A182069 - OEIS

%I #27 Feb 10 2025 20:44:29

%S 1,5,13,29,142,377,757,3785,9841,19685,98422,255905,570862,2795180,

%T 7421206,14901545,74505454,193720085,387440173,1937200865,5036722249,

%U 11235765017,55016504566,146064945221,293292210961,1466461054805,3812798742493,7625597484989

%N Triangle of numbers 2^i*C(n,i) mod 3 converted to decimal.

%C Or, set x=3 in polynomial corresponding to A255285(n). - _N. J. A. Sloane_, Feb 21 2015

%C For k>=1, consider triangle for numbers k^i*C(n,i) with row sums (k+1)^n.

%C If one considers its entries modulo k+1, then we obtain Pascal triangle with the alternating signs within every row:

%C n/k.|..0.....1.....2.....3.....4.....5.....6.....7

%C ==================================================

%C .0..|..1

%C .1..|..1.....-1

%C .2..|..1.....-2.....1

%C .3..|..1.....-3.....3.....-1

%C .4..|..1.....-4.....6.....-4.....1

%C .5..|..1.....-5....10....-10.....5....-1

%C .6..|

%C This is a common basis triangle for our generalizations. Consider the least positive residues of its entries modulo k+1 and read the rows as numbers in base k+1, converting to decimal. In case k=1 we obtain A001317. This sequence corresponds to k=2.

%H G. C. Greubel, <a href="/A182069/b182069.txt">Table of n, a(n) for n = 0..1000</a>

%F For n>=0, a(3*n+2) = 13*a(3*n).

%e Consider the third row of the alternating Pascal triangle in comment: {1,-3,3,-1}={1,0,0,2} mod 3. We have (1002)_3=27+2=29. Thus a(3)=29.

%t Table[FromDigits[Table[Mod[2^i Binomial[n, i], 3], {i, 0, n}], 3], {n, 0, 30}] (* _T. D. Noe_, Apr 10 2012 *)

%Y Cf. A001317, A255285.

%K nonn,base

%O 0,2

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Apr 10 2012