A219840 Irregular triangle R(n,k) = n mod A000040(k), for 1 <= k <= i, where i is the least such that n < A002110(i).

1, 0, 2, 1, 0, 0, 1, 1, 2, 0, 0, 1, 1, 1, 2, 0, 2, 3, 1, 0, 4, 0, 1, 0, 1, 2, 1, 0, 0, 2, 1, 1, 3, 0, 2, 4, 1, 0, 0, 0, 1, 1, 1, 2, 2, 0, 0, 3, 1, 1, 4, 0, 2, 0, 1, 0, 1, 0, 1, 2, 1, 2, 3

Offset: 1

Author

Jason Kimberley, Nov 29 2012

Keywords

• nonn
• easy
• tabf

Comments

By the Chinese Remainder Theorem, x = n is the unique, for 0 <= m < A002110(i), solution to the set of congruences x = R(n,k) (mod A000040(k)), for 1 <= k <= i.

Examples

			1:  1;
2:  0, 2;
3:  1, 0;
4:  0, 1;
5:  1, 2;
6:  0, 0, 1;
  ...
29: 1, 2, 4;
30: 0, 0, 0, 2;
  ...
209:1, 2, 4, 6;
210:0, 0, 0, 0, 1;
		

Links

• Jason Kimberley, Rows n = 1..2310 of irregular triangle, flattened

Crossrefs

The n-th row of this sequence is the length i prefix of the n-th row of A147693.

Programs

Magma

A002110 := func;
A219840_block := funcA002110(i-1)..A002110(i)-1]]>;
[A219840_block(i):i in[1..4]];