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A226871 Triangle read by rows: row n gives the first q divisors d(1), d(2), ..., d(q) of A225110(n) such that Sum_{i = 1..q} 1/d(i) is an integer. 1
1, 1, 2, 3, 6, 1, 2, 3, 6, 1, 2, 4, 7, 14, 28, 1, 2, 3, 6, 1, 2, 3, 6, 1, 2, 3, 6, 1, 2, 3, 6, 1, 2, 3, 6, 1, 2, 3, 6, 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120, 1, 2, 3, 6, 1, 2, 3, 6, 1, 2, 3, 6, 1, 2, 3, 6, 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

Rows 2, 3, 5, 6, 7, ... with the divisors {1, 2, 3, 6} are identical;

rows 4, 18, 62, 67, ... with the divisors {1, 2, 4, 7, 14, 28} are identical;

...

The primitive rows are rows 1, 2, 4, 11, 16, 39, 52, 145, ... corresponding to n = 1, 6, 28, 120, 180, 496, 672, 1890, ... (see A226853).

The irregular triangle of divisors is:

[1]

[1, 2, 3, 6]

[1, 2, 3, 6]

[1, 2, 4, 7, 14, 28]

[1, 2, 3, 6]

...

LINKS

Michel Lagneau, Rows n = 1..2223 of irregular triangle, flattened

EXAMPLE

Row 3 = [1, 2, 3, 6] consists of the first 4 divisors of A225110(3) = 18; 1 + 1/2 + 1/3 + 1/6 = 2 is an integer.

MAPLE

with(numtheory): print({1}):for n from 1 to 5000 do:x:=divisors(n):n1:=nops(x):s:=0:ii:=0:for q from 1 to n1 while(ii=0) do:s:=s+1/x[q]:if s=floor(s) and q>1 then ii:=1: print({seq(x[i], i=1..q)}) else fi:od:od:

CROSSREFS

Cf. A225110, A226853.

Sequence in context: A241293 A107409 A268603 * A178483 A133031 A275732

Adjacent sequences:  A226868 A226869 A226870 * A226872 A226873 A226874

KEYWORD

nonn,tabf

AUTHOR

Michel Lagneau, Jun 20 2013

STATUS

approved

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Last modified February 28 08:28 EST 2020. Contains 332323 sequences. (Running on oeis4.)