A385558 -id:A385558 - cp's OEIS frontend

A022998 If n is odd then n, otherwise 2n.

0, 1, 4, 3, 8, 5, 12, 7, 16, 9, 20, 11, 24, 13, 28, 15, 32, 17, 36, 19, 40, 21, 44, 23, 48, 25, 52, 27, 56, 29, 60, 31, 64, 33, 68, 35, 72, 37, 76, 39, 80, 41, 84, 43, 88, 45, 92, 47, 96, 49, 100, 51, 104, 53, 108, 55, 112, 57, 116, 59, 120, 61, 124, 63, 128, 65, 132, 67

Offset: 0

N. J. A. Sloane

Also for n > 0: numerator of Sum_{i=1..n} 2/(i*(i+1)), denominator=A026741. - Reinhard Zumkeller, Jul 25 2002
For n > 2: a(n) = gcd(A143051((n-1)^2), A143051(1+(n-1)^2)) = A050873(A000290(n-1), A002522(n-1)). - Reinhard Zumkeller, Jul 20 2008
Partial sums give the generalized octagonal numbers A001082. - Omar E. Pol, Sep 10 2011
Multiples of 4 and odd numbers interleaved. - Omar E. Pol, Sep 25 2011
The Pisano period lengths modulo m appear to be A066043(m). - R. J. Mathar, Oct 08 2011
The partial sums a(n)/A026741(n+1) given by R. Zumkeller in a comment above are 2*n/(n+1) (telescopic sum), and thus converge to 2. - Wolfdieter Lang, Apr 09 2013
a(n) = numerator(H(n,1)), where H(n,1) = 2*n/(n+1) is the harmonic mean of 1 and n. a(n) = 2*n/gcd(2n, n+1) = 2*n/gcd(n+1,2). a(n) = A227041(n,1), n>=1. - Wolfdieter Lang, Jul 04 2013
a(n) = numerator of the mean (2n/(n+1), after reduction), of the compositions of n; denominator is given by A001792(n-1). - Clark Kimberling, Mar 11 2014
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n,m)) for all natural numbers n and m. The sequence of convergents of the 2-periodic continued fraction [0; 1, -4, 1, -4, ...] = 1/(1 - 1/(4 - 1/(1 - 1/(4 - ...)))) begins [0/1, 1/1, 4/3, 3/2, 8/5, 5/3, 12/7, ...]. The present sequence is the sequence of numerators. The sequence of denominators of the continued fraction convergents [1, 1, 3, 2, 5, 3, 7, ...] is A026741, also a strong divisibility sequence. Cf. A203976. - Peter Bala, May 19 2014
a(n) is also the length of the n-th line segment of a rectangular spiral on the infinite square grid. The vertices of the spiral are the generalized octagonal numbers. - Omar E. Pol, Jul 27 2018
a(n) is the number of petals of the Rhodonea curve r = a*cos(n*theta) or r = a*sin(n*theta). - Matt Westwood, Nov 19 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Rose
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A026741, A059026, A203976, A345082.
Column 4 of A195151. - Omar E. Pol, Sep 25 2011
Cf. A000034, A001082 (partial sums).
Cf. A227041 (first column). - Wolfdieter Lang, Jul 04 2013
Row 2 of A349593. A385555, A385556, A385557, A385558, A385559, and A385560 are respectively rows 3, 4, 5-6, 7, 8, and 9-10.

a022998 n = a000034 (n + 1) * n
a022998_list = zipWith (*) [0..] $ tail a000034_list
-- Reinhard Zumkeller, Mar 31 2012

[((-1)^n+3)*n/2: n in [0..70]]; // Vincenzo Librandi, Sep 17 2011

A022998 := proc(n) if type(n,'odd') then n ; else 2*n; end if; end proc: # R. J. Mathar, Mar 10 2011

Table[n (3 + (-1)^n)/2, {n, 0, 100}] (* Wesley Ivan Hurt, Dec 13 2013 *)
Table[If[OddQ[n],n,2n],{n,0,150}] (* or *) Riffle[ 2*Range[ 0,150,2], Range[ 1,150,2]] (* Harvey P. Dale, Feb 06 2017 *)

PARI
```
a(n)=if(n%2,n,2*n)
```

def A022998(n): return n if n&1 else n<<1 # Chai Wah Wu, Mar 05 2024

[n*(1+((n+1)%2)) for n in (0..80)] # G. C. Greubel, Jul 31 2022

Denominator of (n+1)*(n-1)*(2*n+1)/(2*n) (for n > 0).
a(n+1) = lcm(n, n+2)/n + lcm(n, n+2)/(n+2) for all n >= 1. - Asher Auel, Dec 15 2000
Multiplicative with a(2^e) = 2^(e+1), a(p^e) = p^e, p > 2.
G.f. x*(1 + 4*x + x^2)/(1-x^2)^2. - Ralf Stephan, Jun 10 2003
a(n) = 3*n/2 + n*(-1)^n/2 = n*(3 + (-1)^n)/2. - Paul Barry, Sep 04 2003
a(n) = A059029(n-1) + 1 = A043547(n+2) - 2.
a(n)*a(n+3) = -4 + a(n+1)*a(n+2).
a(n) = n*(((n+1) mod 2) + 1) = n^2 + 2*n - 2*n*floor((n+1)/2). - William A. Tedeschi, Feb 29 2008
a(n) = denominator((n+1)/(2*n)) for n >= 1; A026741(n+1) = numerator((n+1)/(2*n)) for n >= 1. - Johannes W. Meijer, Jun 18 2009
a(n) = 2*a(n-2) - a(n-4).
Dirichlet g.f. zeta(s-1)*(1+2^(1-s)). - R. J. Mathar, Mar 10 2011
a(n) = n * (2 - n mod 2) = n * A000034(n+1). - Reinhard Zumkeller, Mar 31 2012
a(n) = floor(2*n/(1 + (n mod 2))). - Wesley Ivan Hurt, Dec 13 2013
From Ilya Gutkovskiy, Mar 16 2017: (Start)
E.g.f.: x*(2*sinh(x) + cosh(x)).
It appears that a(n) is the period of the sequence k*(k + 1)/2 mod n. (End) [This is correct; see A349593. - Jianing Song, Jul 03 2025]
a(n) = Sum_{d | n} A345082(d). - Peter Bala, Jan 13 2024

More terms from Michael Somos, Aug 07 2000

A349593 Square array read by downward diagonals: for n >= 0, k >= 1, T(n,k) is the period of {binomial(N,n) mod k: N in Z}.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 3, 4, 1, 1, 5, 8, 9, 8, 1, 1, 6, 5, 8, 9, 8, 1, 1, 7, 12, 5, 16, 9, 8, 1, 1, 8, 7, 36, 5, 16, 9, 8, 1, 1, 9, 16, 7, 72, 25, 16, 9, 16, 1, 1, 10, 9, 16, 7, 72, 25, 16, 9, 16, 1, 1, 11, 20, 27, 32, 7, 72, 25, 32, 27, 16, 1

Offset: 0

Jianing Song, Nov 27 2021

Since binomial(N,n) is defined for all integers N, there is no need to assume that N >= n.
Let Q(N) = 1 if k | binomial(N,n), 0 otherwise. Then T(n,k) is also the period of {Q(N): N in Z}.
By the formula given below, the n-th row is identical to the (n-1)th row if and only if n is not a power of a prime, i.e., n is in A024619. - Jianing Song, Jul 03 2025

			Rows 0..10:
  1,  1,  1,  1,  1,   1,  1,  1,  1,   1, ...
  1,  2,  3,  4,  5,   6,  7,  8,  9,  10, ...
  1,  4,  3,  8,  5,  12,  7, 16,  9,  20, ...
  1,  4,  9,  8,  5,  36,  7, 16, 27,  20, ...
  1,  8,  9, 16,  5,  72,  7, 32, 27,  40, ...
  1,  8,  9, 16, 25,  72,  7, 32, 27, 200, ...
  1,  8,  9, 16, 25,  72,  7, 32, 27, 200, ...
  1,  8,  9, 16, 25,  72, 49, 32, 27, 200, ...
  1, 16,  9, 32, 25, 144, 49, 64, 27, 400, ...
  1, 16, 27, 32, 25, 432, 49, 64, 81, 400, ...
  1, 16, 27, 32, 25, 432, 49, 64, 81, 400, ...
Example showing that T(4,4) = 16: for N == 0, 1, ..., 15 (mod 16), binomial(N,4) == {0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1} (mod 4).
Example showing that T(3,10) = 20: for N == 0, 1, ..., 19 (mod 20), binomial(N,3) == {0, 0, 0, 1, 4, 0, 0, 5, 6, 4, 0, 5, 0, 6, 4, 5, 0, 0, 6, 9} (mod 10).

Jianing Song, Table of n, a(n) for antidiagonals 1..100 (T(n,k) occurs at the ((n+k)*(n+k-1)/2+n)-th place)
Andrew Granville, Arithmetic Properties of Binomial Coefficients I: Binomial Coefficients modulo prime powers
Jianing Song, Proof for my formula for A349593
Jianing Song, A more simple proof of the formula for A349593
Wikipedia, Kummer's_theorem

Cf. A022998 (row n = 2), A385555 (row n = 3), A385556 (row n = 4), A385557 (rows n = 5 and 6), A385558 (row n = 7), A385559 (row n = 8), A385560 (rows n = 9 and 10).
Cf. A062383 (2nd column), A064235 (3rd column if offset 0), A385552 (5th column), A385553 (6th column), A385554 (10th column).
Cf. A349221.

A349593[n_, k_] := If[n == 0 || k == 1, 1, k*Product[p^Floor[Log[p, n]], {p, FactorInteger[k][[All, 1]]}]];
Table[A349593[k - 1, n - k + 2], {n, 0, 15}, {k, n + 1}] (* Paolo Xausa, Jul 07 2025 *)

T(n,k) = if(n==0, 1, my(r=1, f=factor(k)); for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); r *= p^(logint(n,p)+e)); return(r))

The n-th row is multiplicative with T(n,p^e) = 1 if n = 0, p^(e+floor(log(n)/log(p))) otherwise. In other words, for n > 0, T(n,k) = k * Product_{prime p|k} p^(floor(log(n)/log(p))). See my pdf file for a proof.

A385555 Period of {binomial(N,3) mod n: N in Z}.

Original entry on oeis.org

1, 4, 9, 8, 5, 36, 7, 16, 27, 20, 11, 72, 13, 28, 45, 32, 17, 108, 19, 40, 63, 44, 23, 144, 25, 52, 81, 56, 29, 180, 31, 64, 99, 68, 35, 216, 37, 76, 117, 80, 41, 252, 43, 88, 135, 92, 47, 288, 49, 100, 153, 104, 53, 324, 55, 112, 171, 116, 59, 360

Offset: 1

Jianing Song, Jul 03 2025

			For N == 0, 1, ..., 26 (mod 27), binomial(N,3) == {0, 0, 0, 1, 4, 1, 2, 8, 2, 3, 3, 3, 4, 7, 4, 5, 2, 5, 6, 6, 6, 7, 1, 7, 8, 5, 8} (mod 4).

Jianing Song, Table of n, a(n) for n = 1..10000

Row n = 3 of A349593. A022998, A385556, A385557, A385558, A385559, and A385560 are respectively rows n = 2, 4, 5-6, 7, 8, and 9-10.

Cf. A089128.

A385555[n_] := If[n == 1, 1, n*Product[p^Floor[Log[p, 3]], {p, FactorInteger[n][[All, 1]]}]];
Array[A385555, 100] (* Paolo Xausa, Jul 07 2025 *)
a[n_] := n * GCD[n, 6]; Array[a, 100] (* Amiram Eldar, Jul 07 2025 *)

a(n, {choices=3}) = my(r=1, f=factor(n)); for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); r *= p^(logint(choices, p)+e)); return(r)

Multiplicative with a(2^e) = 2^(e+1), a(3^e) = 3^(e+1), and a(p^e) = p^e for primes p >= 5.
From Amiram Eldar, Jul 07 2025: (Start)
a(n) = n * gcd(6, n) = n * A089128(n).
Dirichlet g.f.: zeta(s-1) * (1 + 1/2*(s-1)) * (1 + 2/3*(s-1)).
Sum_{k=1..n} a(k) ~ (5/4) * n^2. (End)

A385556 Period of {binomial(N,4) mod n: N in Z}.

Original entry on oeis.org

1, 8, 9, 16, 5, 72, 7, 32, 27, 40, 11, 144, 13, 56, 45, 64, 17, 216, 19, 80, 63, 88, 23, 288, 25, 104, 81, 112, 29, 360, 31, 128, 99, 136, 35, 432, 37, 152, 117, 160, 41, 504, 43, 176, 135, 184, 47, 576, 49, 200, 153, 208, 53, 648, 55, 224, 171, 232, 59, 720

Offset: 1

Jianing Song, Jul 03 2025

			For N == 0, 1, ..., 31 (mod 32), binomial(N,4) == {0, 0, 0, 0, 1, 5, 7, 3, 6, 6, 2, 2, 7, 3, 1, 5, 4, 4, 4, 4, 5, 1, 3, 7, 2, 2, 6, 6, 3, 7, 5, 1} (mod 8).

Jianing Song, Table of n, a(n) for n = 1..10000

Row n = 4 of A349593. A022998, A385555, A385557, A385558, A385559, and A385560 are respectively rows n = 2, 3, 5-6, 7, 8, and 9-10.

Cf. A000034, A089128.

A385556[n_] := If[n == 1, 1, n*Product[p^Floor[Log[p, 4]], {p, FactorInteger[n][[All, 1]]}]];
Array[A385556, 100] (* Paolo Xausa, Jul 07 2025 *)
a[n_] := n * GCD[n, 6] * (2 - Mod[n, 2]); Array[a, 100] (* Amiram Eldar, Jul 07 2025 *)

a(n, {choices=4}) = my(r=1, f=factor(n)); for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); r *= p^(logint(choices, p)+e)); return(r)

Multiplicative with a(2^e) = 2^(e+2), a(3^e) = 3^(e+1), and a(p^e) = p^e for primes p >= 5.
From Amiram Eldar, Jul 07 2025: (Start)
a(n) = n * gcd(6, n) * (2 - (n mod 2)) = n * A089128(n) * A000034(n-1).
Dirichlet g.f.: zeta(s-1) * (1 + 3/2*(s-1)) * (1 + 2/3*(s-1)).
Sum_{k=1..n} a(k) ~ (25/12) * n^2. (End)

A385557 Period of {binomial(N,5) mod n: N in Z}. Also, period of {binomial(N,6) mod n: N in Z}.

Original entry on oeis.org

1, 8, 9, 16, 25, 72, 7, 32, 27, 200, 11, 144, 13, 56, 225, 64, 17, 216, 19, 400, 63, 88, 23, 288, 125, 104, 81, 112, 29, 1800, 31, 128, 99, 136, 175, 432, 37, 152, 117, 800, 41, 504, 43, 176, 675, 184, 47, 576, 49, 1000, 153, 208, 53, 648, 275, 224, 171, 232, 59, 3600

Offset: 1

Jianing Song, Jul 03 2025

			For N == 0, 1, ..., 71 (mod 72), binomial(N,5) == {0, 0, 0, 0, 0, 1, 0, 3, 2, 0, 0, 0, 0, 3, 4, 3, 0, 2, 0, 0, 0, 3, 0, 1, 0, 0, 2, 0, 0, 3, 0, 3, 4, 0, 0, 2, 0, 3, 0, 3, 0, 4, 0, 0, 2, 3, 0, 3, 0, 0, 4, 0, 0, 5, 0, 3, 0, 0, 0, 4, 0, 3, 2, 3, 0, 0, 0, 0, 4, 3, 0, 5} (mod 6), and binomial(N,6) == {0, 0, 0, 0, 0, 0, 1, 1, 4, 0, 0, 0, 0, 0, 3, 1, 4, 4, 0, 0, 0, 0, 3, 3, 4, 4, 4, 0, 0, 0, 3, 3, 0, 4, 4, 4, 0, 0, 3, 3, 0, 0, 4, 4, 4, 0, 3, 3, 0, 0, 0, 4, 4, 4, 3, 3, 0, 0, 0, 0, 4, 4, 1, 3, 0, 0, 0, 0, 0, 4, 1, 1} (mod 6).

Jianing Song, Table of n, a(n) for n = 1..10000

Rows n = 5 and 6 of A349593. A022998, A385555, A385556, A385558, A385559, and A385560 are respectively rows 2, 3, 4, 7, 8, and 9-10.

A385557[n_] := If[n == 1, 1, n*Product[p^Floor[Log[p, 5]], {p, FactorInteger[n][[All, 1]]}]];
Array[A385557, 100] (* Paolo Xausa, Jul 07 2025 *)
a[n_] := n * GCD[n, 30] * (2 - Mod[n, 2]); Array[a, 100] (* Amiram Eldar, Jul 07 2025 *)

a(n, {choices=5}) = my(r=1, f=factor(n)); for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); r *= p^(logint(choices, p)+e)); return(r)

Multiplicative with a(2^e) = 2^(e+2), a(3^e) = 3^(e+1), a(5^e) = 5^(e+1), and a(p^e) = p^e for primes p >= 7.
From Amiram Eldar, Jul 07 2025: (Start)
a(n) = n * gcd(30, n) * (2 - (n mod 2)).
Dirichlet g.f.: zeta(s-1) * (1 + 3/2*(s-1)) * (1 + 2/3*(s-1)) * (1 + 4/5*(s-1)).
Sum_{k=1..n} a(k) ~ (15/4) * n^2. (End)

A385559 Period of {binomial(N,8) mod n: N in Z}.

Original entry on oeis.org

1, 16, 9, 32, 25, 144, 49, 64, 27, 400, 11, 288, 13, 784, 225, 128, 17, 432, 19, 800, 441, 176, 23, 576, 125, 208, 81, 1568, 29, 3600, 31, 256, 99, 272, 1225, 864, 37, 304, 117, 1600, 41, 7056, 43, 352, 675, 368, 47, 1152, 343, 2000, 153, 416, 53, 1296, 275, 3136, 171, 464, 59, 7200

Offset: 1

Jianing Song, Jul 03 2025

			For N == 0, 1, ..., 63 (mod 64), binomial(N,8) == {0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 5, 7, 7, 3, 3, 6, 6, 6, 6, 2, 2, 2, 2, 7, 7, 3, 3, 1, 1, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 1, 1, 3, 3, 7, 7, 2, 2, 2, 2, 6, 6, 6, 6, 3, 3, 7, 7, 5, 5, 1, 1} (mod 8).

Jianing Song, Table of n, a(n) for n = 1..10000

Row n = 8 of A349593. A022998, A385555, A385556, A385557, A385558, and A385560 are respectively rows 2, 3, 4, 5-6, 7, and 9-10.

A385559[n_] := If[n == 1, 1, n*Product[p^Floor[Log[p, 8]], {p, FactorInteger[n][[All, 1]]}]];
Array[A385559, 100] (* Paolo Xausa, Jul 07 2025 *)
a[n_] := n * GCD[n, 210] * (4 - 3 * Mod[n, 2]); Array[a, 100] (* Amiram Eldar, Jul 07 2025 *)

a(n, {choices=8}) = my(r=1, f=factor(n)); for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); r *= p^(logint(choices, p)+e)); return(r)

Multiplicative with a(2^e) = 2^(e+3), a(3^e) = 3^(e+1), a(5^e) = 5^(e+1), a(7^e) = 7^(e+1), and a(p^e) = p^e for primes p >= 11.
From Amiram Eldar, Jul 07 2025: (Start)
a(n) = n * gcd(210, n) * (4 - 3 * (n mod 2)).
Dirichlet g.f.: zeta(s-1) * (1 + 7/2*(s-1)) * (1 + 2/3*(s-1)) * (1 + 4/5*(s-1)) * (1 + 6/7*(s-1)).
Sum_{k=1..n} a(k) ~ (351/28) * n^2. (End)

A385560 Period of {binomial(N,9) mod n: N in Z}. Also, period of {binomial(N,10) mod n: N in Z}.

Original entry on oeis.org

1, 16, 27, 32, 25, 432, 49, 64, 81, 400, 11, 864, 13, 784, 675, 128, 17, 1296, 19, 800, 1323, 176, 23, 1728, 125, 208, 243, 1568, 29, 10800, 31, 256, 297, 272, 1225, 2592, 37, 304, 351, 1600, 41, 21168, 43, 352, 2025, 368, 47, 3456, 343, 2000, 459, 416, 53, 3888, 275, 3136, 513, 464, 59, 21600

Offset: 1

Jianing Song, Jul 03 2025

			For N == 0, 1, ..., 80 (mod 81), binomial(N,9) == {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 4, 4, 4, 1, 1, 1, 2, 2, 2, 8, 8, 8, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 7, 7, 7, 4, 4, 4, 5, 5, 5, 2, 2, 2, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 1, 1, 1, 7, 7, 7, 8, 8, 8, 5, 5, 5, 8, 8, 8} (mod 9), and binomial(N,10) == {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 7, 2, 6, 7, 8, 0, 2, 4, 6, 5, 4, 3, 5, 7, 0, 3, 6, 0, 3, 6, 0, 3, 6, 0, 4, 8, 3, 1, 8, 6, 1, 5, 0, 5, 1, 6, 8, 1, 3, 8, 4, 0, 6, 3, 0, 6, 3, 0, 6, 3, 0, 7, 5, 3, 4, 5, 6, 4, 2, 0, 8, 7, 6, 2, 7, 3, 2, 1} (mod 9).

Jianing Song, Table of n, a(n) for n = 1..10000

Rows n = 9 and 10 of A349593. A022998, A385555, A385556, A385557, A385558, and A385559 are respectively rows 2, 3, 4, 5-6, 7, and 8.

A385560[n_] := If[n == 1, 1, n*Product[p^Floor[Log[p, 9]], {p, FactorInteger[n][[All, 1]]}]];
Array[A385560, 100] (* Paolo Xausa, Jul 07 2025 *)
a[n_] := n * GCD[n, 6] * GCD[n, 210] * (2 - Mod[n, 2]); Array[a, 100] (* Amiram Eldar, Jul 07 2025 *)

a(n, {choices=10}) = my(r=1, f=factor(n)); for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); r *= p^(logint(choices, p)+e)); return(r)

Multiplicative with a(2^e) = 2^(e+3), a(3^e) = 3^(e+2), a(5^e) = 5^(e+1), a(7^e) = 7^(e+1), and a(p^e) = p^e for primes p >= 11.
From Amiram Eldar, Jul 07 2025: (Start)
a(n) = n * gcd(6, n) * gcd(210, n) * (2 - (n mod 2)).
Dirichlet g.f.: zeta(s-1) * (1 + 7/2*(s-1)) * (1 + 8/3*(s-1)) * (1 + 4/5*(s-1)) * (1 + 6/7*(s-1)).
Sum_{k=1..n} a(k) ~ (3861/140) * n^2. (End)

cp's OEIS Frontend

A022998 If n is odd then n, otherwise 2n.

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A349593 Square array read by downward diagonals: for n >= 0, k >= 1, T(n,k) is the period of {binomial(N,n) mod k: N in Z}.

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A385555 Period of {binomial(N,3) mod n: N in Z}.

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A385556 Period of {binomial(N,4) mod n: N in Z}.

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A385557 Period of {binomial(N,5) mod n: N in Z}. Also, period of {binomial(N,6) mod n: N in Z}.

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A385559 Period of {binomial(N,8) mod n: N in Z}.

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A385560 Period of {binomial(N,9) mod n: N in Z}. Also, period of {binomial(N,10) mod n: N in Z}.

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