A073095 -id:A073095 - 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

A208279 Central terms of Pascal's triangle mod 10 (A008975).

Original entry on oeis.org

1, 2, 6, 0, 0, 2, 4, 2, 0, 0, 6, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 4, 2, 0, 0, 4, 8, 4, 0, 0, 2, 4, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 2, 6, 0, 0, 2, 4, 2, 0, 0, 6, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Reinhard Zumkeller, Feb 25 2012

From Chai Wah Wu, Dec 08 2023: (Start)
Last digit of central binomial coefficient binomial(2n,n) in base 10.
A000984 mod 10.
A073095 are numbers n such that a(n) = A008904(n).
a(n) is even for n>0. (End)
From Robert Israel, Dec 08 2023: (Start)
If at least one base-5 digit of n is 3 or 4, then a(n) = 0.
Otherwise, if k 1's occur in the base-5 expansion of n, then a(n) = 2^k (mod 10) if k > 0, or 6 if k = 0. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A000984, A008904, A008975, A073095.

Haskell
```
a208279 n = a008975 (2*n) n
```

f:= proc(n) local A,v;
   A:= convert(n,base,5);
   if select(`>=`,A,3) <> [] then return 0 fi;
   v:= numboccur(1,A);
   if v > 0 then 2^v mod 10
   else 6
   fi
end proc:
f(0):= 1:
map(f, [$0..200]); # Robert Israel, Dec 08 2023

Array[Mod[Binomial[2#,#],10]&,100,0] (* Paolo Xausa, Dec 09 2023 *)

from sympy.ntheory.factor_ import digits
def A208279(n):
    if n == 0: return 1
    s = digits(n,5)[1:]
    return 0 if any(x>2 for x in s) else ((6,2,4,8)[a&3] if (a:=s.count(1)) else 6) # Chai Wah Wu, Dec 08 2023

a(n) = A008975(2*n,n) = binomial(2n,n) mod 10.

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

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A208279 Central terms of Pascal's triangle mod 10 (A008975).

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Haskell

Maple

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