A022998 -id:A022998 - cp's OEIS frontend

A139570 a(n) = 2n(n+3).

0, 8, 20, 36, 56, 80, 108, 140, 176, 216, 260, 308, 360, 416, 476, 540, 608, 680, 756, 836, 920, 1008, 1100, 1196, 1296, 1400, 1508, 1620, 1736, 1856, 1980, 2108, 2240, 2376, 2516, 2660, 2808, 2960, 3116, 3276, 3440, 3608, 3780, 3956, 4136, 4320, 4508, 4700, 4896

Offset: 0

Omar E. Pol, May 19 2008

Numbers n such that 2*n + 9 is a square. - Vincenzo Librandi, Nov 24 2010

a(n) appears also as the fourth member of the quartet [p0(n), p1(n), p2(n), a(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p1(n) = A046092(n), and p2(n) = A054000(n+1). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A001105, A020739, A022998, A028552, A046092, A054000, A067728, A147973.

CoefficientList[Series[4 x (2 - x)/(1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, May 23 2014 *)

a(n)=2*n*(n+3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 2*A028552(n) = 2*n^2 + 6*n = n*(2*n+6).
a(n) = a(n-1) + 4*n + 4 (with a(0)=0). - Vincenzo Librandi, Nov 24 2010
From Paul Curtz, Mar 27 2011: (Start)
a(n) = A022998(n)*A022998(n+3).
a(n) = 4*A000096(n). (End)
G.f.: 4*x*(2 - x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 31 2011
From Amiram Eldar, Dec 23 2022: (Start)
Sum_{n>=1} 1/a(n) = 11/36.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/3 - 5/36. (End)
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: 2*exp(x)*x*(4 + x).
a(n) = n*A020739(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A178415 Array T(n,k) of odd Collatz preimages read by antidiagonals.

Original entry on oeis.org

1, 3, 5, 9, 13, 21, 7, 37, 53, 85, 17, 29, 149, 213, 341, 11, 69, 117, 597, 853, 1365, 25, 45, 277, 469, 2389, 3413, 5461, 15, 101, 181, 1109, 1877, 9557, 13653, 21845, 33, 61, 405, 725, 4437, 7509, 38229, 54613, 87381, 19, 133, 245, 1621, 2901, 17749, 30037

Offset: 1

T. D. Noe, May 28 2010

Every odd number occurs uniquely in this array. See A178414.

			Array T begins:
.    1    5   21    85   341   1365    5461   21845    87381   349525
.    3   13   53   213   853   3413   13653   54613   218453   873813
.    9   37  149   597  2389   9557   38229  152917   611669  2446677
.    7   29  117   469  1877   7509   30037  120149   480597  1922389
.   17   69  277  1109  4437  17749   70997  283989  1135957  4543829
.   11   45  181   725  2901  11605   46421  185685   742741  2970965
.   25  101  405  1621  6485  25941  103765  415061  1660245  6640981
.   15   61  245   981  3925  15701   62805  251221  1004885  4019541
.   33  133  533  2133  8533  34133  136533  546133  2184533  8738133
.   19   77  309  1237  4949  19797   79189  316757  1267029  5068117
- _L. Edson Jeffery_, Mar 11 2015
From _Bob Selcoe_, Apr 09 2015 (Start):
n=5, j=13: T(5,3) = 277 = (13*4^3 - 1)/3;
n=6, j=17: T(6,4) = 725 = (17*2^7 - 1)/3.
(End)

T. D. Noe, T(n,k) for n = 1..50, by antidiagonals
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Rows of array: -A007583(k-1) (n=0), A002450 (n=1), A072197(k-1) (n=2), A206374(n=3), A072261 (n=4), A323824 (n=5), A072262 (n=6), A330246 (n=7), A072201 (n=8), ...
Columns of array: A022998(n-1)/2 (k=0), A178414 (k=1), ...
Cf. A347834 (permuted rows of the array).

t[n_,1] := t[n,1] = If[OddQ[n],4n-3,2n-1]; t[n_,k_] := t[n,k] = 4*t[n,k-1]+1; Flatten[Table[t[n-i+1,i], {n,20}, {i,n}]]

From Bob Selcoe, Apr 09 2015 (Start):
T(n,k) = 4*T(n,k-1) + 1.
T(n,k) = T(1,k) + 2^(2k+1)*(n-1)/2 when n is odd;
T(n,k) = T(2,k) + 4^k*(n-2)/2 when n >= 2 and n is even. So equivalently:
T(n,k) = T(n-2,k) + 2^(2k+1) when n is odd; and
T(n,k) = T(n-2,k) + 4^k when n is even.
Let j be the n-th positive odd number coprime with 3. Then:
T(n,k) = (j*4^k - 1)/3 when j == 1 (mod 3); and
T(n,k) = (j*2^(2k-1) - 1)/3 when j == 2 (mod 3).
(End)
From Wolfdieter Lang, Sep 18 2021: (Start)
T(n, k) = ((3*n - 1)*4^k - 2)/6 if n is even, and ((3*n - 2)*4^k - 1)/3 if n is odd, for n >= 1 and  k >= 1. Also for n = 0: -A007583(k-1), with A007583(-1) = 1/2, and for k = 0: A022998(n-1)/2, with A022998(-1) = -1.
O.g.f. for array T (with row n = 0 and column k = 0; z for rows and x for columns): G(z, x) = (1/(2*(1-x)*(1-4*x)*(1-z^2)^2)) * ((2*x-4)*z^3 + (3-5*x)*z^2 + 2*x*z + 3*x - 1). (End)

A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

Original entry on oeis.org

1, 2, 7, 5, 13, 8, 19, 11, 25, 14, 31, 17, 37, 20, 43, 23, 49, 26, 55, 29, 61, 32, 67, 35, 73, 38, 79, 41, 85, 44, 91, 47, 97, 50, 103, 53, 109, 56, 115, 59, 121, 62, 127, 65, 133, 68, 139, 71, 145, 74, 151, 77, 157, 80, 163, 83, 169, 86, 175, 89, 181, 92, 187, 95, 193, 98

Offset: 0

Paul Curtz, Sep 16 2009

Second trisection of A026741.
A111329(n+1) = A000041(a(n)). - Reinhard Zumkeller, Nov 19 2009
We observe that this sequence is a particular case of sequence for which there exists q: a(n+3) = (a(n+2)*a(n+1)+q)/a(n) for every n >= n0. Here q=-9 and n0=0. - Richard Choulet, Mar 01 2010
The entries are also encountered via the bilinear transform approximation to the natural log (unit circle). Specifically, evaluating 2(z-1)/(z+1) at z = 2, 3, 4, ..., A165355 entries stem from the pair (sums) seen 2 ahead of each new successive prime. For clarity, the evaluation output is 2, 3, 1, 1, 6, 5, 4, 3, 10, 7, 3, 2, 14, 9, 8, 5, 18, 11, ..., where (1+1), (4+3), (3+2), (8+5), ... generate the A165355 entries (after the first). As an aside, the same mechanism links A165355 to A140777. - Bill McEachen, Jan 08 2015
As a follow-up to the previous comment, it appears that the numerators and denominators of 2(z-1)/(z+1) are respectively given by A145979 and A060819, but with different offsets. - Michel Marcus, Jan 14 2015
Odd parts of the terms give A067745. E.g.: 1, 2/2, 7, 5, 13, 8/8 .... - Joe Slater, Nov 30 2016

G. C. Greubel, Table of n, a(n) for n = 0..1000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A165351, A165367, A067745, A007814.

f[n_] := If[ OddQ@ n, (3n +1)/2, (3n +1)]; Array[f, 66, 0] (* Robert G. Wilson v, Jan 26 2015 *)
f[n_] := (3 (-1)^(2n) + (-1)^(1 + n)) (-2 + 3n)/4; Array[f, 66] (* or *)
CoefficientList[ Series[(x^3 + 5x^2 + 2x + 1)/(x^2 - 1)^2, {x, 0, 65}], x] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {1, 2, 7, 5}, 66] (* Robert G. Wilson v, Apr 13 2017 *)

a(n)=n+=2*n+1; if(n%2,n,n/2) \\ Charles R Greathouse IV, Jan 13 2015

a(n) = A026741(3*n+1).
a(n)*A026741(n) = A005449(n).
a(n)*A022998(n+1) = A000567(n+1).
a(n) = A026741(n+1) + A022998(n).
a(2n) = A016921(n). a(2n+1) = A016789(n).
a(2n+1)*A026741(2n) = A045944(n).
G.f.: (1+2*x+5*x^2+x^3)/((x-1)^2 * (1+x)^2). - R. J. Mathar, Sep 26 2009
a(n) = (3+9*n)/4 + (-1)^n*(1+3*n)/4. - R. J. Mathar, Sep 26 2009
a(n) = 2*(3n+1)/(4-((2n+2) mod 4)). - Bill McEachen, Jan 09 2015
If a(2n-1) = x then a(2n) = 2x+3. - Robert G. Wilson v, Jan 26 2015
Let the reduced Collatz procedure be defined as Cr(n) = (3*n+1)/2. For odd n, a(n) = Cr(n). For even n, a(n) = Cr(4*n+1)/2. - Joe Slater, Nov 29 2016
a(n) = A067745(n+1) * 2^A007814((3n+1)/2). - Joe Slater, Nov 30 2016
a(n) = 2*a(n-2) - a(n-4). - G. C. Greubel, Apr 13 2017

All comments changed to formulas by R. J. Mathar, Sep 26 2009
New name from Charles R Greathouse IV, Jan 13 2015
Name corrected by Joe Slater, Nov 29 2016

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.

A043547 Odd numbers interspersed with double the previous odd number.

Original entry on oeis.org

1, 2, 3, 6, 5, 10, 7, 14, 9, 18, 11, 22, 13, 26, 15, 30, 17, 34, 19, 38, 21, 42, 23, 46, 25, 50, 27, 54, 29, 58, 31, 62, 33, 66, 35, 70, 37, 74, 39, 78, 41, 82, 43, 86, 45, 90, 47, 94, 49, 98, 51, 102, 53, 106, 55, 110, 57, 114, 59, 118, 61, 122, 63, 126, 65, 130, 67, 134

Offset: 1

Jim Cook (jcook(AT)halcyon.com), Mar 01 2000

As pointed out by E. Angelini on the SeqFan list (cf. link), this is the lexicographically earliest sequence of positive integers without repetitions such that the sum of four consecutive terms is always a multiple of 4. - M. F. Hasler, Mar 22 2013

			a(1)=1 because n is odd. a(2)=2 because a(1)*2=2.

E. Angelini, k-chunks sum and division by k, post to the SeqFan list, Mar 22 2013
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

For n>3, a(n) = A059029(n-3)+3. For n>1, a(n) = A022998(n-2)+2.

[n*(-1)^n/2-(-1)^n+3*n/2-1 : n in [1..50]]; // Wesley Ivan Hurt, Nov 22 2015

A043547:=n->n*(-1)^n/2-(-1)^n+3*n/2-1: seq(A043547(n), n=1..100); # Wesley Ivan Hurt, Nov 22 2015

Flatten[Table[Accumulate[{2 n - 1, 2 n - 1}], {n, 40}]] (* Wesley Ivan Hurt, Nov 22 2015 *)
With[{o=Range[1,71,2]},Riffle[o,2o]] (* Harvey P. Dale, Sep 17 2019 *)

A043547(n)=n+!bittest(n,0)*(n-2) \\ M. F. Hasler, Mar 22 2013

Vec(x*(1+2*x)*(1+x^2)/(1-x^2)^2 + O(x^100)) \\ Altug Alkan, Nov 22 2015

a(n) = (2 - n) * (n - floor(n/2) * 2) + 2 * (n - 1).
G.f.: x*(1+2*x)*(1+x^2)/(1-x^2)^2. - Ralf Stephan, Jun 10 2003
a(2n-1) = 2n-1, a(2n) = 4n-2. - M. F. Hasler, Mar 22 2013
From Wesley Ivan Hurt, Nov 22 2015: (Start)
a(n) = 2*a(n-2) - a(n-4) for n>4.
a(n) = n*(-1)^n/2 - (-1)^n + 3*n/2 - 1. (End)

More terms from James Sellers, Mar 01 2000

A059029 a(n) = n if n is even, 2*n + 1 if n is odd.

Original entry on oeis.org

0, 3, 2, 7, 4, 11, 6, 15, 8, 19, 10, 23, 12, 27, 14, 31, 16, 35, 18, 39, 20, 43, 22, 47, 24, 51, 26, 55, 28, 59, 30, 63, 32, 67, 34, 71, 36, 75, 38, 79, 40, 83, 42, 87, 44, 91, 46, 95, 48, 99, 50, 103, 52, 107, 54, 111, 56, 115, 58, 119, 60, 123, 62, 127, 64, 131, 66, 135

Offset: 0

Asher Auel, Dec 15 2000

a(n-1) = n^k - 1 mod 2*n, n >= 1, for any k >= 2, also for k = n. - Wolfdieter Lang, Dec 21 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A059026, A059030, A059031.

a(n) = A022998(n+1) - 1 = A043547(n+3) - 3. Partial sums in A032438.

[n+((n+1)/2)*(1-(-1)^n): n in [0..70]]; // Vincenzo Librandi, Aug 14 2011

B := (n,m) -> lcm(n,m)/n + lcm(n,m)/m - 1: seq(B(m+2,m),m=1..90);

Table[n +(n+1)*(1-(-1)^n)/2, {n,0,70}] (* G. C. Greubel, Nov 08 2018 *)
Table[If[EvenQ[n],n,2n+1],{n,0,70}] (* or *) LinearRecurrence[{0,2,0,-1},{0,3,2,7},70] (* Harvey P. Dale, Jul 23 2025 *)

PARI
```
a(n)=if(n%2,2*n+1,n)
```

G.f.: x*(x^2 + 2*x + 3)/(1 - x^2)^2. - Ralf Stephan, Jun 10 2003
Third main diagonal of A059026: a(n) = B(n+2, n) = lcm(n+2, n)/(n+2) + lcm(n+2, n)/n - 1 for all n >= 1.
a(2*n) + a(2*n+1) = A016945(n). - Paul Curtz, Aug 29 2008
E.g.f.: 2*x*cosh(x) + (1 + x)*sinh(x). - Franck Maminirina Ramaharo, Nov 08 2018

New description from Ralf Stephan, Jun 10 2003

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)

A385558 Period of {binomial(N,7) mod n: N in Z}.

Original entry on oeis.org

1, 8, 9, 16, 25, 72, 49, 32, 27, 200, 11, 144, 13, 392, 225, 64, 17, 216, 19, 400, 441, 88, 23, 288, 125, 104, 81, 784, 29, 1800, 31, 128, 99, 136, 1225, 432, 37, 152, 117, 800, 41, 3528, 43, 176, 675, 184, 47, 576, 343, 1000, 153, 208, 53, 648, 275, 1568, 171, 232, 59, 3600

Offset: 1

Jianing Song, Jul 03 2025

			For N == 0, 1, ..., 48 (mod 49), binomial(N,7) == {0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6} (mod 7).

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

Row n = 7 of A349593. A022998, A385555, A385556, A385557, A385559, and A385560 are respectively rows 2, 3, 4, 5-6, 8, and 9-10.

A385558[n_] := If[n == 1, 1, n*Product[p^Floor[Log[p, 7]], {p, FactorInteger[n][[All, 1]]}]];
Array[A385558, 100] (* Paolo Xausa, Jul 07 2025 *)
a[n_] := n * GCD[n, 210] * (2 - Mod[n, 2]); Array[a, 100] (* Amiram Eldar, Jul 07 2025 *)

a(n, {choices=7}) = 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), 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) * (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)) * (1 + 6/7*(s-1)).
Sum_{k=1..n} a(k) ~ (195/28) * n^2. (End)

cp's OEIS Frontend

A139570 a(n) = 2*n*(n+3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A178415 Array T(n,k) of odd Collatz preimages read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

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}.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A043547 Odd numbers interspersed with double the previous odd number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Formula

Extensions

A059029 a(n) = n if n is even, 2*n + 1 if n is odd.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

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

Views

Author

A139570 a(n) = 2n(n+3).