

A290041


Set b(n) = A290040(n). Then a(n) is the least divisor d > 1 of b(n) with binomial(b(n)+d, d) == 1 mod b(n).


2



10, 264, 10, 10, 55, 18, 20, 10, 52, 10, 18, 34, 10, 20, 34, 18, 10, 10, 18, 10, 2525, 10, 20, 10, 38, 18, 10, 10, 10, 20, 10, 55, 272, 10, 26, 10, 57, 10, 68, 18, 10, 10, 68, 18, 10, 18, 20, 10, 36, 10, 22, 10, 18, 33, 10, 38, 10, 18, 10, 10, 20, 10, 18, 10, 33, 10, 10, 10, 10, 18, 10, 34, 10, 50, 10, 20, 10, 33, 18, 10, 10
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OFFSET

1,1


COMMENTS

Sondow (2017) shows that if d is any divisor of m > 0 with d > 1 and binomial(m+d, d) == 1 mod m, then d is composite. Can d be a prime power?
The first term of A290040 for which more than one such d exists is A290040(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



FORMULA



EXAMPLE

b(1) = 260 and binomial(260+10,10) = 479322759878148681 == 1 mod 260 gives the least d = 10, so a(1) = 10.


MATHEMATICA

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


PROG

(PARI) terms(n) = my(i=0); for(k=1, oo, my(d=divisors(k)); for(x=2, #d, if(Mod(binomial(k+d[x], d[x]), k)==1, print1(d[x], ", "))); if(i==n, break))
/* Print initial 40 terms as follows */


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



