A107755 -id:A107755 - cp's OEIS frontend

A108784 Difference between A107757 and A107755.

1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -1, 1, -1, -1, 1, -1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, 1, -1, 1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -1, 1, -1, -1, 1, -1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, 1

Offset: 1

Robert G. Wilson v, Jun 14 2005

sign

Of the 255 terms less than 10^4, 128 are positive.

R. J. Mathar, Table of n, a(n) for n = 1..120

Cf. A107755, A107756, A107757.

Maple code from R. J. Mathar, Feb 25 2008:
A000108 := proc(n) option remember ; binomial(2*n,n)/(n+1) ; end:
A107757 := proc(n) option remember ; local a; if n = 1 then 3; else for a from A107757(n-1)+1 do if add(A000108(k),k=1..a) mod 3 = 2 then RETURN(a) ; fi ; od: fi ; end:
A107755 := proc(n) option remember ; local a; if n = 1 then 2; else for a from A107755(n-1)+1 do if add(A000108(k),k=1..a) mod 3 = 0 then RETURN(a) ; fi ; od: fi ; end:
A108784 := proc(n) A107757(n)-A107755(n) ; end: seq(A108784(n),n=1..120) ;

s0 = s2 = {}; s = 0; Do[s = Mod[s + (2 n)!/n!/(n + 1)!, 3]; Switch[ Mod[s, 3], 0, AppendTo[s0, n], 2, AppendTo[s2, n]], {n, 10^4}]; s2 - s0

It appears that a(n) = A076826(2n)-1. - T. D. Noe, Jun 14 2007

a(n) = A107757(n) - A107755(n).

Corrected by T. D. Noe, Jun 14 2007

A137822 First differences of A137821 (numbers such that sum( Catalan(k), k=1..2n) = 0 (mod 3)).

Original entry on oeis.org

1, 3, 2, 7, 2, 3, 1, 21, 2, 3, 1, 8, 1, 3, 2, 61, 2, 3, 1, 8, 1, 3, 2, 21, 1, 3, 2, 7, 2, 3, 1, 183, 2, 3, 1, 8, 1, 3, 2, 21, 1, 3, 2, 7, 2, 3, 1, 62, 1, 3, 2, 7, 2, 3, 1, 21, 2, 3, 1, 8, 1, 3, 2, 547, 2, 3, 1, 8, 1, 3, 2, 21, 1, 3, 2, 7, 2, 3, 1, 62, 1, 3, 2, 7, 2, 3, 1, 21, 2, 3, 1, 8, 1, 3, 2, 183, 1, 3, 2

Offset: 1

M. F. Hasler, Feb 25 2008, revised Mar 15 2008

nonn

For the initial term, we use A137821(0)=0 (cf. formula).
Sequence A122983 lists record values of this one, which occur at index 2^j (cf. formula). The fact that these values roughly grow by a factor 3 is explained by the fact that these values are given as the sum of all preceding terms (up to +1 or +2 according to the parity of j, cf. formula).
The only values occurring in this sequence are { 1, 2, 3, 7, 8, 21, 61, 62, 183, 547, 548, 1641,... } = A137823, consisting of the record values a(2^j) and, for every other one of these (i.e. for even j), its successor a(2^j)+1, occurring first as a(3*2^j).
The remarkably simple sequence A137824 (= 1,3,2, 4,12,8,...: pattern 1,3,2 multiplied by powers of 4) gives the index at which the value A137823(m) first occurs. - M. F. Hasler, Mar 15 2008
The PARI code given here (function A137822(n)) allows one to calculate hundreds of terms of A107755 in a few microseconds. - M. F. Hasler, Mar 15 2008

			Record values are a(1)=1, a(2)=3, a(4)=7, a(8)=21, a(16)=61, ...
Apart from these values, the only other values occurring in the sequence are:
2=a(1)+1=a(3*1), 8=a(4)+1=a(3*4), 62=a(16)+1=a(3*16), ...

M. F. Hasler, Table of n, a(n) for n = 1..499.

Cf. A000108, A107755, A137821-A137824.

Cf. A122983 (record values of this).

Join[{1},Differences[Flatten[Position[Accumulate[CatalanNumber[Range[3000]]],?(Mod[#,3]==0&)]]/2]] (* _Harvey P. Dale, Jun 19 2025 *)

A137822 = D( A137821 ) /* where D(v)=vector(#v-1,i,v[i+1]-v[i]) or D(v)=vecextract(v, "^1")-vecextract(v,"^-1") */

n=0; A137822=vector(499,i,{ o=n; if( bitand(i,i-1), while(n++ && s+=binomial(4*n-2, 2*n-1)/(2*n)*(10*n-1)/(2*n+1),),s=Mod(0,3); n=2*n+1+log(i+.5)\log(2)%2 ); n-o})

A137822(n)= local( L=log(n+.5)\log(2) ); while( n>0 || ((n+=2^L) && L=log(n+.5)\log(2)), (n-=2^L) || return( 3^(L+1)\4+1 ); (n-=2^(L-1)) || return( 3^L\4+1+L%2 );n<0 && n+=2<M. F. Hasler, Mar 15 2008

a(m) = A137821(m)-A137821(m-1), A137821(m)=sum( a(j), j=1..m).
a(2^j) = A122983(j-1) = A137821(2^j-1) + 1 (resp. +2) for j even (resp. odd).
a(3*2^j) = a(2^j) (resp. = a(2^j)+1) for j odd (resp. j even).

A137821 Numbers k such that Sum_{j=1..2k} Catalan(j) == 0 (mod 3).

Original entry on oeis.org

1, 4, 6, 13, 15, 18, 19, 40, 42, 45, 46, 54, 55, 58, 60, 121, 123, 126, 127, 135, 136, 139, 141, 162, 163, 166, 168, 175, 177, 180, 181, 364, 366, 369, 370, 378, 379, 382, 384, 405, 406, 409, 411, 418, 420, 423, 424, 486, 487, 490, 492, 499, 501, 504, 505

Offset: 1

M. F. Hasler, Feb 25 2008

nonn

It would be natural to prepend an initial term a(1)=0 (for which the sum is to be considered empty, thus zero), but we omit it to avoid confusion w.r.t. indices of A107755.

M. F. Hasler, Table of n, a(n) for n = 1..499.

Cf. A107755 (twice this), A137822-A137824.

Flatten[Position[Accumulate[CatalanNumber[Range[1100]]],?(Mod[#,3]==0&)]]/2 (* _Harvey P. Dale, Jun 19 2025 *)

n=0; A137821=vector(499,i,{ if( bitand(i,i-1), while(n++ & s+=binomial(4*n-2,2*n-1)/(2*n)*(10*n-1)/(2*n+1),),s=Mod(0,3); n=2*n+1+log(i+.5)\log(2)%2 ); n})

a(n) = A107755(n)/2 = Sum_{k=0..n} A137822(k).
a(2^j) = 2 a(2^j-1) + 1 (resp. +2) for j even (resp. odd).
Sum_{k=1..2n} Catalan(k) = Sum_{k=1..n} Catalan(2k-1) * (10k-1)/(2k+1), thus:
{ a(m) } = { n>0 | Sum_{k=1..n} Catalan(2k-1) * (10k-1)/(2k+1) == 0 (mod 3) }.

A137824 Index at which A137823(n) occurs first in A137822 (gaps in numbers m such that 3 | sum( Catalan(k), k=1..2m)).

Original entry on oeis.org

1, 3, 2, 4, 12, 8, 16, 48, 32, 64, 192, 128, 256, 768, 512, 1024, 3072, 2048, 4096, 12288, 8192, 16384, 49152, 32768, 65536, 196608, 131072, 262144, 786432, 524288, 1048576, 3145728, 2097152, 4194304, 12582912, 8388608, 16777216, 50331648

Offset: 1

M. F. Hasler, May 15 2008

Other characterization of the sequence: concatenate pattern (1,3,2) multiplying it by 4 after each concatenation step. Or: Start with 1,3,2, then iteratively append the whole sequence obtained so far multiplied by 4^(length of the sequence divided by 3)

See A137822 and A137823 for more comments and formulas.

Index entries for linear recurrences with constant coefficients, signature (0,0,4).

Cf. A107755, A122983, A137821, A137822, A137823.

A137824(n) = if( n%3==2,3,1)<<(2*(n-1)\3)

A137824(n) = for( i=1,#A137822, A137822[i]==A137823[n] & return(i))

a=[1,3,2]; for( i=1,5, a=concat( a, 4^(#a/3)*a )); a

If n==2 (mod 3) then a(n) = 3*2^[2*(n-1)/3]; else a(n) = 2^[2*(n-1)/3].

a(n) = 4*a(n-3) for n>3. G.f.: x*(1+x)*(1+2*x)/(1-4*x^3). - Colin Barker, Aug 19 2012

A137823 Numbers occurring in A137822 : first differences of numbers n such that 3 | sum( Catalan(k), k=1..2n).

Original entry on oeis.org

1, 2, 3, 7, 8, 21, 61, 62, 183, 547, 548, 1641, 4921, 4922, 14763, 44287, 44288, 132861, 398581, 398582, 1195743, 3587227, 3587228, 10761681, 32285041, 32285042, 96855123, 290565367, 290565368, 871696101, 2615088301, 2615088302

Offset: 1

M. F. Hasler, May 15 2008

This sequence consists of A122983 (which are record values of A137822, i.e. record gaps of A137821), plus the successor of every other term thereof (i.e. A122983(2k+1)+1). See A137822 for more details & formulas.

The (very simple) sequence A137824 gives the index of A137822 at which A137823(n) occurs for the first time.

Cf. A107755, A122983, A137821-A137824.

A137823(n) = (n%3==2)+3^((n-1)*2\3+1)\4+1

A107756 Numbers k such that Sum_{j=1..k} Catalan(j) == 1 (mod 3).

Original entry on oeis.org

1, 4, 5, 6, 7, 10, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 28, 31, 32, 33, 34, 37, 40, 41, 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, 82, 85, 86, 87, 88, 91, 94, 95

Offset: 1

N. J. A. Sloane, Jun 11 2005

Y. More, Problem 11165, Amer. Math. Monthly, 112 (2005), 568.

Cf. A000108, A107755, A107757.

Equals A074940 - 1.

Select[ Range[ 100], Mod[ Sum[(2 n)!/n!/(n + 1)!, {n, #}], 3] == 1 &] (* Robert G. Wilson v, Jun 14 2005 *)
Position[Accumulate[CatalanNumber[Range[100]]],?(Mod[#,3]==1&)]//Flatten (* _Harvey P. Dale, Aug 26 2022 *)

Equivalently, numbers k such that base-3 expansion of k+1 contains a 2.

A107757 Numbers k such that Sum_{j=1..k} Catalan(j) == 2 (mod 3).

Original entry on oeis.org

3, 9, 11, 27, 29, 35, 39, 81, 83, 89, 93, 107, 111, 117, 119, 243, 245, 251, 255, 269, 273, 279, 281, 323, 327, 333, 335, 351, 353, 359, 363, 729, 731, 737, 741, 755, 759, 765, 767, 809, 813, 819, 821, 837, 839, 845, 849, 971, 975, 981, 983, 999, 1001, 1007, 1011

Offset: 1

N. J. A. Sloane, Jun 11 2005

Y. More, Problem 11165, Amer. Math. Monthly, 112 (2005), 568.

Cf. A000108, A107755, A107756.

Equals A074939 - 1.

c:=n->binomial(2*n,n)/(n+1): s:=0: for n from 1 to 1500 do s:=s+c(n): a[n]:=s mod 3: od: A:=[seq(a[n],n=1..1500)]: p:=proc(n) if A[n]=2 then n else fi end: seq(p(n),n=1..1500); # Emeric Deutsch, Jun 12 2005

s0 = s2 = {}; s = 0; Do[s = Mod[s + (2 n)!/n!/(n + 1)!, 3]; Switch[ Mod[s, 3], 0, AppendTo[s0, n], 2, AppendTo[s2, n]], {n, 1055}]; s2 (* Robert G. Wilson v, Jun 14 2005 *)

More terms from Emeric Deutsch, Jun 12 2005

A137992 A014137 (= partial sums of Catalan numbers A000108) mod 3.

Original entry on oeis.org

1, 2, 1, 0, 2, 2, 2, 2, 1, 0, 2, 0, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 0, 2, 0, 1, 2, 2, 2, 2, 0, 1, 2, 1, 0, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 0

M. F. Hasler, Mar 16 2008

As usual, "mod 3" means to choose the unique representative in { 0,1,2 } of the equivalence class modulo 3Z.

Cf. A014137, A000108, A137821-A137824, A107755; A014138(n)+1 = a(n+1) (mod 3).

A137992(n) = lift( sum( k=0,n, binomial( 2*k,k )/(k+1), Mod(0,3) ))

a(n) = sum( k=0..n, C(k) ) (mod 3), where C(k) = binomial(2k,k)/(k+1).

a(n) = 1 <=> n = 2 A137821(m) for some m (with A137821(0)=0).

A137993 A014138 (= partial sums of Catalan numbers starting with 1,2,5) mod 3.

Original entry on oeis.org

1, 0, 2, 1, 1, 1, 1, 0, 2, 1, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 2, 1, 2, 0, 1, 1, 1, 1, 2, 0, 1, 0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0

Offset: 0

M. F. Hasler, Mar 16 2008

As usual, "mod 3" means to choose the unique representative in { 0,1,2 } of the equivalence class modulo 3Z.
Here the conventions of A014138 are used, but it seems somehow unnatural to start with offset 0 corresponding to the Catalan number A000108(1).
For m>1, the length of the m-th block of nonzero elements (and thus the approximate length of the m-th string of consecutive 1's) is given by 2 A137822(m)-1.

Cf. A014138, A000108, A137821-A137824, A107755, A137992, A014137(n+1) = a(n)+1 (mod 3).

A137993(n) = lift( sum( k=1,n+1, binomial( 2*k,k )/(k+1), Mod(0,3) ))

a(n) = sum( k=1..n+1, C(k) ) (mod 3), where C(k) = binomial(2k,k)/(k+1) = A000108(k).

a(n) = 0 <=> n+1 = 2 A137821(m) for some m.

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A108784 Difference between A107757 and A107755.

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A137822 First differences of A137821 (numbers such that sum( Catalan(k), k=1..2n) = 0 (mod 3)).

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A137821 Numbers k such that Sum_{j=1..2k} Catalan(j) == 0 (mod 3).

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A137824 Index at which A137823(n) occurs first in A137822 (gaps in numbers m such that 3 | sum( Catalan(k), k=1..2m)).

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A137823 Numbers occurring in A137822 : first differences of numbers n such that 3 | sum( Catalan(k), k=1..2n).

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A107756 Numbers k such that Sum_{j=1..k} Catalan(j) == 1 (mod 3).

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A107757 Numbers k such that Sum_{j=1..k} Catalan(j) == 2 (mod 3).

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A137992 A014137 (= partial sums of Catalan numbers A000108) mod 3.

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