A290040 Numbers m > 0 that have a divisor d > 1 with binomial(m+d, d) == 1 mod m.

260, 1056, 1060, 3460, 3905, 4428, 5000, 5060, 5512, 5860, 6372, 6596, 7460, 8200, 8908, 9612, 9860, 10660, 11556, 12260, 12625, 13060, 14600, 14660, 14744, 14796, 15460, 16260, 17060, 17800, 17860, 18425, 18496, 18660, 19396, 20260, 21717, 21860, 22168, 22248, 22660, 24260, 24616, 25164, 26660, 27108, 27400, 27460, 28872, 29060, 29128, 29860

Offset

1,1

Comments

Sondow (2017) shows that d must be composite. Can d be a prime power?

The corresponding sequence of smallest such d is A290041.

The first term a(n) for which more than one d exists is a(165) = 101000, where d = 20 or d = 100.

d cannot be a prime power p^r | m if p^(r+1) does not divide m. Can d = 6? - Jonathan Sondow, Dec 26 2017

Links

Giovanni Resta, Table of n, a(n) for n = 1..654 (terms < 400000)

C. Babbage, Demonstration of a theorem relating to prime numbers, Edinburgh Philosophical Journal, 1 (1819), 46-49.

J. Sondow, Extending Babbage's (non-)primality tests, in Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, 269-277, CANT 2015 and 2016, New York, 2017; arXiv:1812.07650 [math.NT], 2018.

J. Sondow, Problem 12030, Amer. Math. Monthly, 125 (2018), 276.

Formula

binomial(a(n) + A290041(n), A290041(n)) == 1 mod a(n).

Example

The first case is binomial(260+10,10) = 479322759878148681 == 1 mod 260, so a(1) = 260.

Mathematica

Select[ Range[ 30000], (d = Divisors[#]; Product[ Mod[ Binomial[# + d[[k]], d[[k]]], #] - 1, {k, 2, Length[d]}] == 0) &]

Prog

(PARI) is(n) = my(d=divisors(n)); for(k=2, #d, if(Mod(binomial(n+d[k], d[k]), n)==1, return(1))); 0 \\ Felix Fröhlich, Dec 26 2017

Crossrefs

Cf. A290041, A299115.

Sequence in context: A237631 A202585 A037219 * A061224 A108109 A235905

Adjacent sequences: A290037 A290038 A290039 * A290041 A290042 A290043

Keyword

nonn

Author

Jonathan Sondow, Jul 23 2017

Status

approved