A038503 Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 0".
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
Offset: 0
Examples
a(3;0,0)=1 since the one binary string of trace 0, subtrace 0 and length 3 is { 000 }.
References
- 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.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..3000
- Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5, pp. 1-24.
- John B. Dobson, A matrix variation on Ramus's identity for lacunary sums of binomial coefficients, arXiv preprint arXiv:1610.09361 [math.NT], 2016.
- Frank Ruskey, Strings over Z_2 with given trace and subtrace
- Frank 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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Maple
A038503_list := proc(n) local i; series(exp(z)*(cosh(z)+cos(z))/2,z,n+2): seq(i!*coeff(%,z,i),i=0..n) end: A038503_list(32); # Peter Luschny, Jul 10 2012 a := n -> hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4], [1/4, 1/2, 3/4], 1): seq(simplify(a(n)), n = 0..36); # Peter Luschny, Mar 18 2023
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Mathematica
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 *) -
PARI
a(n) = sum(k=0, n\4, binomial(n, 4*k)); \\ Michel Marcus, Mar 13 2019
Formula
From Paul Barry, Mar 18 2004: (Start)
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
From Vladimir Shevelev, Aug 01 2017: (Start)
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
Comments