

A299115


Irregular triangle read by rows: nth row gives binomial(n+d,d) mod n for prime divisors d of n > 1.


2



0, 2, 3, 2, 4, 0, 2, 5, 4, 6, 3, 2, 7, 11, 2, 8, 10, 6, 9, 9, 2, 10, 16, 2, 11, 10, 8, 18, 12, 14, 2, 13, 21, 6, 14, 16, 10, 15, 12, 2, 16, 26, 2, 2, 17, 12, 4, 18, 3, 8, 13, 19, 31, 2, 20, 22, 14, 4, 21, 39, 2, 22, 36, 0, 2, 23, 38, 16, 25, 24, 26, 2, 25, 41, 8, 26, 11, 18, 21, 27, 44, 2, 28, 46, 12, 28, 29, 23, 20, 4
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OFFSET

2,2


COMMENTS

Sondow (2017) proves that 1 is not a member. See A290040 and A290041 for binomial(n+d,d) == 1 (mod n) with d a composite divisor of n.
All integers > 1 are members, since if n is even, say n = 2*k, then binomial(n+2,2) = (k+1)*(2*k+1) == k+1 (mod n).


LINKS

Table of n, a(n) for n=2..91.
J. Sondow, Extending Babbage's (non)primality tests, in Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, 269277, CANT 2015 and 2016, New York, 2017; arXiv:1812.07650 [math.NT], 2018.


EXAMPLE

For n = 2 and d = 2, binomial(n+d,d) = binomial(4,2) = 6 == 0 (mod n), so 0 is the first member. The rows for n = 2, 3, 4, 5, 6, 7 are
0,
2,
3,
2,
4, 0,
2,


MATHEMATICA

Table[ Mod[ Binomial[m + d, d], m], {m, 2, 60}, {d, Select[ Divisors[m], PrimeQ]}] // Flatten


CROSSREFS

Cf. A290040, A290041.
Sequence in context: A326991 A131580 A087038 * A179590 A167504 A286013
Adjacent sequences: A299112 A299113 A299114 * A299116 A299117 A299118


KEYWORD

nonn,tabf


AUTHOR

Jonathan Sondow, Feb 23 2018


STATUS

approved



