A134660 Number of odd coefficients in (1 + x + x^2 + x^3)^n.
1, 4, 4, 4, 4, 16, 4, 8, 4, 16, 16, 4, 4, 16, 8, 16, 4, 16, 16, 16, 16, 64, 4, 8, 4, 16, 16, 8, 8, 32, 16, 32, 4, 16, 16, 16, 16, 64, 16, 32, 16, 64, 64, 4, 4, 16, 8, 16, 4, 16, 16, 16, 16, 64, 8, 16, 8, 32, 32, 16, 16, 64, 32, 64, 4, 16, 16, 16, 16, 64, 16, 32, 16, 64, 64, 16, 16, 64
Offset: 0
Keywords
Examples
From _Omar E. Pol_, Mar 01 2015: (Start) Written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins: 1; 4; 4,4; 4,16,4,8; 4,16,16,4,4,16,8,16; 4,16,16,16,16,64,4,8,4,16,16,8,8,32,16,32; 4,16,16,16,16,64,16,32,16,64,64,4,4,16,8,16,4,16,16,16,16,64,8,16,8,32,32,16,16,64,32,64; ... (End)
Links
- G. C. Greubel, Table of n, a(n) for n = 0..10000
- S. R. Finch, P. Sebah and Z.-Q. Bai, Odd Entries in Pascal's Trinomial Triangle, arXiv:0802.2654 [math.NT], 2008.
Programs
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Maple
seq(igcd(4^n,binomial(4*n,n)),n=0..77); # Peter Luschny, Nov 08 2011
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Mathematica
PolynomialMod[(1+x+x^2+x^3)^n, 2] /. x->1 A036555 = Total /@ IntegerDigits[3 Range[0, 100], 2]; Table[2^A036555[[n]], {n, 1, 20}] (* or *) Table[GCD[4^n, Binomial[4*n, n]], {n, 0, 50}] (* G. C. Greubel, Dec 31 2017 *)
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PARI
a(n) = {my(pol= Pol([1,1,1,1], xx)*Mod(1,2)); subst(lift(pol^n), xx, 1);} \\ Michel Marcus, Mar 01 2015
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PARI
a(n) = 2^hammingweight(3*n); \\ Joerg Arndt, Mar 10 2015
Formula
a(n) = 2^A036555(n).
a(n) = gcd(4^n, C(4*n, n)). - Peter Luschny, Nov 08 2011
Comments