A050607 -id:A050607 - cp's OEIS frontend

A037453 Positive numbers whose base-5 representation contains no 3 or 4.

1, 2, 5, 6, 7, 10, 11, 12, 25, 26, 27, 30, 31, 32, 35, 36, 37, 50, 51, 52, 55, 56, 57, 60, 61, 62, 125, 126, 127, 130, 131, 132, 135, 136, 137, 150, 151, 152, 155, 156, 157, 160, 161, 162, 175, 176, 177, 180, 181, 182, 185, 186, 187

Offset: 1

Clark Kimberling

5 divides neither 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.
k such that binomial(2k,k) != 0 (mod 10). - Benoit Cloitre, Aug 18 2002
Let us recall the plan of Apery's irrationality proof. Consider the recurrence (n+1)^3 * u_(n+1) = (34n^3 + 51n^2 + 27n + 5)u_n - n^3 * u_(n-1). The solution with starting values u_0 = 1; u_1 = 5 has the peculiar property that it has integral terms, despite the fact that at every recursion step we divide by (n+1)^3. The n-th term is given by f(n) = Sum_{i=0..n} binomial(n+i,i)^2 * binomial(n,i)^2 = A005259(n) (see Beukers link) and m such that f(m) mod 5 <> 0 equals 2*a(m). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 08 2004
Numbers k such that A208279(k) <> 0. A073095 is a subsequence. - Chai Wah Wu, Dec 08 2023

			From _David A. Corneth_, Dec 23 2023: (Start)
27_10 = 102_5 is a term since its base-5 representation contains no 3 and no 4.
28_10 = 103_5 is not a term since its base-5 representation contains a 3.
(End)

Clark Kimberling, Table of n, a(n) for n = 1..1000
Frits Beukers Consequences of Apery's work on zeta(3), Rencontres Arithmétiques de Caen, zeta(3) irrationnel: les retombées, 1995.
Barry Brent, On the Constant Terms of Certain Laurent Series, Preprints (2023) 2023061164.
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. A050607, A005836, A007091.

Cf. A000984, A073095, A208279.

function a(n)
    m, r, b = n, 0, 1
    while m > 0
        m, q = divrem(m, 3)
        r += b * q
        b *= 5
    end
r end; [a(n) for n in 1:53] |> println # Peter Luschny, Jan 03 2021

a:= proc(t) option remember; 5*procname(floor(t/3))+ (t mod 3) end proc:
a(0):= 0:
seq(a(n),n=1..100); # Robert Israel, Sep 02 2014

Table[FromDigits[IntegerDigits[k,3],5], {k,60}] (* T. D. Noe, Apr 18 2007 *)
Rest[FromDigits[#,5]&/@Tuples[{0,1,2},4]] (* Harvey P. Dale, Aug 31 2016 *)
Select[Range[187], !Divisible[Binomial[2#, #], 10]&] (* Stefano Spezia, Dec 09 2023 *)

f(n)=sum(i=0,n,binomial(n+i,i)^2*binomial(n,i)^2); for (i=1,1000,if(Mod(f(i),5)<>0,print1(i/2,",")))

isok(k) = binomial(2*k, k) % 10; \\ Michel Marcus, Dec 08 2023

is(n) = my(s = Set(digits(n, 5))); s[#s] < 3 \\ David A. Corneth, Dec 23 2023

a(n) = fromdigits(digits(n, 3), 5) \\ David A. Corneth, Dec 23 2023

from itertools import count, islice
from sympy.ntheory.factor_ import digits
def A037453_gen(startvalue=1): # generator of terms >= startvalue
    if startvalue <= 0: yield 0
    yield from filter(lambda n: all(x<3 for x in digits(n, 5)[1:]), count(max(startvalue, 1)))
A037453_list = list(islice(A037453_gen(), 30)) # Chai Wah Wu, Dec 08 2023

from gmpy2 import digits
def A037453(n): return int(digits(n,3),5) # Chai Wah Wu, Aug 10 2025

a(3n)=5a(n), a(3n+1)=5a(n)+1, a(3n+2)=5a(n)+2, where by definition a(0)=0. - Emeric Deutsch, Mar 23 2004

G.f. satisfies g(x) = 5*(1+x+x^2)*g(x^3) + (x + 2*x^2)/(1-x^3). - Robert Israel, Sep 02 2014

Better definition from T. D. Noe, Apr 18 2007

A046097 Values of n for which binomial(2n-1, n) is squarefree.

Original entry on oeis.org

1, 2, 3, 4, 6, 9, 10, 12, 36

Offset: 1

Eric W. Weisstein

No more terms up to 2^300. The sequence is finite by results of Sander and of Granville and Ramaré (see links). - Robert Israel, Dec 10 2015

Eric Weisstein's World of Mathematics, Binomial Coefficient.
A. Granville and O. Ramaré, Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients, Mathematika 43 (1996), 73-107.
J. W. Sander, Prime power divisors of binomial coefficients, Journal für die reine und angewandte Mathematik 430 (1992), 1-20.

Cf. A001700.

For a term to be here, it needs to be at least in the intersection of A048645, A051382, A050607, A050608 and an infinitude of similar sequences. The corresponding location in next-to-center column should be nonzero in A034931 (Pascal's triangle mod 4) and all similarly constructed fractal triangles (Pascal's triangle mod p^2).

[n: n in [1..150] | IsSquarefree(Binomial(2*n-1,n))]; // Vincenzo Librandi, Dec 10 2015

select(n -> numtheory:-issqrfree(binomial(2*n-1,n)), [$1..2000]); # Robert Israel, Dec 09 2015
N:= 300: # to find all terms <= 2^N
carries:= proc(n,m,p)
# number of carries when adding n + m in base p.
local A,B,C,j,nc, t;
   A:= convert(m,base,p);
   B:= convert(n,base,p);
C:= 0; nc:= 0;
   if nops(A) < nops(B) then A = [op(A),0$(nops(B)-nops(A))]
   elif nops(A) > nops(B) then B:= [op(B), 0$(nops(A)-nops(B))]
   fi;
for j from 1 to nops(A) do
    t:= C + A[j] + B[j];
    if t >= p then
       nc:= nc+1;
       C:= 1;
    else
       C:= 0
    fi
od:
nc;
end proc:
Cands:=  {seq(2^j,j=0..N), seq(seq(2^j + 2^k, k=0..j-1),j=1..N-1)}:
for i from 2 to 10 do
  Cands:= select(n -> carries(n-1,n,ithprime(i)) <= 1, Cands)
od:
select(n -> numtheory:-issqrfree(binomial(2*n-1,n)),Cands); # Robert Israel, Dec 10 2015

Select[ Range[1500], SquareFreeQ[ Binomial[ 2#-1, #]] &] (* Jean-François Alcover, Oct 25 2012 *)

is(n)=issquarefree(binomial(2*n-1,n)) \\ Anders Hellström, Dec 09 2015

James Sellers reports no further terms below 1500.
Michael Somos checked to 99999. Probably there are no more terms.
Mauro Fiorentini checked up to 2^64, as for n = 545259520, the binomial coefficient is a multiple of 5^4 and other possible exceptions have been checked (see Weisstein page for details).

A050608 Numbers k such that base 7 expansion matches (0|1|2|3)((0|1|2)(4|5|6))?(0|1|2|3).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 28, 29, 30, 31, 35, 36, 37, 38, 42, 43, 44, 45, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 77, 78, 79, 80, 84, 85, 86, 87, 91, 92, 93, 94, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113

Offset: 1

Antti Karttunen, 1999

49 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 7-automatic sequences.

Cf. A037461, A046097, A050607, A051382.

# For conv_x_base_n function, see A050607.
for($i=0; $i <= 1000; $i++) { if(("0" . conv_x_base_n($i,7)) =~ /^(0|1|2|3)*((0|1|2)(4|5|6))?(0|1|2|3)*$/) { print $i, ","; } }

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

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A037453 Positive numbers whose base-5 representation contains no 3 or 4.

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A046097 Values of n for which binomial(2n-1, n) is squarefree.

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A050608 Numbers k such that base 7 expansion matches (0|1|2|3)*((0|1|2)(4|5|6))?(0|1|2|3)*.

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A050608 Numbers k such that base 7 expansion matches (0|1|2|3)((0|1|2)(4|5|6))?(0|1|2|3).