A027746 Irregular triangle in which first row is 1, n-th row (n>1) gives prime factors of n with repetition.
1, 2, 3, 2, 2, 5, 2, 3, 7, 2, 2, 2, 3, 3, 2, 5, 11, 2, 2, 3, 13, 2, 7, 3, 5, 2, 2, 2, 2, 17, 2, 3, 3, 19, 2, 2, 5, 3, 7, 2, 11, 23, 2, 2, 2, 3, 5, 5, 2, 13, 3, 3, 3, 2, 2, 7, 29, 2, 3, 5, 31, 2, 2, 2, 2, 2, 3, 11, 2, 17, 5, 7, 2, 2, 3, 3, 37, 2, 19, 3, 13, 2, 2, 2, 5, 41, 2, 3, 7, 43, 2, 2, 11, 3, 3, 5
Offset: 1
Examples
Triangle begins 1; 2; 3; 2, 2; 5; 2, 3; 7; 2, 2, 2; 3, 3; 2, 5; 11; 2, 2, 3; ...
Links
- N. J. A. Sloane, First 2048 rows of triangle, flattened
- S. von Worley (?), Animated Factorization Diagrams, Oct. 2012.
- Brent Yorgey, Factorization diagrams, The Math Less Traveled, Oct 05 2012.
Crossrefs
Programs
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Haskell
import Data.List (unfoldr) a027746 n k = a027746_tabl !! (n-1) !! (k-1) a027746_tabl = map a027746_row [1..] a027746_row 1 = [1] a027746_row n = unfoldr fact n where fact 1 = Nothing fact x = Just (p, x `div` p) where p = a020639 x -- Reinhard Zumkeller, Aug 27 2011
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Maple
P:=proc(n) local FM: FM:=ifactors(n)[2]: seq(seq(FM[j][1],k=1..FM[j][2]),j=1..nops(FM)) end: 1; for n from 2 to 45 do P(n) od; # yields sequence in triangular form; Emeric Deutsch, Feb 13 2005
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Mathematica
row[n_] := Flatten[ Table[#[[1]], {#[[2]]}] & /@ FactorInteger[n]]; Flatten[ Table[ row[n], {n, 1, 45}]] (* Jean-François Alcover, Dec 01 2011 *)
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PARI
A027746_row(n,o=[1])=if(n>1,concat(apply(t->vector(t[2],i,t[1]), Vec(factor(n)~))),o) \\ Use %(n,[]) if you want the more natural [] for the first row. - M. F. Hasler, Jul 29 2015
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Python
def factors(n: int) -> list[int]: p = n L:list[int] = [] for f in range(2, p + 1): if f * f > p: break while True: q, r = divmod(p, f) if r != 0: break L.append(f) p = q if p == 1: return L L.append(p) return L # Peter Luschny, Jul 18 2024
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Sage
v=[1] for k in [2..45]: v.extend(p for (p, m) in factor(k) for _ in range(m)) print(v) # Giuseppe Coppoletta, Dec 29 2017
Formula
Extensions
More terms from James Sellers
Comments