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

Author

Clark Kimberling

Keywords

• nonn
• easy
• base

Comments

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

Examples

			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)
		

Links

• 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).

Crossrefs

Cf. A050607, A005836, A007091.

Cf. A000984, A073095, A208279.

Programs

Julia

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

Maple

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

Mathematica

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 *)

PARI

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,",")))

PARI

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

PARI

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

PARI

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

Python

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

Python

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

Formula

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

Extensions

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