A219840 - OEIS

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

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

OFFSET

1,3

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.

LINKS

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

EXAMPLE

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;

PROG

(Magma) A002110 := func<n|&*[Integers()|NthPrime(j):j in[1..n]]>;

A219840_block := func<i|[[n mod NthPrime(k):k in[1..i]]:n in[A002110(i-1)..A002110(i)-1]]>;

[A219840_block(i):i in[1..4]];

CROSSREFS

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

Sequence in context: A254605 A335664 A269518 * A343220 A264893 A340653

Adjacent sequences: A219837 A219838 A219839 * A219841 A219842 A219843

KEYWORD

nonn,easy,tabf

AUTHOR

Jason Kimberley, Nov 29 2012

STATUS

approved