A118113 -id:A118113 - cp's OEIS frontend

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A263132 Positive values of m such that binomial(4*m - 1, m) is odd.

1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 32, 43, 44, 48, 64, 86, 88, 96, 128, 171, 172, 176, 192, 256, 342, 344, 352, 384, 512, 683, 684, 688, 704, 768, 1024, 1366, 1368, 1376, 1408, 1536, 2048, 2731, 2732, 2736, 2752, 2816, 3072, 4096, 5462, 5464, 5472, 5504

Offset: 1

Peter Bala, Oct 10 2015

This sequence, when viewed as a set, equals the set of numbers of the form 4^n * ceiling(2^k/3) for n >= 0, k >= 1, i.e., the product subset in Z of A000302 and A005578 regarded as sets. See the example below.
Equivalently, this sequence, when viewed as a set, equals the set of numbers of the form 2^n * (2^(2*k + 1) + 1)/3 for n,k >= 0, i.e., the product subset in Z of A000079 and A007583 regarded as sets. See the example below.
2*a(n) gives the values of m such that binomial(4*m - 2,m) is odd. 4*a(n) gives the values of m such that binomial(4*m - 3,m) is odd (other than m = 1) and also the values of m such that binomial(4*m - 4,m) is odd.

			1) Notice how this sequence can be read from Table 1 below by moving through the table in a sequence of 'knight moves' (1 down and 2 to the left) starting from the first row. For example, starting at 11 on the top row we move in a series of knights moves 11 -> 12 -> 16, then return to the top row at 22 and move 22 -> 24 -> 32, return to the top row at 43 and move 43 -> 44 -> 48 -> 64, then return to top row at 86 and so on.
........................................................
.   Table 1: 4^n * ceiling(2^k/3) for n >= 0, k >= 1   .
........................................................
n\k|   1    2    3    4     5     6    7    8     9
---+----------------------------------------------------
0  |   1    2    3    6    11    22   43   86   171 ...
1  |   4    8   12   24    44    88  172  ...
2  |  16   32   48   96   176    ...
3  |  64  128  192  ...
4  | 256  ...
...
2) Notice how this sequence can be read from Table 2 below in a sequence of 'knight moves' (2 down and 1 to the left) starting from the first two rows. For example, starting at 43 in the first row we jump 43 -> 44 -> 48 -> 64, then return to the second row at 86 and jump 86 -> 88 -> 96 -> 128, followed by 171 -> 172 -> 176 -> 192 -> 256, and so on.
....................................................
.   Table 2: 2^n * (2^(2*k + 1) + 1)/3, n,k >= 0   .
....................................................
n\k|   0    1     2     3      4      5
---+----------------------------------------------
0  |   1    3    11    43    171    683  ...
1  |   2    6    22    86    342   1366  ...
2  |   4   12    44   172    684   2732  ...
3  |   8   24    88   344   1368   5464  ...
4  |  16   48   176   688   2736  10928  ...
5  |  32   96   352  1376   5472  21856  ...
6  |  64  192   704  2752  10944  43712  ...
7  | 128  384  1408  5504  21888  87424  ...
8  | 256 ...

Cf. A000079, A000267, A000302, A005578, A007583, A048716, A384688.

Other odd binomials: A002450 (4*m+1,m), A020988 (4*m+2,m), A263133 (4*m+3,m), A080674 (4*m+4,m), A118113 (3*m-2,m), A003714 (3*m,m).

[n: n in [1..6000] | Binomial(4*n-1, n) mod 2 eq 1]; // Vincenzo Librandi, Oct 12 2015

for n from 1 to 5000 do if mod(binomial(4*n-1, n), 2) = 1 then print(n) end if end do;

Select[Range[6000],OddQ[Binomial[4#-1,#]]&] (* Harvey P. Dale, Dec 26 2015 *)

for(n=1, 1e4, if (binomial(4*n-1, n) % 2 == 1, print1(n", "))) \\ Altug Alkan, Oct 11 2015

a(n) = my(r,s=sqrtint(4*n-3,&r)); (1<Kevin Ryde, Jun 14 2025

Python

A263132_list = [m for m in range(1,10**6) if not ~(4*m-1) & m] # Chai Wah Wu, Feb 07 2016

Formula

a(n) = A263133(n) + 1.

m is a term if and only if m AND NOT (4*m-1) = 0 where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 07 2016

a(n) = (2^A000267(n-1) + 2^A384688(n-1)) / 3. - Kevin Ryde, Jun 14 2025

Extensions

More terms from Vincenzo Librandi, Oct 12 2015

A263075 G.f. satisfies: [x^(n-1)] A(x)^(n^2) = n^n * (n-1)! for n>=1.

Original entry on oeis.org

1, 1, 2, 31, 1150, 68713, 5914776, 692005074, 105932315154, 20617891510063, 4984425649932314, 1467604324373250545, 517561005098562714944, 215501019188749426210440, 104642607303457024105207408, 58625315029802441203026824094, 37541542090285460025870424920666

Offset: 0

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Author

Paul D. Hanna, Oct 08 2015

Keywords

nonn

Comments

It appears that for k>0, a(k) is odd iff k = 2*A003714(n)+1 for n>=0, where A003714 is the fibbinary numbers (integers whose binary representation contains no consecutive ones); this is true for at least the first 531 terms. [See also A263190 and A171791.]

Examples

G.f.: A(x) = 1 + x + 2*x^2 + 31*x^3 + 1150*x^4 + 68713*x^5 + 5914776*x^6 + 692005074*x^7 + 105932315154*x^8 +... The coefficients in A(x)^(n^2) begin: n=1: [1, 1, 2, 31, 1150, 68713, 5914776, 692005074, 105932315154, ...]; n=2: [1, 4, 14, 152, 5021, 289824, 24532494, 2841972672, 432284291486, ...]; n=3: [1, 9, 54, 507, 13356, 715635, 58722228, 6685822296, 1005887241243, ...]; n=4: [1, 16, 152, 1536, 31500, 1468016, 114260704, 12668897920, ...]; n=5: [1, 25, 350, 4275, 75000, 2840855, 202155100, 21547156900, ...]; n=6: [1, 36, 702, 10776, 184725, 5598720, 344795598, 34598389248, ...]; n=7: [1, 49, 1274, 24647, 456386, 11753973, 592950960, 54103596918, ...]; n=8: [1, 64, 2144, 51712, 1092016, 26366656, 1071635712, 84557168640, ...]; n=9: [1, 81, 3402, 100791, 2482650, 61309629, 2096140032, 135856780686, ...]; ... where the terms along the main diagonal begin: [1, 4, 54, 1536, 75000, 5598720, 592950960, 84557168640, ..., n^n*(n-1)!, ...]. Note that odd terms a(n) occur at positions n starting with: [0, 1, 3, 5, 9, 11, 17, 19, 21, 33, 35, 37, 41, 43, 65, 67, 69, 73, 75, 81, 83, 85, 129, 131, 133, 137, 139, 145, 147, 149, 161, 163, 165, 169, 171, 257, ...], which seems to equal A118113, the even Fibbinary numbers + 1, with an initial zero included.

Links

Paul D. Hanna, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A171791, A263190, A003714, A118113, A244589.

Programs

PARI

{a(n) = local(A=[1,1]); for(i=1,n+1, A=concat(A,0); m=#A; A[m] = ( m^m*(m-1)! - Vec(Ser(A)^(m^2))[m] )/m^2 );A[n+1]} for(n=0,20,print1(a(n),", "))

Formula

a(n) ~ exp(1-exp(-1)) * n! * n^(n-1). - Vaclav Kotesovec, Oct 20 2020

A263190 G.f. satisfies: [x^(n-1)] A(x)^(n^2) = (n!)^2 for n>=1.

Original entry on oeis.org

1, 1, 0, 1, 46, 1723, 81104, 4793304, 349869074, 31080492631, 3318717525832, 421195540320465, 62871475566985208, 10927921934497456588, 2191068343727736627744, 502384409006686040020572, 130687814451798554601790746, 38294333521028379285810681487, 12557951067433973525611840784048, 4581888866092825667058378205370595

Offset: 0

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Author

Paul D. Hanna, Oct 12 2015

Keywords

nonn

Comments

CONJECTURES.

(1) Limit a(n)/(n!)^2 = 1/exp(1).

(2) There are no negative terms.

(3) ODD TERMS: It appears that for k>0, a(k) is odd iff k = 2*A003714(n)+1 for n>=0, where A003714 is the fibbinary numbers (integers whose binary representation contains no consecutive ones); this is true for at least the first 531 terms. [See also A171791 and A263075.]

Conjectures hold to at least a(1000). - Sean A. Irvine, Oct 21 2015

Examples

G.f.: A(x) = 1 + x + x^3 + 46*x^4 + 1723*x^5 + 81104*x^6 + 4793304*x^7 +... The coefficients in A(x)^(n^2) begin: n=1: [1, 1, 0, 1, 46, 1723, 81104, 4793304, 349869074, ...]; n=2: [1, 4, 6, 8, 197, 7456, 345654, 20167888, 1458010566, ...]; n=3: [1, 9, 36, 93, 612, 19197, 866208, 49440834, 3515499819, ...]; n=4: [1, 16, 120, 576, 2796, 44656, 1803872, 99433344, ...]; n=5: [1, 25, 300, 2325, 14400, 130705, 3606800, 183492150, ...]; n=6: [1, 36, 630, 7176, 61821, 518400, 8260086, 332807184, ...]; n=7: [1, 49, 1176, 18473, 216482, 2154775, 25401600, 655445812, ...]; n=8: [1, 64, 2016, 41728, 642352, 8045248, 95405312, 1625702400, ...]; n=9: [1, 81, 3240, 85401, 1673946, 26315199, 360707040, 5266837404, 131681894400, ...]; ... where the terms along the main diagonal begin: [1, 4, 36, 576, 14400, 518400, 25401600, 1625702400, ..., (n!)^2, ...]. LOCATION OF ODD TERMS. Note that odd terms a(n) occur at positions n starting with: [0, 1, 3, 5, 9, 11, 17, 19, 21, 33, 35, 37, 41, 43, 65, 67, 69, 73, 75, 81, 83, 85, 129, 131, 133, 137, 139, 145, 147, 149, 161, 163, 165, 169, 171, 257, ...], which seems to equal A118113, the even fibbinary numbers + 1, with an initial zero included.

Links

Paul D. Hanna, Table of n, a(n) for n = 0..265

Crossrefs

Cf. A263075, A294360, A171791, A003714, A118113.

Programs

PARI

{a(n) = local(A=[1, 1]); for(i=1, n+1, A=concat(A, 0); m=#A; A[m] = ( m!^2 - Vec(Ser(A)^(m^2))[m] )/m^2 ); A[n+1]} for(n=0, 20, print1(a(n), ", "))

A039744 Number of ways n*(n-1) can be partitioned into the sum of 2*(n-1) integers in the range 0..n.

Original entry on oeis.org

1, 1, 2, 5, 18, 73, 338, 1656, 8512, 45207, 246448, 1371535, 7764392, 44585180, 259140928, 1521967986, 9020077206, 53885028921, 324176252022, 1962530559999, 11947926290396, 73108804084505, 449408984811980, 2774152288318052, 17190155366056138, 106894140685782646

Offset: 0

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Author

Bill Daly (bill.daly(AT)tradition.co.uk)

Keywords

nonn

Comments

An upper bound on A007878.

The indices of the odd terms appear to be A118113. - T. D. Noe, Dec 19 2006

Links

Alois P. Heinz, Table of n, a(n) for n = 0..150 (terms n=1..65 from T. D. Noe)

Crossrefs

Cf. A007878, A118113, A362967.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(t*i n, 0, b(n-i, i, t-1)))) end: a:= n-> b(n*(n-1), n, 2*(n-1)): seq(a(n), n=0..25); # Alois P. Heinz, May 15 2016

Mathematica

T[0,p_,m_]=1; T[k_,0,m_]=0; T[k_,p_,m_]:=T[k,p,m]=Sum[T[k+i,p-1,-i], {i,-m,-1}]; Table[T[n(n-1),2n-2,n], {n,40}] (* T. D. Noe, Dec 19 2006 *)

PARI

a039744(n) = polcoef(matpascal(3*n-1, x)[3*n-1, n+1], n*(n-1)); \\ Max Alekseyev, Jun 16 2023

Sage

def a039744(n): return gaussian_binomial(3*n-2, n)[n*(n-1)] # Max Alekseyev, Jun 16 2023

Formula

a(n) = T(n(n+1),2n-2,n), where T(k,p,m) is a recursive function that gives the number of partitions of k into p parts of 0..m. It is defined T(k,p,m) = sum_{i=1..m} T(k-i,p-1,i), with the boundary conditions T(0,p,m)=1 and T(k,0,m)=0 for all positive k, p and m. - T. D. Noe, Dec 19 2006

a(n) = coefficient of q^(n*(n-1)) in q-binomial(3*n-2, n). - Max Alekseyev, Jun 16 2023

a(n) ~ 3^(3*n - 3/2) / (Pi * n^2 * 2^(2*n - 1)). - Vaclav Kotesovec, Jun 17 2023

Extensions

Definition corrected by Jozsef Pelikan (pelikan(AT)cs.elte.hu), Dec 05 2006

More terms from T. D. Noe, Dec 19 2006

a(0)=1 prepended by Alois P. Heinz, May 15 2016

A300994 G.f. A(x) satisfies: [x^n] A(x)^n / (x*A(x))' = 0 for n > 1.

Original entry on oeis.org

1, 1, 1, 2, 10, 78, 787, 9640, 138098, 2257718, 41409511, 841370530, 18753127340, 454916534472, 11932152340884, 336529124983248, 10156870523418506, 326668220673870022, 11154514687950236767, 403044704522955726502, 15364691402538733828594, 616307960867208404852222, 25948893032918425391596907

Offset: 0

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Author

Paul D. Hanna, Mar 29 2018

Keywords

nonn

Comments

Odd terms seem to occur only at positions 0, 1, and 2*A118113(k) for k >= 0.

Compare to identity: [x^n] (x*F(x))' / F(x)^(n+1) = 0 holds when F(0) = 1.

More generally, [x^n] G(x,k)^(k*(n+1)-1) / (x*G(x,k)^k)' = 0 is satisfied by an integer series G(x,k) when k is a fixed positive integer.

Examples

G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 10*x^4 + 78*x^5 + 787*x^6 + 9640*x^7 + 138098*x^8 + 2257718*x^9 + 41409511*x^10 + ... such that [x^n] A(x)^n / (x*A(x))' = 0 for n > 1. ILLUSTRATION OF DEFINITION. The table of coefficients in A(x)^n / (x*A(x))' begins: n=0: [1, -2, 1, -4, -29, -306, -3892, -57436, -961833, -17990518, ...]; n=1: [1, -1, 0, -3, -26, -279, -3594, -53588, -904770, -17034879, ...]; n=2: [1, 0, 0, -2, -21, -240, -3196, -48690, -834546, -15887984, ...]; n=3: [1, 1, 1, 0, -13, -185, -2674, -42548, -749180, -14525506, ...]; n=4: [1, 2, 3, 4, 0, -108, -1997, -34928, -646377, -12919990, ...]; n=5: [1, 3, 6, 11, 21, 0, -1124, -25545, -523467, -11040387, ...]; n=6: [1, 4, 10, 22, 54, 152, 0, -14048, -377328, -8851500, ...]; n=7: [1, 5, 15, 38, 104, 366, 1449, 0, -204288, -6313320, ...]; n=8: [1, 6, 21, 60, 177, 666, 3322, 17148, 0, -3380224, ...]; n=9: [1, 7, 28, 89, 280, 1083, 5750, 38090, 240717, 0, ...]; ... in which the main diagonal consists of all zeros after the initial terms, illustrating that [x^n] A(x)^n / (x*A(x))' = 0 for n > 1. RELATED SERIES. (x*A(x))' = 1 + 2*x + 3*x^2 + 8*x^3 + 50*x^4 + 468*x^5 + 5509*x^6 + 77120*x^7 + 1242882*x^8 + 22577180*x^9 + 455504621*x^10 + ... log(A(x)) = x + x^2/2 + 4*x^3/3 + 33*x^4/4 + 341*x^5/5 + 4252*x^6/6 + 61916*x^7/7 + 1026865*x^8/8 + 19065307*x^9/9 + 391349641*x^10/10 + ...

Links

Paul D. Hanna, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A300995, A300627, A302059, A302060.

Programs

PARI

{a(n) = my(A=[1,1]); for(i=1,n, A = concat(A,0); A[#A] = Vec( Ser(A)^(#A-1)/(x*Ser(A))' )[#A]); A[n+1]} for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) satisfies: [x^n] A(x)^n / (A(x) + x*A'(x)) = 0 for n > 1.

a(n) ~ c * 2^n * (n-1)!, where c = 0.1261880758068409567445... - Vaclav Kotesovec, Oct 20 2020

A118112 a(n) = binomial(3n,n) mod (n+1).

Original entry on oeis.org

1, 0, 0, 0, 3, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 9, 0, 0, 0, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 17, 0, 0, 0, 19, 0, 0, 0, 21, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 33, 0, 0, 0, 35, 0, 0, 0, 37, 0, 0, 0, 0, 0, 0, 0, 41, 0, 0, 0, 43, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

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Author

Labos Elemer, Apr 13 2006

Keywords

nonn

Comments

These divisibilities are analogous to those of Catalan numbers. For rather long sequences of consecutive integers, a(n)=0. For the first 10000 integers 9678 residues equals zero. See A118113.

If n+1 is in A061345, a(n)=0. This follows from Kummer's theorem. - Robert Israel, May 09 2018

Examples

For n=9, binomial(27,7) = 4686825; 4686825 mod 10 = 5.

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Wikipedia, Kummer's theorem

Crossrefs

Cf. A000108, A061345, A118113.

Programs

Maple

seq(binomial(3*n,n) mod (n+1), n=1..200); # Robert Israel, May 09 2018

Mathematica

Table[Mod[Binomial[3*k,k],k+1],{k,500}]

PARI

a(n) = binomial(3*n, n) % (n+1); \\ Michel Marcus, May 10 2018

Formula

a(n) = binomial(3n,n) mod (n+1).

Extensions

Mathematica program corrected by Harvey P. Dale, Dec 28 2012

A263133 Numbers m such that binomial(4*m + 3, m) is odd.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 10, 11, 15, 21, 23, 31, 42, 43, 47, 63, 85, 87, 95, 127, 170, 171, 175, 191, 255, 341, 343, 351, 383, 511, 682, 683, 687, 703, 767, 1023, 1365, 1367, 1375, 1407, 1535, 2047, 2730, 2731, 2735, 2751, 2815, 3071, 4095, 5461, 5463, 5471, 5503

Offset: 1

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Author

Peter Bala, Oct 11 2015

Keywords

nonn
easy

Comments

The even terms in the sequence are A020988. If m is a term in the sequence then 2*m + 1 is also a term in the sequence. Repeatedly applying the transformation m -> 2*m + 1 to the terms of A020988 produces all the terms of this sequence. See the example below.

2*a(n) gives the values of m such that binomial(4*m + 6, m) is odd.

Examples

1) This sequence can be read from Table 1 below in a sequence of 'knight moves' (2 down and 1 to the left) starting from the first two rows. For example, starting at 42 in the first row we jump 42 -> 43 -> 47 -> 63, then return to the second row at 85 and jump 85 -> 87 -> 95 -> 127, followed by 170 -> 171 -> 175 -> 191 -> 255, and so on. ........................................................... . Table 1. 2^n*ceiling((2^(2*k + 1) - 1)/3) - 1, n,k >= 0 . ........................................................... n\k| 0 1 2 3 4 5 ---+--------------------------------- 0 | 0 2 10 42 170 682 ... 1 | 1 5 21 85 341 ... 2 | 3 11 43 171 683 ... 3 | 7 23 87 343 ... 4 | 15 47 175 687 ... 5 | 31 95 351 ... 6 | 63 191 703 ... 7 | 127 383 ... 8 | 255 767 ... 9 | 511 ... ... The first row of the table is A020988. The columns of the table are obtained by repeatedly applying the transformation m -> 2*m + 1 to the entries in the first row. 2) Alternatively, this sequence can be read from Table 2 below by starting with a number on the top row and moving in a series of 'knight moves' (1 down and 2 to the left) through the table as far as you can, before returning to the next number in the top row and repeating the process. For example, starting at 10 in the first row we move 10 -> 11 -> 15, then return to the top row at 21 and move 21 -> 23 -> 31, before returning to the top row at 42 and so on. ........................................................ . Table 2. (4^n)*ceiling(2^k/3) - 1 for n >= 0, k >= 1 . ........................................................ n\k| 1 2 3 4 5 6 7 8 9 10 ---+--------------------------------------------------------- 0| 0 1 2 5 10 21 42 85 170 682... 1| 3 7 11 23 43 87 171 343 683 ... 2| 15 31 47 95 175 351 687 1375 ... 3| 63 127 191 383 703 1407 2751 5503 ... 4| 255 511 767 1535 2815 5631 11007 22015 ... 5| 1023 2047 3071 6143 11263 22527 44031 88063 ... 6| 4095 ... ... The first row of the table is A000975. The columns of the table are obtained by repeatedly applying the transformation m -> 4*m + 3 to the entries in the first row.

Crossrefs

Cf. A000975.

Other odd binomials: A263132 (4*m-1,m), A002450 (4*m+1,m), A020988 (4*m+2,m), A080674 (4*m+4,m), A118113 (3*m-2,m), A003714 (3*m,m).

Programs

Magma

[n: n in [0..6000] | Binomial(4*n+3, n) mod 2 eq 1]; // Vincenzo Librandi, Oct 12 2015

Maple

for n from 1 to 4096 do if mod(binomial(4*n+3, n), 2) = 1 then print(n) end if end do;

Mathematica

Select[Range[0,5600],OddQ[Binomial[4#+3,#]]&] (* Harvey P. Dale, Apr 15 2019 *)

PARI

for(n=0, 1e4, if (binomial(4*n+3, n) % 2 == 1, print1(n", "))) \\ Altug Alkan, Oct 11 2015

PARI

a(n) = my(r,s=sqrtint(4*n-3,&r)); (1<~~Kevin Ryde, Jul 06 2025~~

Python

A263133_list = [m for m in range(10**6) if not ~(4*m+3) & m] # Chai Wah Wu, Feb 07 2016

Formula

a(n) = A263132(n) - 1.

m is a term if and only if m AND NOT (4*m+3) = 0 where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 07 2016

a(n) = (2^A000267(n) + 2^A384688(n))/3 - 1, for n >= 1. - Kevin Ryde, Jul 06 2025

Extensions

More terms from Vincenzo Librandi, Oct 12 2015

A118114 a(n) = binomial(3n,n) mod((n+1)(n+2)).

Original entry on oeis.org

3, 3, 4, 15, 21, 28, 0, 81, 55, 99, 0, 0, 84, 120, 0, 153, 171, 285, 0, 231, 253, 0, 360, 0, 0, 0, 0, 522, 0, 496, 0, 561, 833, 945, 0, 703, 741, 156, 0, 861, 903, 1419, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2016, 1664, 2145, 2211, 3417, 0, 2415, 2485, 2556, 0

Offset: 1

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Author

Labos Elemer, Apr 13 2006

Keywords

nonn

Comments

Compared with A118112: larger nonzero value more often and in non-monotonic order.

Examples

For n=5, binomial(15,5) = 3003 = (5+1)*(5+2)*71 + 21, a(5) = 21, the residue. Interestingly, a very large zone of zeros occurs between about n=5460 and n=7800, uninterrupted by nonzero residues.

Links

Michel Marcus, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000108, A118112, A118113.

Programs

Maple

seq(binomial(3*n,n) mod((n+1)*(n+2)),n=1..71); # Emeric Deutsch, Apr 15 2006

Mathematica

Table[Mod[Binomial[3*k, k], (k + 1)*(k + 2)], {k, 1, 1000}]

PARI

a(n) = binomial(3*n,n) % ((n+1)*(n+2)); \\ Michel Marcus, Jan 05 2017

A163618 a(2*n) = 2 * a(n). a(2*n - 1) = 2 * a(n) - 2 - (-1)^n, for all n in Z.

Original entry on oeis.org

0, 1, 2, 1, 4, 1, 2, 5, 8, 1, 2, 1, 4, 9, 10, 13, 16, 1, 2, 1, 4, 1, 2, 5, 8, 17, 18, 17, 20, 25, 26, 29, 32, 1, 2, 1, 4, 1, 2, 5, 8, 1, 2, 1, 4, 9, 10, 13, 16, 33, 34, 33, 36, 33, 34, 37, 40, 49, 50, 49, 52, 57, 58, 61, 64, 1, 2, 1, 4, 1, 2, 5, 8, 1, 2, 1, 4, 9, 10, 13, 16, 1, 2, 1, 4, 1, 2, 5, 8, 17

Offset: 0

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Author

Michael Somos, Aug 01 2009

Keywords

nonn

Comments

Integers n>=0 such that a(n) = 1 is A118113.

Fibbinary numbers (A003714) give all integers n>=0 for which a(n+1) = 1 or 2. - Michael Somos, Feb 21 2016

Examples

G.f. = x + 2*x^2 + x^3 + 4*x^4 + x^5 + 2*x^6 + 5*x^7 + 8*x^8 + x^9 + 2*x^10 + ...

Links

G. C. Greubel, Table of n, a(n) for n = 0..5000

Crossrefs

Cf. A163617.

Programs

Mathematica

Table[(-1)*BitOr[-n, -2*n], {n, 0, 50}] (* G. C. Greubel, Jul 30 2017 *)

PARI

{a(n) = n=-n; -bitor(n, n<<1)};

PARI

{a(n) = if( n==0 || n==1, n, 2 * a((n+1) \ 2) - (n%2) * (2 + (-1)^((n+1) \ 2)))};

Formula

a(n) = -A163617(-n) for all n in Z.

A215124 Number of solid standard Young tableaux of shape [[(n-2)*2,2],[n-2]].

Original entry on oeis.org

0, 0, 0, 8, 174, 2084, 21025, 194064, 1694224, 14232672, 116228871, 928763000, 7294771770, 56497996620, 432520209420, 3278863236544, 24649138276800, 183964353480832, 1364323157872947, 10061883449658936, 73839952091271730, 539488089621673500

Offset: 0

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Author

Alois P. Heinz, Aug 03 2012

Keywords

nonn

Comments

a(n) is odd if and only if n = 2*k and k >= 3 and k in { A118113 }.

Links

Alois P. Heinz, Table of n, a(n) for n = 0..1000

S. B. Ekhad, D. Zeilberger, Computational and Theoretical Challenges on Counting Solid Standard Young Tableaux, arXiv:1202.6229v1 [math.CO], 2012

Wikipedia, Young tableau

Crossrefs

Column k=2 of A215122.

Programs

Maple

a:= proc(n) option remember; `if`(n<4, [0, 0, 0, 8][n+1], 3*(n-1)*(3*n-8)*(3*n-10)*(937*n-486*n^2+81*n^3-576)*a(n-1) /(2*(n-2)^2*(2*n-3)*(2152*n-729*n^2+81*n^3-2080))) end: seq(a(n), n=0..30);

Mathematica

Flatten[{0, 0, 0, 8, Table[3*(n-1) * (3*n-8) * (-576 + 937*n - 486*n^2 + 81*n^3) * (3*n-10)! / (2 * (n-4)! * (2*n-3)!), {n, 4, 20}]}] (* Vaclav Kotesovec, Sep 02 2014 *)

Formula

For n > 3, a(n) = 3*(n-1) * (3*n-8) * (-576 + 937*n - 486*n^2 + 81*n^3) * (3*n-10)! / (2 * (n-4)! * (2*n-3)!). - Vaclav Kotesovec, Sep 02 2014

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A263132 Positive values of m such that binomial(4*m - 1, m) is odd.

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A263075 G.f. satisfies: [x^(n-1)] A(x)^(n^2) = n^n * (n-1)! for n>=1.

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A263190 G.f. satisfies: [x^(n-1)] A(x)^(n^2) = (n!)^2 for n>=1.

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A039744 Number of ways n*(n-1) can be partitioned into the sum of 2*(n-1) integers in the range 0..n.

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Sage

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A300994 G.f. A(x) satisfies: [x^n] A(x)^n / (x*A(x))' = 0 for n > 1.

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A118112 a(n) = binomial(3n,n) mod (n+1).

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A263133 Numbers m such that binomial(4*m + 3, m) is odd.

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A039744 Number of ways n(n-1) can be partitioned into the sum of 2(n-1) integers in the range 0..n.

A163618 a(2n) = 2 a(n). a(2n - 1) = 2 a(n) - 2 - (-1)^n, for all n in Z.