A038003 -id:A038003 - cp's OEIS frontend

A352509 Numbers k such that k and k+1 are both Catalan-Niven numbers (A352508).

1, 4, 5, 9, 32, 44, 55, 56, 134, 144, 145, 146, 155, 184, 234, 324, 329, 414, 426, 429, 434, 455, 511, 512, 603, 636, 930, 1004, 1014, 1160, 1183, 1215, 1287, 1308, 1448, 1472, 1505, 1562, 1595, 1808, 1854, 1967, 1985, 1995, 2051, 2075, 2096, 2135, 2165, 2255

Offset: 1

Amiram Eldar, Mar 19 2022

			4 is a term since 4 and 5 are both Catalan-Niven numbers: the Catalan representation of 4, A014418(20) = 20, has the sum of digits 2+0 = 2 and 4 is divisible by 2, and the Catalan representation of 5, A014418(5) = 100, has the sum of digits 1+0+0 = 1 and 5 is divisible by 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000108, A014418, A014420.
Subsequence of A352508.
Subsequences: A038003, A352510, A352511.
Similar sequences: A330927, A328205, A328209, A328213, A330931, A331086, A333427, A334309, A331820, A342427, A344342, A351715, A351720, A352090, A352108, A352321, A352343.

c[n_] := c[n] = CatalanNumber[n]; q[n_] := Module[{s = {}, m = n, i}, While[m > 0, i = 1; While[c[i] <= m, i++]; i--; m -= c[i]; AppendTo[s, i]]; Divisible[n, Plus @@ IntegerDigits[Total[4^(s - 1)], 4]]]; Select[Range[2300], q[#] && q[#+1] &]

A259667 Catalan numbers mod 6.

Original entry on oeis.org

1, 1, 2, 5, 2, 0, 0, 3, 2, 2, 2, 4, 4, 4, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 4, 4, 1, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 4, 4, 4, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 2, 2, 2, 4

Offset: 0

M. F. Hasler, Nov 08 2015

nonn

The only odd terms are those with indices n = 2^k - 1 (k = 0, 1, 2, 3, ...); see also A038003.
It is conjectured that the only k which yield a(2^k-1) = 1 are k = 0, 1 and 5. Are there other k than 2 and 8 that yield a(2^k-1) = 5? Otherwise said, is a(2^k-1) = 3 for all k > 8?
The question is equivalent to: does 2^k - 1 always contain a digit 2 when converted into base 3 for all k > 8? Similar conjecture has been proposed for 2^k, see A004642. - Jianing Song, Sep 04 2018

M. Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems, sect. III: Binomial coefficients modulo integers, binomod.gp (v.1.4, 11/2015).
V. Reshetnikov, A000108(n) ≡ 1 (mod 6), SeqFan list, Nov. 8, 2015.

Cf. A000108, A004642, A038003.

Mod[CatalanNumber[Range[0,120]],6] (* Harvey P. Dale, Oct 24 2020 *)

PARI
```
a(n)=binomial(2*n,n)/(n+1)%6
```

A259667(n)=lift(if(n%3!=1,binomod(2*n+1,n,6)/(2*n+1), if(bittest(n,0),binomod(2*n,n-1,6)/n,binomod(2*n,n,6)/(n+1)))) \\ using binomod.gp by M. Alekseyev, cf. Links.

a(n) = A000108(n) mod 6.

A152670 Even Catalan numbers.

Original entry on oeis.org

2, 14, 42, 132, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152, 69533550916004

Offset: 1

Omar E. Pol, Dec 10 2008

G. C. Greubel, Table of n, a(n) for n = 1..541 [duplicated terms at n=343 and n=393 removed by _Georg Fischer_, Oct 19 2024]

Cf. A000108, A038003, A152671.

Select[CatalanNumber[Range[30]], EvenQ] (* Vladimir Reshetnikov, Nov 02 2015 *)

lista(nn) = for (n=1, nn, if (((c=binomial(2*n,n)/(n+1)) % 2) == 0, print1(c, ", "))); \\ Michel Marcus, Nov 02 2015

A379152 The binary weights of the odd Catalan numbers.

Original entry on oeis.org

1, 1, 2, 6, 16, 25, 60, 127, 244, 494, 1010, 2015, 4076, 8086, 16281, 32818, 65518, 131059, 262348, 524448, 1047643, 2097675, 4194133, 8386693, 16776916, 33554390, 67114125, 134214652, 268452748

Offset: 0

Amiram Eldar, Dec 17 2024

Florian Luca and Paul Thomas Young, On the binary expansion of the odd Catalan numbers, Proceedings of the XIVth International Conference on Fibonacci Numbers, Morelia, Mexico, 2010, pp. 185-190; ResearchGate link.

Cf. A000108, A000120, A038003.

Similar sequences: A011373, A079584, A082481, A379151, A379153.

a[n_] := DigitCount[CatalanNumber[2^n-1], 2, 1]; Array[a, 23, 0]

a(n) = my(m = -1 + 1 << n); hammingweight(binomial(2*m, m)/(m+1));

from itertools import count, islice
def A379152_gen(): # generator of terms
    yield from [1,1]
    c, s = 1, 3
    for n in count(2):
        c = (c*((n<<2)-2))//(n+1)
        if n == s:
            yield c.bit_count()
            s = (s<<1)|1
A379152_list = list(islice(A379152_gen(),10)) # Chai Wah Wu, Dec 17 2024

a(n) = A000120(A038003(n)) = A000120(A000108(2^n-1)).

a(n) = A379151(2^n-1).

A135759 Least Catalan number divisible by 2^n: a(n) = A000108(2^(n+1)-2).

Original entry on oeis.org

1, 2, 132, 2674440, 3814986502092304, 24139737743045626825711458546273312, 2861304849265668492891140780463352404986232263244287143198790516197234752

Offset: 0

Paul D. Hanna, Dec 02 2007

nonn

The next term has 150 digits. - Harvey P. Dale, Jan 09 2017

Cf. A038003 (odd Catalan numbers).

Table[Binomial[2^(n + 2) - 4, 2^(n + 1) - 2]/(2^(n + 1) - 1), {n,0,10}] (* G. C. Greubel, Nov 07 2016 *)
Table[SelectFirst[CatalanNumber[Range[300]],Divisible[#,2^n]&],{n,0,7}] (* Harvey P. Dale, Jan 09 2017 *)

{a(n) = binomial(2^(n+2)-4, 2^(n+1)-2) / (2^(n+1)-1)}
for(n=0,8,print1(a(n),", "))

a(n) = C(2^(n+2)-4, 2^(n+1)-2) / (2^(n+1)-1).

A178854 Asymptotic value of odd Catalan numbers mod 2^n.

Original entry on oeis.org

0, 1, 1, 5, 13, 29, 29, 93, 221, 221, 733, 1757, 3805, 7901, 7901, 24285, 57053, 122589, 122589, 384733, 384733, 384733, 2481885, 2481885, 10870493, 10870493, 10870493, 10870493, 145088221

Offset: 0

David A. Madore, Jun 18 2010

nonn

For every n, the odd Catalan numbers C(2^m-1) are eventually constant mod 2^n (namely for m >= n-1): then a(n) is the asymptotic value of the remainder.

			The odd Catalan numbers mod 2^6=64 are 1,5,45,61,29,29,29, so a(6)=29.

Shu-Chung Liu and Jean C.-C. Yeh, Catalan numbers modulo 2^k, J. Int. Seq. 13 (2010), article 10.5.4.

Cf. A038003 (odd Catalan numbers).

A000108 := proc(n) binomial(2*n,n)/(n+1) ; end proc:
A038003 := proc(n) A000108(2^n-1) ; end proc:
A178854 := proc(n) if n = 0 then 0; else modp(A038003(n-1),2^n) ; end if; end proc:
for n from 0 do printf("%d,\n",A178854(n)) ; end do: # R. J. Mathar, Jun 28 2010

(* first do *) Needs["DiscreteMath`CombinatorialFunctions`"] (* then *) f[n_] := Mod[ CatalanNumber[2^n - 1], 2^n]; Array[f, 25, 0] (* Robert G. Wilson v, Jun 28 2010 *)

a(n) = remainder(Catalan(2^m-1), 2^n) for any m >= n-1.

a(12)-a(24) from Robert G. Wilson v, Jun 28 2010

a(25)-a(28) from Robert G. Wilson v, Jul 23 2010

A060318 Powers of 3 in the odd Catalan numbers Catalan(2^n - 1).

Original entry on oeis.org

0, 0, 1, 2, 0, 1, 3, 0, 3, 3, 3, 6, 2, 2, 9, 5, 5, 4, 8, 5, 9, 10, 5, 4, 4, 4, 9, 9, 8, 11, 13, 13, 10, 11, 10, 8, 6, 12, 13, 14, 13, 11, 14, 15, 16, 13, 11, 10, 12, 18, 20, 19, 20, 11, 13, 19, 22, 18, 15, 26, 20, 17, 17, 26, 21, 22, 18, 18, 23, 26, 20, 19, 23, 21, 22, 19, 27, 17, 35

Offset: 1

Wouter Meeussen, Mar 28 2001

nonn

Conjecture: all odd Catalan numbers have smallest prime factor 3, except Catalan(3), which has smallest prime factor 5, and Catalan(31) and Catalan(255), which have smallest prime factor 7 (checked up to Catalan(-1 + 2^2048)).

			a(5)=0 because 2^5 -1 = 31 and Catalan(31) = 7*11*17*19*37*41*43*47*53*59*61 so the power of 3 is zero.

Cf. A007949, A038003.

Cf. A000108, A048896, A048881.

pow3[ nfac_ ] := (nfac - Plus @@ IntegerDigits[ nfac, 3 ])/(3-1); powcat3[ n_ ] := pow3[ 2n ]-pow3[ n+1 ]-pow3[ n ]; Table[ powcat3[ 2^n-1 ], {n, 2048} ]

a(n) = A007949(A038003(n)). - Michel Marcus, Feb 02 2020

A116943 Number of 4s digits plus non-final 3s digits 3 base 5 expansion of 2^n.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 3, 2, 2, 2, 1, 1, 3, 4, 5, 3, 3, 1, 4, 4, 7, 2, 7, 7, 4, 6, 9, 9, 6, 5, 5, 7, 4, 9, 4, 7, 7, 7, 10, 8, 6, 8, 6, 9, 8, 9, 8, 10, 11, 11, 8, 13, 5, 11, 15, 13, 10, 10, 8, 12, 9, 14, 11, 8, 11, 12, 10, 13, 13, 13, 10, 10, 12, 6, 10, 15, 8, 17, 17, 16, 16, 12, 16, 15, 13

Offset: 0

Jonathan Vos Post, Mar 23 2006

In his comment on A038003 Frank Adams-Watters conjectures "that 2^n contains such a base 5 digit for n>=9. This is almost certainly true." That is equivalent to a(n) > 0 for n>=9, which is also equivalent to A094389(n) = 5 where A094389 is last decimal digit of the odd Catalan number A038003(n).

			a(7) = 0 because 2^7 (modulo 5) = 1003, which contains 0 digits 4 plus 0 non-final digits 3 (it has a digit 3, but that digit is finial, meaning rightmost).
a(10) = 3 because 2^10 mod 5 = 13044, which contains 2 digits 4 plus 1 non-final digits 3, so 2 + 1 = 3.
a(60) = 10 because 2^60 mod 5 = 34132411211412413323100401, which contains 5 digits 4 plus 5 non-final digits 3, so 5 + 5 = 10.

Cf. A038003, A094389.

f[n_] := Block[{id = IntegerDigits[2^n, 5]}, Count[id, 4] + Count[Most@id, 3]]; Table[ f[n], {n, 0, 88}] (* Robert G. Wilson v *)

More terms from Robert G. Wilson v, Apr 01 2006

A152753 Last digit of even Catalan number A152670(n).

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Dec 12 2008

Even values of A152669.

Cf. A000108, A038003, A094389, A152669, A152670.

A201823 Terms of A001764 not divisible by 3, where A001764 enumerates ternary trees.

Original entry on oeis.org

1, 1, 55, 300830572, 1414282077098335379544565517191

Offset: 0

Paul D. Hanna, Dec 05 2011

nonn

The number of digits for the terms begin:

[1, 1, 2, 9, 31, 97, 298, 902, 2715, 8155, 24478, 73446, 220354, ...].

			Sequence A001764 begins: [1, 1, 3, 12, 55, 273, 1428, 7752, 43263, 246675, ...],
where A001764(n) == 1 (mod 3) at positions: [0, 1, 4, 13, 40, 121, 364, 1093, ...].

Cf. A038003, A001764.

a(n)=binomial(3*(3^n-1)/2,(3^n-1)/2)/3^n

a(n) = A001764( (3^n-1)/2 ) = binomial( 3*(3^n-1)/2, (3^n-1)/2 ) / 3^n.

a(n) == 1 (mod 3).

cp's OEIS Frontend

A352509 Numbers k such that k and k+1 are both Catalan-Niven numbers (A352508).

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A259667 Catalan numbers mod 6.

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A152670 Even Catalan numbers.

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A379152 The binary weights of the odd Catalan numbers.

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A135759 Least Catalan number divisible by 2^n: a(n) = A000108(2^(n+1)-2).

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A178854 Asymptotic value of odd Catalan numbers mod 2^n.

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A060318 Powers of 3 in the odd Catalan numbers Catalan(2^n - 1).

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A116943 Number of 4s digits plus non-final 3s digits 3 base 5 expansion of 2^n.

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