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A062730
Rows of Pascal's triangle which contain 3 terms in arithmetic progression.
5
7, 12, 14, 19, 21, 23, 32, 34, 45, 47, 60, 62, 77, 79, 96, 98, 117, 119, 140, 142, 165, 167, 192, 194, 221, 223, 252, 254, 285, 287, 320, 322, 357, 359, 396, 398, 437, 439, 480, 482, 525, 527, 572, 574, 621, 623, 672, 674, 725, 727, 780, 782, 837, 839
OFFSET
1,1
COMMENTS
Except for n=19, all n < 1000 have the form k^2-2 or k^2-4. When n=k^2-2, the three terms in AP are consecutive binomial coefficients C(n,k(k-1)/2-2), C(n,k(k-1)/2-1), and C(n,k(k-1)/2). When n=k^2-4, the three terms in AP differ by two: C(n,k(k-1)/2-4), C(n,k(k-1)/2-2), and C(n,k(k-1)/2). When n=19, the three terms in AP are C(19,4), C(19,6), and C(19,7). [From T. D. Noe, Mar 23 2009]
LINKS
FORMULA
G.f.: (-5x^8+3x^7+7x^6-3x^5+5x^4-5x^3-12x^2+5x+7)/[(1-x)(1-x^2)^2] (conjectured). - Ralf Stephan, May 08 2004
a(n)=(n^2+8*n+8)/4 for n>4 and even; a(n)=(n^2+10*n+9)/4 for n>4 and odd (conjectured). - Colin Barker, Aug 29 2013
EXAMPLE
12 is in the list since the 12th row of Pascal's triangle starts 1 12 (66) 220 (495) 792 (924) and 66 495 924 are in arithmetic progression.
MATHEMATICA
kmax = 30; row[n_] := Table[Binomial[n, k], {k, 0, Floor[n/2]}]; Reap[Do[r = row[n]; If[ (r /. {___, a_, ___, b_, ___, c_, ___} /; b - a == c - b -> {}) == {}, Print[n]; Sow[n]], {n, Union[{19}, Range[2, kmax]^2 - 2, Range[2, kmax]^2 - 4]}]][[2, 1]] (* Jean-François Alcover, Jul 11 2012, after T. D. Noe *)
PROG
(Haskell)
-- import Data.List (intersect)
a062730 n = a062730_list !! (n-1)
a062730_list = filter f $ [3..] where
f x = not $ all null $ zipWith
(\us (v:vs) -> map (v -) us `intersect` map (subtract v) vs)
(tail $ init $ inits bns) (tail $ init $ tails bns)
where bns = a034868_row x
-- Reinhard Zumkeller, Jun 10 2013
CROSSREFS
Sequence in context: A353444 A153245 A317670 * A287562 A341092 A351474
KEYWORD
nice,nonn
AUTHOR
Erich Friedman, Jul 11 2001
EXTENSIONS
More terms from Naohiro Nomoto, Oct 01 2001
Offset corrected by Reinhard Zumkeller, Jun 10 2013
STATUS
approved