A027746 - OEIS

A027746

Irregular triangle in which first row is 1, n-th row (n>1) gives prime factors of n with repetition.

282

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

n-th row has length A001222(n) (n>1).

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.

FORMULA

Product_{k=1..A001222(n)} T(n,k) = n.

From Reinhard Zumkeller, Aug 27 2011: (Start)

A001414(n) = Sum_{k=1..A001222(n)} T(n,k), n>1;

A006530(n) = T(n,A001222(n)) = Max_{k=1..A001222(n)} T(n,k);

A020639(n) = T(n,1) = Min_{k=1..A001222(n)} T(n,k). (End)

EXAMPLE

Triangle begins

2, 2;

2, 3;

2, 2, 2;

3, 3;

2, 5;

11;

2, 2, 3;

...

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

MATHEMATICA

row[n_] := Flatten[ Table[#[[1]], {#[[2]]}] & /@ FactorInteger[n]]; Flatten[ Table[ row[n], {n, 1, 45}]] (* Jean-François Alcover, Dec 01 2011 *)

PROG

(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

(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

(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

(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