

A249722


Numbers n such that there is a multiple of 4 on row n of Pascal's triangle with property that all multiples of 9 on the same row (if they exist) are larger than it.


5



4, 6, 8, 12, 14, 16, 17, 20, 22, 24, 25, 26, 28, 30, 32, 33, 34, 35, 38, 40, 41, 42, 44, 48, 49, 50, 51, 52, 53, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 74, 76, 77, 78, 80, 84, 86, 88, 89, 92, 94, 96, 97, 98, 100, 101, 102, 104, 105, 106, 107, 112, 113, 114, 115, 116, 120, 121, 122, 124, 125
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OFFSET

1,1


COMMENTS

All n such that on row n of A034931 (Pascal's triangle reduced modulo 4) there is at least one zero and the distance from the edge to the nearest zero is shorter than the distance from the edge to the nearest zero on row n of A095143 (Pascal's triangle reduced modulo 9), the latter distance taken to be infinite if there are no zeros on that row in the latter triangle.


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..10000


EXAMPLE

Row 4 of Pascal's triangle (A007318) is {1,4,6,4,1}. The least multiple of 4 occurs as C(4,1) = 4, and there are no multiples of 9 present, thus 4 is included among the terms.
Row 12 of Pascal's triangle is {1,12,66,220,495,792,924,792,495,220,66,12,1}. The least multiple of 4 occurs as C(12,1) = 12, which is less than the least multiple of 9 present at C(12,4) = 495 = 9*55, thus 12 is included among the terms.


PROG

(PARI)
A249722list(upto_n) = { my(i=0, n=0); while(i<upto_n, for(k=0, n\2, if(!(binomial(n, k)%9), break, if(!(binomial(n, k)%4), i++; write("b249722.txt", i, " ", n); break))); n++); }


CROSSREFS

A subsequence of A249724.
Natural numbers (A000027) is a disjoint union of the sequences A048278, A249722, A249723 and A249726.
Cf. A007318, A034931, A095143, A048645, A051382.
Sequence in context: A110606 A241124 A117247 * A047407 A090697 A225508
Adjacent sequences: A249719 A249720 A249721 * A249723 A249724 A249725


KEYWORD

nonn


AUTHOR

Antti Karttunen, Nov 04 2014


STATUS

approved



