A182069 Triangle of numbers 2^i*C(n,i) mod 3 converted to decimal.

1, 5, 13, 29, 142, 377, 757, 3785, 9841, 19685, 98422, 255905, 570862, 2795180, 7421206, 14901545, 74505454, 193720085, 387440173, 1937200865, 5036722249, 11235765017, 55016504566, 146064945221, 293292210961, 1466461054805, 3812798742493, 7625597484989

Offset

0,2

Comments

Or, set x=3 in polynomial corresponding to A255285(n). - N. J. A. Sloane, Feb 21 2015

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

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

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

==================================================

.0..|..1

.1..|..1.....-1

.2..|..1.....-2.....1

.3..|..1.....-3.....3.....-1

.4..|..1.....-4.....6.....-4.....1

.5..|..1.....-5....10....-10.....5....-1

.6..|

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.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Formula

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

Example

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.

Mathematica

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

Crossrefs

Cf. A001317, A255285.

Sequence in context: A147288 A146420 A147492 * A085555 A224888 A256314

Adjacent sequences: A182066 A182067 A182068 * A182070 A182071 A182072

Keyword

nonn,base

Author

Vladimir Shevelev and Peter J. C. Moses, Apr 10 2012

Status

approved