of the 4 terms = 2^n. Example: M^6 = [16, 20, 16, 12], sum of terms = 64 = 2^6. - Gary W. Adamson, Mar 13 2009
a(n) is the number of generalized compositions of n when there are i^2/2 - 5i/2 + 3 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010
{A038503, A038504, A038505, A000749} is the difference analog of the hyperbolic functions {h_1(x), h_2(x), h_3(x), h_4(x)} of order 4. For the definitions of {h_i(x)} and the difference analog {H_i (n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Aug 01 2017
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REFERENCES
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A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, 2nd ed., Problem 38, p. 70, gives an explicit formula for the sum.
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LINKS
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FORMULA
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G.f.: (1-x)^3/((1-x)^4-x^4);
a(n) = Sum_{k=0..floor(n/4)} binomial(n, 4k); a(n) = 2^(n-1) + 2^((n-2)/2)(cos(Pi*n/4) - sin(Pi*n/4)). (End)
Binomial transform of 1/(1-x^4). a(n) = 4a(n-1) - 6a(n-2) + 4a(n-3); a(n) = Sum_{k=0..n} binomial(n, k)(sin(Pi*(k+1)/2)/2 + (1+(-1)^k)/4); a(n) = Sum_{k=0..floor(n/4)} binomial(n, 4k). - Paul Barry, Jul 25 2004
a(n) = Sum_{k=0..n} binomial(n, 4(n-k)). - Paul Barry, Aug 30 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)(1+(-1)^k)/2. - Paul Barry, Nov 29 2004
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.
E.g.f.: exp(z)*(cosh(z) + cos(z))/2. - Peter Luschny, Jul 10 2012
For n >= 1, {H_i(n)} are linearly dependent sequences: a(n) = H_1(n) = H_2(n) - H_3(n) + H_4(n);
a(n+m) = a(n)*a(m) + H_4(n)*H_2(m) + H_3(n)*H_3(m) + H_2(n)*H_4(m), where H_2 = A038504, H_3 = A038505, H_4 = A000749.
For proofs, see Shevelev's link, Theorems 2, 3. (End)
a(n) = hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4], [1/4, 1/2, 3/4], 1). - Peter Luschny, Mar 18 2023
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EXAMPLE
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a(3;0,0)=1 since the one binary string of trace 0, subtrace 0 and length 3 is { 000 }.
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MAPLE
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A038503_list := proc(n) local i; series(exp(z)*(cosh(z)+cos(z))/2, z, n+2):
a := n -> hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4], [1/4, 1/2, 3/4], 1):
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MATHEMATICA
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nn = 18; a = Sum[x^(4 i)/(4 i)!, {i, 0, nn}]; b = Exp[x]; Range[0, nn]! CoefficientList[Series[a b, {x, 0, nn}], x] (* Geoffrey Critzer, Dec 27 2011 *)
Join[{1}, LinearRecurrence[{4, -6, 4}, {1, 1, 1}, 40]] (* Harvey P. Dale, Dec 02 2014 *)
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PROG
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(PARI) a(n) = sum(k=0, n\4, binomial(n, 4*k)); \\ Michel Marcus, Mar 13 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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STATUS
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approved
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