A222266 Irregular triangle which lists the bi-unitary divisors of n in row n.
1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 2, 3, 6, 1, 7, 1, 2, 4, 8, 1, 9, 1, 2, 5, 10, 1, 11, 1, 3, 4, 12, 1, 13, 1, 2, 7, 14, 1, 3, 5, 15, 1, 2, 8, 16, 1, 17, 1, 2, 9, 18, 1, 19, 1, 4, 5, 20, 1, 3, 7, 21, 1, 2, 11, 22, 1, 23, 1, 2, 3, 4, 6, 8, 12, 24, 1, 25, 1, 2, 13, 26, 1, 3, 9, 27, 1, 4, 7, 28, 1, 29, 1, 2, 3, 5, 6, 10, 15, 30, 1, 31, 1, 2, 4, 8, 16, 32, 1, 3, 11, 33, 1, 2, 17, 34, 1, 5, 7, 35
Offset: 1
Examples
The table starts 1; 1, 2; 1, 3; 1, 4; 1, 5; 1, 2, 3, 6; 1, 7; 1, 2, 4, 8; 1, 9; 1, 2, 5, 10; 1, 11; 1, 3, 4, 12; 1, 13; 1, 2, 7, 14; 1, 3, 5, 15; 1, 2, 8, 16; 1, 17;
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..13171 (rows 1 <= n <= 2000).
- D. Suryanarayana, The number of bi-unitary divisors of an integer, in: A. A. Gioia and D. L. Goldsmith (eds.), The Theory of Arithmetic Functions, Lecture Notes in Mathematics, Vol 251, Springer, Berlin, Heidelberg, 1972.
Programs
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Maple
# Return set of unitary divisors of n. A077610_row := proc(n) local u,d ; u := {} ; for d in numtheory[divisors](n) do if igcd(n/d,d) = 1 then u := u union {d} ; end if; end do: u ; end proc: # true if d is a bi-unitary divisor of n. isbiudiv := proc(n,d) if n mod d = 0 then A077610_row(d) intersect A077610_row(n/d) ; if % = {1} then true; else false; end if; else false; end if; end proc: # Return set of bi-unitary divisors of n biudivs := proc(n) local u,d ; u := {} ; for d in numtheory[divisors](n) do if isbiudiv(n,d) then u := u union {d} ; end if; end do: u ; end proc: for n from 1 to 35 do print(op(biudivs(n))) ; end do:
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Mathematica
f[n_] := Select[Divisors[n], Function[d, CoprimeQ[d, n/d]]]; Table[Function[d, Union@ Flatten@ Select[Transpose@ {d, n/d}, Last@ Intersection[f@ #1, f@ #2] == 1 & @@ # &]]@ Select[Divisors@ n, # <= Floor@ Sqrt@ n &], {n, 35}] (* Michael De Vlieger, May 07 2017 *)
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PARI
isbdiv(f, d) = {for (i=1, #f~, if(f[i, 2]%2 == 0 && valuation(d, f[i, 1]) == f[i, 2]/2, return(0))); 1;} row(n) = {my(d = divisors(n), f = factor(n), bdiv = []); for(i=1, #d, if(isbdiv(f, d[i]), bdiv = concat(bdiv, d[i]))); bdiv; } \\ Amiram Eldar, Mar 24 2023
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