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A182069 Triangle of numbers 2^i*C(n,i) mod 3 converted to decimal. 4
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified July 29 11:07 EDT 2021. Contains 346344 sequences. (Running on oeis4.)