A037453 -id:A037453 - cp's OEIS frontend

A030979 Numbers k such that binomial(2k,k) is not divisible by 3, 5 or 7.

0, 1, 10, 756, 757, 3160, 3186, 3187, 3250, 7560, 7561, 7651, 20007, 59548377, 59548401, 45773612811, 45775397187, 237617431723407, 24991943420078301, 24991943420078302, 24991943420078307, 24991943715007536, 24991943715007537

Offset: 1

Shawn Godin (sgodin(AT)onlink.net)

nonn

By Lucas's theorem, binomial(2k,k) is not divisible by a prime p iff all base-p digits of k are smaller than p/2.
Ronald L. Graham offered $1000 to the first person who could settle the question of whether this sequence is finite or infinite. He remarked that heuristic arguments show that it should be infinite, but finite if it is required that binomial(2k,k) is prime to 3, 5, 7 and 11, with k = 3160 probably the last k which has this property.
The Erdős et al. paper shows that for any two odd primes p and q there are an infinite number of k for which gcd(p*q,binomial(2k,k))=1; i.e., p and q do not divide binomial(2k,k). The paper does not deal with the case of three primes. - T. D. Noe, Apr 18 2007
Pomerance gives a heuristic suggesting that there are on the order of x^0.02595... terms up to x. - Charles R Greathouse IV, Oct 09 2015

R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B33.

Christopher E. Thompson, Table of n, a(n) for n = 1..1374 (complete up to 10^70, extends first 62 terms computed by Max Alekseyev).
Christian Ballot, Divisibility of the middle Lucasnomial coefficient, Fib. Q., 55 (2017), 297-308.
Ernie Croot, Hamed Mousavi and Maxie Schmidt, On a conjecture of Graham on the p-divisibility of central binomial coefficients, arXiv:2201.11274 [math.NT], 2022.
P. Erdős, R. L. Graham, I. Z. Russa and E. G. Straus, On the prime factors of C(2n,n), Math. Comp. 29 (1975), 83-92, doi:10.2307/2005464.
Gennady Eremin, Factoring Middle Binomial Coefficients, arXiv:2003.01494 [math.CO], 2020.
R. D. Mauldin and S. M. Ulam, Mathematical problems and games, Adv. Appl. Math. 8 (3) (1987) 281-344.
C. Pomerance, Divisors of the middle binomial coefficient, Amer. Math. Monthly, 112 (2015), 636-644.
Wikipedia, Lucas' theorem
Han Yu, Fractal projections with an application in number theory, arXiv:2004.05924 [math.NT], 2020.

Cf. A129488, A129489, A129508, A151750.

lim=10000; Intersection[Table[FromDigits[IntegerDigits[k,2],3], {k,0,lim}], Table[FromDigits[IntegerDigits[k,3],5], {k,0,lim}], Table[FromDigits[IntegerDigits[k,4],7], {k,0,lim}]] (* T. D. Noe, Apr 18 2007 *)

fval(n,p)=my(s);while(n\=p,s+=n);s
is(n)=fval(2*n,3)==2*fval(n,3) && fval(2*n,5)==2*fval(n,5) && fval(2*n,7)==2*fval(n,7) \\ Charles R Greathouse IV, Oct 09 2015

Intersection of A005836, A037453 and A037461. - T. D. Noe, Apr 18 2007

More terms from Naohiro Nomoto, May 06 2002
Additional comments from R. L. Graham, Apr 25 2007
Additional comments and terms up 3^41 in b-file from Max Alekseyev, Nov 23 2008
Additional terms up to 10^70 in b-file from Christopher E. Thompson, Nov 06 2015

A129508 Numbers k such that 3 and 5 do not divide binomial(2*k, k).

Original entry on oeis.org

0, 1, 10, 12, 27, 30, 31, 36, 37, 252, 255, 256, 280, 282, 756, 757, 760, 810, 811, 3160, 3162, 3186, 3187, 3250, 3252, 3276, 3277, 3280, 6561, 6562, 6885, 6886, 6912, 6925, 7536, 7537, 7560, 7561, 7626, 7627, 7650, 7651, 19686, 19687, 20007, 20010, 20011

Offset: 1

T. D. Noe, Apr 18 2007

nonn

The Erdos paper proves that for any two odd primes p and q, there are an infinite number of k for which gcd(p*q, binomial(2*k, k)) = 1; i.e., p and q do not divide binomial(2*k, k).

T. D. Noe, Table of n, a(n) for n = 1..1000
P. Erdos, R. L. Graham, I. Z. Russa, and E. G. Straus, On the prime factors of C(2n,n), Math. Comp. 29 (1975), 83-92.

Cf. A030979 (k such that 3, 5 and 7 do not divide binomial(2*k, k)).

lim=10000; Intersection[Table[FromDigits[IntegerDigits[k,2],3], {k,0,lim}], Table[FromDigits[IntegerDigits[k,3],5], {k,0,lim}]]

valp(n, p)=my(s); while(n\=p, s+=n); s
is(n)=valp(2*n, 3)==2*valp(n, 3) && valp(2*n, 5)==2*valp(n, 5) \\ Charles R Greathouse IV, Feb 03 2016

Intersection of A005836 and A037453.

A050607 Numbers k such that base 5 expansion matches (0|1|2)((0|1)(3|4))?(0|1|2).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 16, 17, 20, 21, 22, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 40, 41, 42, 45, 46, 47, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 75, 76, 77, 80, 81, 82, 85, 86, 87, 100, 101, 102, 105, 106, 107, 110, 111, 112, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 140, 141, 142, 145

Offset: 1

Antti Karttunen, Oct 24 1999

25 does not divide C(2s-1,s) = A001700(s) (nor C(2s,s) = A000984(s), central column of Pascal's triangle) if and only if s is one of the terms in this sequence.

Index entries for 5-automatic sequences.

Cf. A037453, A046097, A050608, A051382.

sub conv_x_base_n { my($x,$b) = @_; my ($r,$z) = (0,'');
do { $r = $x % $b; $x = ($x - $r)/$b; $z = "$r" . $z; } while(0 != $x);
return($z); }
for($i=0; $i <= 201; $i++) { if(("0" . conv_x_base_n($i,5)) =~ /^(0|1|2)*((0|1)(3|4))?(0|1|2)*$/) { print $i, ","; } }

a(1)=0 inserted by Georg Fischer, Jun 26 2021

A268082 Numbers n such that gcd(binomial(2*n-1,n), n) is equal to 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, 29, 31, 32, 37, 39, 41, 43, 47, 49, 53, 55, 59, 61, 64, 67, 71, 73, 79, 81, 83, 89, 93, 97, 101, 103, 107, 109, 111, 113, 119, 121, 125, 127, 128, 131, 137, 139, 149, 151, 155, 157, 161, 163, 167, 169, 173, 179

Offset: 1

Michel Marcus, Jan 26 2016

nonn

Or numbers n such that A088218(n) is coprime to n.
The power of primes (A000961) are terms of this sequence.
From Robert Israel, Jan 26 2016: (Start)
By Lucas's theorem, these are the n such that for every prime p dividing n, no base-p digit of n is greater than the corresponding base-p digit of 2n-1. Equivalently (Kummer's theorem), there are no carries in base-p addition of n and n-1. Thus if p is odd, each base-p digit of n is less than p/2.
The only even terms are powers of 2.
All terms divisible by 3 are in A005836, and all terms divisible by 5 are in A037453. (End)
A082916 (after 0) lists the odd terms of this sequence. - Bruno Berselli, Jan 26 2015

			For n=3, binomial(2*n-1, n) = binomial(5, 3) = 10 and 10 is coprime to 3, so 3 is in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000
Victor J.W. Guo and Jiang Zeng, Factors of binomial sums from the Catalan triangle, Journal of Number Theory 130 (2010) 172-186.
Wikipedia, Kummer's theorem
Wikipedia, Lucas's theorem

Cf. A000961, A005836, A037453, A082916 , A088218, A268083.

[n: n in [1..200] | Gcd(Binomial(2*n-1,n), n) eq 1]; // Vincenzo Librandi, Jan 26 2016

filter:= proc(n) local F,p;
if n::even then evalb(n = 2^padic:-ordp(n,2))
else
   F:= numtheory:-factorset(n);
   for p in F do
     if max(convert(n,base,p)) > p/2 then return false fi;
   od;
   true
fi
end proc:
select(filter, [$1..1000]); # Robert Israel, Jan 26 2016

Select[Range@ 180, GCD[Binomial[2 # - 1, #], #] == 1 &] (* Michael De Vlieger, Jan 26 2016 *)

isok(n) = gcd(binomial(2*n-1,n), n) == 1;

lista(nn) = for(n=1, nn, if(gcd(binomial(2*n-1, n), n) == 1, print1(n, ", "))); \\ Altug Alkan, Jan 26 2016

A073094 Final digit of C(2k,k) when not equal to zero.

Original entry on oeis.org

2, 6, 2, 4, 2, 6, 2, 6, 2, 4, 2, 4, 8, 4, 2, 4, 2, 6, 2, 6, 2, 4, 2, 6, 2, 6, 2, 4, 2, 4, 8, 4, 2, 4, 2, 4, 8, 4, 8, 6, 8, 4, 8, 4, 2, 4, 2, 4, 8, 4, 2, 4, 2, 6, 2, 6, 2, 4, 2, 6, 2, 6, 2, 4, 2, 4, 8, 4, 2, 4, 2, 6, 2, 6, 2, 4, 2, 6, 2, 6, 2, 4, 2, 4, 8, 4, 2, 4, 2, 4, 8, 4, 8, 6, 8, 4, 8, 4, 2, 4, 2, 4, 8, 4, 2

Offset: 1

Benoit Cloitre, Aug 18 2002

Sequence contains 2,4,6,8, only.

Robert Israel, Table of n, a(n) for n = 1..10000

f:= proc(n) local L,r;
  L:= convert(n,base,3);
  r:= 2 &^ numboccur(1,L) mod 5;
  r + 5*(r mod 2)
end proc:
map(f, [$1..200]); # Robert Israel, Jun 11 2018

Mod[#,10]&/@Table[Binomial[2n,n],{n,800}]/.(0->Nothing) (* Harvey P. Dale, Apr 14 2018 *)

a(n) = C(2*A037453(n), A037453) reduced modulo 10.

A218394 Numbers such that sum(i<=n) binomial(n,i)binomial(2n-2*i, n-i) is not divisible by 5.

Original entry on oeis.org

1, 5, 7, 11, 25, 27, 31, 35, 37, 51, 55, 57, 61, 125, 127, 131, 135, 137, 151, 155, 157, 161, 175, 177, 181, 185, 187, 251, 255, 257, 261, 275, 277, 281, 285, 287, 301, 305, 307, 311, 625, 627, 631, 635, 637, 651, 655, 657, 661, 675, 677, 681, 685, 687, 751

Offset: 1

Michel Marcus, Oct 28 2012

nonn

a(n) = A037453(2*n-1) (proved by Schur, see link).

W. Shur, The last digit of C(2*n,n) and Sigma C(n,i)*C(2*n-2*i,n-i), The Electronic Journal of Combinatorics, #R16, Volume 4, Issue 2 (1997).

Cf. A037453.

lista(nb) = {for (n=1, nb, if (sum(i=1,n, binomial(n, i)*binomial(2*n-2*i,n-i)) % 5 != 0, print1(n, ", ")););}

a(n) = {2*n-1+2*sum(i=1,n, 5^(i-1)*floor((2*n-1)/3^i))}

from gmpy2 import digits
def A218394(n): return int(digits((n<<1)-1,3),5) # Chai Wah Wu, Aug 10 2025

a(n) = 2*n - 1 + 2*sum{i=1,n} 5^(i-1)*floor((2*n-1)/3^i).

A261691 Change of base from fractional base 3/2 to base 3.

Original entry on oeis.org

0, 1, 2, 6, 7, 8, 21, 22, 23, 63, 64, 65, 69, 70, 71, 192, 193, 194, 207, 208, 209, 213, 214, 215, 579, 580, 581, 621, 622, 623, 627, 628, 629, 642, 643, 644, 1737, 1738, 1739, 1743, 1744, 1745, 1866, 1867, 1868, 1881, 1882, 1883, 1887, 1888, 1889, 1929, 1930

Offset: 0

Tom Edgar, Aug 28 2015

To obtain a(n), we interpret A024629(n) as a base 3 representation (instead of base 3/2). More precisely, if A024629(n) = A007089(m), then a(n) = m.

The digits used in fractional base 3/2 are 0, 1, and 2, which are the same as the digits used in base 3.

			The base 3/2 representation of 7 is (2,1,1); i.e., 7 = 2*(3/2)^2 + 1*(3/2) + 1. Since 2*(3^2) + 1*3 + 1*1 = 22, we have a(7) = 22.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Index entries for sequences related to fractional bases.

Cf. A024629, A007089.
Cf. A005836, A023717, A000695, A037453, A037454, A037455, A037456, A037314.
Cf. A003462, A087088.

a[n_] := a[n] = If[n == 0, 0, 3 * a[2 * Floor[n/3]] + Mod[n, 3]]; Array[a, 100, 0] (* Amiram Eldar, Aug 04 2025 *)

a(n) = { my (v=0, t=1); while (n, v+=t*(n%3); n=(n\3)*2; t*=3); v } \\ Rémy Sigrist, Apr 06 2021

def changebase(n):
    L=[n]
    i=1
    while L[i-1]>2:
        x=L[i-1]
        L[i-1]=x.mod(3)
        L.append(2*floor(x/3))
        i+=1
    return sum([L[i]*3^i for i in [0..len(L)-1]])
[changebase(n) for n in [0..100]]

For n = Sum_{i=0..m} c_i*(3/2)^i with each c_i in {0,1,2}, a(n) = Sum_{i=0..m} c_i*3^i.
From Rémy Sigrist, Apr 06 2021: (Start)
Apparently:
- a(3*n) = a(3*n-1) + A003462(1+A087088(n)) for any n > 0,
- a(3*n+1) = a(3*n) + 1 for any n >= 0,
- a(3*n+2) = a(3*n+1) + 1 for any n >= 0,
(End)

cp's OEIS Frontend

A030979 Numbers k such that binomial(2k,k) is not divisible by 3, 5 or 7.

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A129508 Numbers k such that 3 and 5 do not divide binomial(2*k, k).

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A050607 Numbers k such that base 5 expansion matches (0|1|2)*((0|1)(3|4))?(0|1|2)*.

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A268082 Numbers n such that gcd(binomial(2*n-1,n), n) is equal to 1.

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A073094 Final digit of C(2k,k) when not equal to zero.

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A218394 Numbers such that sum(i<=n) binomial(n,i)*binomial(2*n-2*i, n-i) is not divisible by 5.

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A261691 Change of base from fractional base 3/2 to base 3.

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A050607 Numbers k such that base 5 expansion matches (0|1|2)((0|1)(3|4))?(0|1|2).

A218394 Numbers such that sum(i<=n) binomial(n,i)binomial(2n-2*i, n-i) is not divisible by 5.