Shreevatsa R - cp's OEIS frontend

A357677 Powers of either 3 or 5 or 7 (and 0).

0, 1, 3, 5, 7, 9, 25, 27, 49, 81, 125, 243, 343, 625, 729, 2187, 2401, 3125, 6561, 15625, 16807, 19683, 59049, 78125, 117649, 177147, 390625, 531441, 823543, 1594323, 1953125, 4782969, 5764801, 9765625, 14348907, 40353607, 43046721, 48828125, 129140163

Offset: 1

Shreevatsa R, Oct 08 2022

Block groups in which copies of the superblock are stored, when the "sparse superblock feature" is enabled, in the ext2/ext3/ext4 filesystems. (That is, copies of the superblock are stored in every block group when the feature is disabled, or in block groups 0, 1, 3, 5, 7, 9, 25, ... when the feature is enabled.)

The Linux Kernel documentation, The Second Extended Filesystem (ext2)
The Linux Kernel documentation, ext4 Data Structures and Algorithms
Unix.StackExchange, Superblock replicas in ext4

Union of (0 and) A000244, A000351, A000420.

With[{max = 1.3*10^8}, Sort @ Flatten @ Join[{0, 1}, Table[p^Range[1, Floor[Log[p, max]]], {p, {3, 5, 7}}]]] (* Amiram Eldar, Oct 09 2022 *)

A347291 Multiplicative function defined by a(p) = 2 and a(p^k) = p^(k-1) for k >= 2.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 4, 3, 4, 2, 4, 2, 4, 4, 8, 2, 6, 2, 4, 4, 4, 2, 8, 5, 4, 9, 4, 2, 8, 2, 16, 4, 4, 4, 6, 2, 4, 4, 8, 2, 8, 2, 4, 6, 4, 2, 16, 7, 10, 4, 4, 2, 18, 4, 8, 4, 4, 2, 8, 2, 4, 6, 32, 4, 8, 2, 4, 4, 8, 2, 12, 2, 4, 10, 4, 4, 8, 2, 16, 27, 4, 2, 8, 4

Offset: 1

Shreevatsa R, Jan 22 2022

a(n) is the least number of distinct values mod n attained by a polynomial p(x) such that p(x) == 0 (mod n) if and only if x == 0 (mod n).

			For n = 8, a(8) = a(2^3) = 2^2 = 4. Also, any polynomial p(x) that is 0 only for x == 0 (mod 8) takes at least a(8)=4 distinct values. (The polynomial p(x) = x^3 + x is an example.)
When n is a prime number, the polynomial p(x) = x^(n-1), by Fermat's little theorem, is 0 (mod n) when x == 0 (mod n), and 1 (mod n) otherwise. So it takes only two distinct values, and a(p) = 2.
a(360) = a(2^3 * 3^2 * 5) = 2^2 * 3 * 2 = 24.

Math StackExchange, Smallest number of residue classes for a polynomial function mod n that separates 0.

f[p_, e_] := If[e == 1, 2, p^(e - 1)]; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Jan 23 2022 *)

a(n) = { my(f = factor(n)); for(i = 1, #f~, if(f[i, 2] == 1, f[i, 1] = 2; , f[i, 2]--; ) ); factorback(f) } \\ David A. Corneth, Jan 22 2022

from sympy import factorint, prod
def a(n):
    return prod((2 if k == 1 else p**(k-1)) for (p, k) in factorint(n).items())

If n = p1^k1 p2^k2 ... pr^kr, then a(n) = a(p1^k1) a(p2^k2) ... a(pr^kr), where a(p^k) is 2 if k=1 and p^(k-1) if k>=2.

Dirichlet g.f.: zeta(s-1) * Product_{p prime} (1 - 1/p^(s-1) + 2/p^s - 1/p^(2*s-1)). - Amiram Eldar, Oct 01 2023

Corrected and extended by David A. Corneth, Jan 22 2022

A323425 Number of ways n people in a line can each choose two others both on the same side of them.

Original entry on oeis.org

1, 0, 0, 0, 9, 648, 57600, 6615000, 972504225, 179499220992, 40789783609344, 11212877910528000, 3671848787797265625, 1413385410212064432000, 632129969391038455873536, 325176984737061807515098752, 190691488202627199302740850625, 126479088749202444199526400000000

Offset: 0

Shreevatsa R, Jan 14 2019

nonn

			Example: For n = 4, with four people ABCD, A can choose any two of {B, C, D} (3 choices), B can choose {C, D} (1 choice), C can choose {A, B} (1 choice), and D can choose any two of {A, B, C} (3 choices), so there are 3*1*1*3=9 possible overall choices and a(4) = 9.
Example: For n = 5, with five people ABCDE, A can choose any two of {B, C, D, E} (6 choices), B can choose any two of {C, D, E} (3 choices), C can choose either {A, B} or {D, E} (2 choices), D can choose any two of {A, B, C} (3 choices), E can choose any two of {A, B, C, D} (6 choices), so a(5) = 6*3*2*3*6 = 648.

a n = product [(k-1)*(k-2) `div` 2 + (n-k)*(n-k-1) `div` 2 | k<-[1..n]]

Table[Product[Binomial[k-1, 2] + Binomial[n-k, 2], {k, 1, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jan 15 2019 *)

a(n) = Product_{k = 1..n} ( binomial(k-1, 2) + binomial(n-k, 2) ).

a(n) ~ exp(Pi*(n/2 - 1) - 2*n) * n^(2*n) / 2^n. - Vaclav Kotesovec, Jan 15 2019

More terms from Vaclav Kotesovec, Jan 15 2019

A260249 Month numbers in the Gregorian calendar, in alphabetical order of their English names.

Original entry on oeis.org

4, 8, 12, 2, 1, 7, 6, 3, 5, 11, 10, 9

Offset: 1

Shreevatsa R, Jul 20 2015

			In alphabetical order, the months of the Gregorian calendar are: April (4), August (8), December (12), February (2), January (1), July (7), June (6), March (3), May (5), November (11), October (10), September (9).

Wikipedia, Gregorian calendar

Cf. A033986.

A227480 Triangular numbers of the form p^2 - 1 where p is a prime.

Original entry on oeis.org

3, 120, 528, 139128, 160554240, 812293020528, 379188080252621270252095320, 2816640597719456759751165488439681098364673273236637971046370174350128

Offset: 1

Shreevatsa R, Jul 12 2013

nonn

			139128 is both 527*528/2 and 373^2 - 1 (and 373 is prime).

Subsequence of A006454 and of A225115.

Select[Prime[Range[100000]]^2-1,OddQ[Sqrt[8#+1]]&] (* The program generates the first six terms of the sequence. *) (* Harvey P. Dale, Sep 17 2023 *)

a(7)-a(8) from Giovanni Resta, Jul 13 2013

A161169 Triangle read by rows: T(n,k) = number of permutations of {1..n} with at most k inversions.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 5, 6, 1, 4, 9, 15, 20, 23, 24, 1, 5, 14, 29, 49, 71, 91, 106, 115, 119, 120, 1, 6, 20, 49, 98, 169, 259, 360, 461, 551, 622, 671, 700, 714, 719, 720, 1, 7, 27, 76, 174, 343, 602, 961, 1416, 1947, 2520, 3093, 3624, 4079, 4438, 4697, 4866, 4964, 5013

Offset: 0

Shreevatsa R, Jun 04 2009

T(n,k) is also the number of permutations with Kendall tau distance ("bubble-sort distance") to the identity permutation being at most k. This is the number of swaps performed by the bubble-sort algorithm to sort the sequence.

The above only gives T(n,k) for k<=n(n-1)/2, but T(n,k)=n! for all k>=n(n-1)/2.

			Triangle begins:
  1;
  1;
  1, 2;
  1, 3,  5,  6;
  1, 4,  9, 15, 20,  23,  24;
  1, 5, 14, 29, 49,  71,  91, 106, 115, 119, 120;
  1, 6, 20, 49, 98, 169, 259, 360, 461, 551, 622, 671, 700, 714, 719, 720;
  ...
T(3,2)=5 because there are 5 permutations of {1,2,3} with at most 2 inversions: (1,2,3) with 0 inversions, (1,3,2), (2,1,3) with 1 inversion each, (2,3,1), (3,1,2) with 2 inversions each.
T(n,0)=1 because there is exactly 1 permutation (the identity permutation) with no inversions,
T(n,k) = n! for all k >= n(n-1)/2 because all permutations have at most n(n-1)/2 inversions.

Alois P. Heinz, Rows n = 0..50, flattened
D. Wang, A. Mazumdar and G. W. Wornell, A Rate-Distortion Theory for Permutation Spaces, 2013.

Partial sums of A008302.

ct = {(0,0): 1}
def c(n,k):
    if k<0: return 0
    k = min(k, n*(n-1)/2)
    if (n, k) in ct: return ct[(n, k)]
    ct[(n, k)] = c(n, k-1) + c(n-1, k) - c(n-1, k-n)
    return ct[(n, k)]

T(n,k) = Sum_{i s.t. n-i<=k} T(n-1, k-(n-i));
T(n,k) = Sum_{i=max(1,n-k)..n} T(n-1, k-n+i);
T(n,k) = Sum_{j=max(k-n+1,0)..n} T(n-1, j).
T(n,k) = T(n,k-1) + T(n-1,k) - T(n-1,k-n), taking T(n,k)=0 for k<0.
Also, T(n,k) = n! - T(n, n(n-1)/2-k-1).
For k<=n, T(n,k) = A008302(n+1,k).
G.f.: (1/(1-x))*Product_{j=1..n} (1-x^j)/(1-x) = Sum_{k>=0} T(n,k) x^k. - Alejandro H. Morales, Apr 04 2024

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User: Shreevatsa R

A357677 Powers of either 3 or 5 or 7 (and 0).

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A347291 Multiplicative function defined by a(p) = 2 and a(p^k) = p^(k-1) for k >= 2.

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A323425 Number of ways n people in a line can each choose two others both on the same side of them.

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A260249 Month numbers in the Gregorian calendar, in alphabetical order of their English names.

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A227480 Triangular numbers of the form p^2 - 1 where p is a prime.

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A161169 Triangle read by rows: T(n,k) = number of permutations of {1..n} with at most k inversions.

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