A222310 Array read by antidiagonals: first row is 1, 2, 3, 4, ...; for subsequent rows, write i*j/gcd(i,j)^2 under ...i.j... in previous row.

1, 2, 2, 3, 6, 3, 6, 2, 12, 4, 5, 30, 15, 20, 5, 15, 3, 10, 6, 30, 6, 105, 7, 21, 210, 35, 42, 7, 70, 6, 42, 2, 420, 12, 56, 8, 1, 70, 105, 10, 5, 84, 63, 72, 9, 5, 5, 14, 30, 3, 15, 1260, 20, 90, 10, 33, 165, 33, 462, 385, 1155, 77, 1980, 99, 110, 11, 55, 15, 11, 3, 154, 10, 462, 6, 330, 30, 132, 12, 65, 143, 2145, 195, 65, 10010, 1001, 78

Offset: 1

Author

N. J. A. Sloane, Feb 16 2013

Keywords

• nonn
• tabl

Examples

			Array begins:
1...2...3.....4......5......6.....7.....8.....9.....10
..2...6....12....20.....30....42.....56...72.....90
....3...2....15......6.....35....12....63....20
......6....30....10....210...420.....84..1260
........5.....3.....21......2.....5....15
...........15.....7.....42....10......3
.............105.....6.....105...30
........
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..8001
• Cristian Cobeli, Mihai Prunescu, Alexandru Zaharescu, A growth model based on the arithmetic Z-game, arXiv:1511.04315 [math.NT], 2015.
• C. Cobeli and A. Zaharescu, Promenade around Pascal Triangle-Number Motives, Bull. Math. Soc. Sci. Math. Roumanie, Tome 56(104) No. 1, 2013, 73-98.

Crossrefs

Cf. A036262. Leading diagonal is A222311 (cf. A222313).

Similar array with primes in the starting row is A255483.

Programs

Maple

# To get first M rows of the array (s0 is A222311):
g:=(i,j)->i*j/gcd(i,j)^2;
M:=50;
s0:=[1]:
s1:=[seq(n,n=1..M)]:
for i1 from 1 to M-1 do
lprint(s1);
s2:=[seq(g(s1[i],s1[i+1]),i=1..nops(s1)-1)];
s0:=[op(s0),s2[1]];
s1:=[seq(s2[i],i=1..nops(s2))];
od:
# To produce A222310 (i.e., to read the array by antidiagonals):
g:=(i,j)->i*j/gcd(i,j)^2;
M:=15;
b1:=Array(1..M);
s0:=[1]:
s1:=[seq(n,n=1..M)]:
b1[1]:=s1;
for i1 from 1 to M-1 do
#lprint(s1);
s2:=[seq(g(s1[i],s1[i+1]),i=1..nops(s1)-1)];
b1[i1+1]:=s2;
s0:=[op(s0),s2[1]];
s1:=[seq(s2[i],i=1..nops(s2))];
od:
#[seq(s0[i],i=1..nops(s0))]; (that gives A222311)
lis:=[]:
for i from 1 to M do for j from 1 to i do
lis:=[op(lis),b1[i-j+1][j]];
od: od:
[seq(lis[k],k=1..nops(lis))];

Mathematica

a = r = {1}; Do[a = Join[a, Reverse[r = FoldList[#1*#2/GCD[#1, #2]^2 &, n, r]]], {n, 2, 13}]; a (* Ivan Neretin, May 14 2015 *)