A096304 -id:A096304 - cp's OEIS frontend

A019469 Numbers k such that k does not divide binomial(2*k-4, k-2).

2, 3, 4, 6, 8, 9, 12, 15, 16, 27, 30, 32, 33, 36, 39, 42, 64, 81, 84, 87, 90, 93, 96, 108, 111, 114, 117, 120, 123, 128, 243, 246, 249, 252, 255, 256, 258, 270, 273, 276, 279, 282, 285, 324, 327, 330, 333, 336, 339

Offset: 1

Wouter Meeussen

nonn

Previous name was: Numbers n such that (n-2)-nd Catalan number is not divisible by n.

Conjecture (confirmed, see links): sequence is union of powers of two > 1 (A000079) and 3 * A096304.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Kevin Ryde, Proof of Wouter Meeussen's conjectured union

Cf. A000079, A096304, A000984.

Complement of A019470.

A019469:=n->`if`(binomial(2*n-4,n-2) mod n <> 0,n,NULL): seq(A019469(n), n=1..400); # Wesley Ivan Hurt, Sep 13 2014

Select[Range[400],!Divisible[Binomial[2#-4,#-2],#]&] (* Harvey P. Dale, Aug 13 2015 *)

valp(n, p)=my(s); while(n\=p, s+=n); s
bin(n, p)=valp(2*n, p)-2*valp(n, p)
is(n)=my(f=factor(n)); for(i=1, #f~, if(bin(n-2, f[i, 1])Charles R Greathouse IV, Nov 04 2016

Name changed by Wesley Ivan Hurt, Sep 16 2014

A340051 Mixed-radix representation of n where the least significant digit is in base 3 and other digits are in base 2.

Original entry on oeis.org

0, 1, 2, 10, 11, 12, 100, 101, 102, 110, 111, 112, 1000, 1001, 1002, 1010, 1011, 1012, 1100, 1101, 1102, 1110, 1111, 1112, 10000, 10001, 10002, 10010, 10011, 10012, 10100, 10101, 10102, 10110, 10111, 10112, 11000, 11001, 11002, 11010, 11011, 11012, 11100, 11101

Offset: 0

Kevin Ryde, Apr 24 2021

Kevin Ryde, Table of n, a(n) for n = 0..3072
Index entries for 10-automatic sequences.

Cf. A007088 (base 2).

Cf. A096304 (interpreted as ternary).

a(n) = my(r); [n,r]=divrem(n,3); fromdigits(concat(binary(n),r));

a(n) = 10*A007088(floor(n/3)) + (n mod 3).

G.f.: x*(1+2*x)/(1-x^3) + 10/(1-x) * Sum_{k>=0} 10^k*x^(3*2^k)/(1 + x^(3*2^k)).

A119955 Numbers n such that denominator of n-th Harmonic Number equals denominator of n-th Alternative Harmonic Number.

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 10, 11, 12, 13, 14, 27, 49, 50, 51, 52, 53, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 289, 290, 291, 292, 293, 841, 842, 843, 844

Offset: 1

Alexander Adamchuk, Aug 02 2006

nonn

Up to n=14 A002805[n] coincides with A058312[n]. a(n) up to a(12)=27 coincides with A096304[n].

			Denominators of Harmonic Number (H[n] = Sum[1/i, {i, n}]) are A002805[n] = {1,2,6,12,60,20,140,280,2520,2520,27720,27720,360360,360360,360360,...}.
Denominators of Alternative Harmonic Number (H'[n] = Sum[(-1)^(i+1)*1/i, {i, n}]) are A058312[n] = {1,2,6,12,60,60,420,840,2520,2520,27720,27720,360360,360360,72072,...}.
a(1) = 1 because A002805[1] = A058312[1].
15 is not in a(n) because A002805[15] = 360360 is not equal to A058312[15] = 72072.

Eric Weisstein's World of Mathematics, Harmonic Number.

Cf. A002805, A058312, A096304.

Do[s1=Denominator[Sum[(-1)^(i+1)*1/i, {i, n}]]; s2=Denominator[Sum[1/i, {i, n}]]; If[Equal[s2, s1], Print[n]], {n, 1, 1500}]

A272344 Positive integers n where the number of parts function on the set of 3-ary partitions of n is equidistributed mod 3.

Original entry on oeis.org

6, 7, 8, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 33, 34, 35, 42, 43, 44, 45, 46, 47, 48, 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, 74, 75, 76, 77, 78, 79, 80, 87, 88, 89, 96, 97, 98, 99, 100, 101

Offset: 1

Tom Edgar, Apr 26 2016

nonn

An integer n is in the list if and only if n_i=2 for some index i>0 where n = Sum_{i>=0}n_i3^i is the base 3 representation of n.

Appears to be the complement of A096304.

			There are three 3-ary partitions of 6: one has 2 parts (3+3), one has 4 parts (3+1+1+1), and one has 6 parts (1+1+1+1+1+1); thus, modulo 3, the number of parts function is equidistributed mod 3 and so 6 is a term.
There are five 3-ary partitions of 9 so the number of parts function cannot be equidistributed mod 3. Thus, 9 is not a term.

Tom Edgar, The distribution of the number of parts of m-ary partitions modulo m., arXiv:1603.00085 [math.CO], 2016.

Cf. A062051, A096304.

M=[n for n in [1..105] if (2) in n.digits(3)[1:]]

A356447 Integers k such that (k+1)(2k-1) does not divide the central binomial coefficient B(k) = binomial(2*k,k) = A000984(k).

Original entry on oeis.org

2, 5, 8, 11, 14, 26, 29, 32, 35, 38, 41, 80, 83, 86, 89, 92, 95, 107, 110, 113, 116, 119, 122, 242, 245, 248, 251, 254, 257, 269, 272, 275, 278, 281, 284, 323, 326, 329, 332, 335, 338, 350, 353, 356, 359, 362, 365, 728, 731, 734, 737, 740, 743, 755, 758, 761

Offset: 1

Valerio De Angelis, Aug 07 2022

It is well known that B(k) divided by (k+1) is an integer (the Catalan numbers A000108). It is also easy to see that (2k-1) divides B(k). So we ask when the product (k+1)*(2k-1) divides B(k). The terms of this sequence are the positive integers k such that (k+1)*(2k-1) does not divide B(k).
A necessary and sufficient condition for an integer k to be a term of this sequence is: k is congruent to 2 (mod 3), and at least one of (k+1) or (k-1) has no 2's in its base-3 expansion. In particular, this sequence has density 0. This is proved in the Stack Exchange post cited below.
Other equivalent conditions are:
1) k is congruent to 2 (mod 3), and its base-3 expansion either has no 2's, or is of form u12, or u02^i for some i>=1, where u has no 2's and 2^i means a string of i consecutive 2's.
2) the base 3 expansion of k+1 is either u0 or u20, where u has no 2's.

			k = 95 is a term, since it is k == 2 (mod 3) and k-1 = 94 has base-3 expansion 10111 which has no digit 2's. It can be checked that B(k) = binomial(190,95) is not divisible by (k+1)*(2*k-1) = 18144.
As another example, a(18)=107, also congruent to 2 (mod 3), and 107+1=108 has base-3 expansion 11000. It can be checked that binomial(2*107,107) is not divisible by (107+1)*(2*107-1).
125 is not a term of the sequence, because even though it's congruent to 2 (mod 3), the base-3 expansions of 125+1=126 and 125-1=124 are 11200 and 11121. It can be checked that binomial(2*125,125) is divisible by (125+1)*(2*125-1).

Mathematics Stack Exchange, Factors of central binomial coefficient

Cf. A000108, A000984, A073076, A096304,

kmax=762; a={}; For[k=1, k<=kmax, k++, If[Not[Divisible[Binomial[2k, k], (k+1)(2k-1)]], AppendTo[a, k]]]; a (* Stefano Spezia, Aug 12 2022 *)

isok(k) = Mod(binomial(2*k,k), (k+1)*(2*k-1)) != 0; \\ Michel Marcus, Aug 10 2022

def A356447(n):
    a, b = divmod(n,3)
    return (int(bin(a)[2:],3)*3+b)*3-1 # Chai Wah Wu, Jul 29 2025

a(n) = 3*A096304(n) - 1.

cp's OEIS Frontend

A019469 Numbers k such that k does not divide binomial(2*k-4, k-2).

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A340051 Mixed-radix representation of n where the least significant digit is in base 3 and other digits are in base 2.

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A119955 Numbers n such that denominator of n-th Harmonic Number equals denominator of n-th Alternative Harmonic Number.

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A272344 Positive integers n where the number of parts function on the set of 3-ary partitions of n is equidistributed mod 3.

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A356447 Integers k such that (k+1)*(2*k-1) does not divide the central binomial coefficient B(k) = binomial(2*k,k) = A000984(k).

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A356447 Integers k such that (k+1)(2k-1) does not divide the central binomial coefficient B(k) = binomial(2*k,k) = A000984(k).