Darío Clavijo - cp's OEIS frontend

A385029 a(n) = Sum_{-n <= a, b, c <= n} (b^2 - 4ac).

18, 250, 1372, 4860, 13310, 30758, 63000, 117912, 205770, 339570, 535348, 812500, 1194102, 1707230, 2383280, 3258288, 4373250, 5774442, 7513740, 9648940, 12244078, 15369750, 19103432, 23529800, 28741050, 34837218, 41926500, 50125572, 59559910, 70364110, 82682208, 96668000

Offset: 1

Darío Clavijo, Jun 15 2025

There are (2*n + 1)^3 combinations of a, b, c.

Paolo Xausa, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000384, A002378, A014105, A016755, A016754, A055112, A384666.

A385029[n_] := (n*(n + 1)*(2*n + 1)^3)/3;
Array[A385029, 50] (* Paolo Xausa, Jun 18 2025 *)

a = lambda n: ((n*n+n)*((n << 1)+1)**3)//3
print([a(n) for n in range(1, 11)])

a(n) = (n*(n+1)*(2*n+1)^3)/3.
a(n) = (A055112(n)*A016754(n))/3.
a(n) = (A002378(n)*A016755(n))/3.
G.f.: 2*x*(9 + 71*x + 71*x^2 + 9*x^3)/(1 - x)^6. - Stefano Spezia, Jun 15 2025
From Amiram Eldar, Jun 18 2025; (Start)
Sum_{n>=1} 1/a(n) = 21*(1 - zeta(3)/2) - 12*log(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*Pi^3/8 + 3*Pi - 21. (End)

A384716 The totient of the product of unitary divisors of n.

Original entry on oeis.org

1, 1, 2, 2, 4, 12, 6, 4, 6, 40, 10, 48, 12, 84, 120, 8, 16, 108, 18, 160, 252, 220, 22, 192, 20, 312, 18, 336, 28, 216000, 30, 16, 660, 544, 840, 432, 36, 684, 936, 640, 40, 889056, 42, 880, 1080, 1012, 46, 768, 42, 1000, 1632, 1248, 52, 972, 2200, 1344, 2052

Offset: 1

Darío Clavijo, Jun 11 2025

nonn

a(n) is the totient of the product over all unitary divisors d of n; i.e., those divisors satisfying gcd(d, n/d) = 1.
Growth rate of a(n) is ~ n^O(log log n).
Also, a(n) is lower bounded by A000010(n).

Cf. A000010, A000225, A006093, A061537, A384763.

f[p_, e_, m_] := (p-1)*p^(e*m-1); a[n_] := Module[{fct = FactorInteger[n]}, Times @@ (f[#1, #2, 2^(Length[fct]-1)] & @@@ fct)]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Jun 11 2025 *)

from sympy import factorint
def a(n):
    if n == 1: return 1
    factors = factorint(n)
    phi, w = 1, len(factors)
    for p, e in factors.items():
        phi *= (p - 1) * p**(e - 1)
    return n**((1 << (w-1)) - 1) * phi
print([a(n) for n in range(1, 58)])

a(n) = phi(Product_{d|n} d if gcd(d, n/d) = 1).
a(n) = n^(2^(omega(n)-1)-1) * phi(n).
a(n) = A000010(A061537(n)).
a(p) = p-1 for p prime.

A384763 Product of the Euler totients of the unitary divisors of n.

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 6, 4, 6, 16, 10, 16, 12, 36, 64, 8, 16, 36, 18, 64, 144, 100, 22, 64, 20, 144, 18, 144, 28, 4096, 30, 16, 400, 256, 576, 144, 36, 324, 576, 256, 40, 20736, 42, 400, 576, 484, 46, 256, 42, 400, 1024, 576, 52, 324, 1600, 576, 1296, 784, 58, 65536

Offset: 1

Darío Clavijo, Jun 09 2025

nonn

a(n) is the product of phi(d) over all unitary divisors d of n; i.e., those divisors satisfying gcd(d, n/d) = 1.
a(n) is upper bounded by A061537(n) (product of phi(d) over all divisors d of n).
The function is not multiplicative.
The sum of the totients over all unitary divisors d of n is A055653(n).

			For n = 6, a(6) = phi(1) * phi(2) * phi(3) * phi(6) = 1*1*2*2 = 4.

Cf. A000010, A034444, A055653, A061537, A077610.

a[n_] := EulerPhi[n]^(2^(PrimeNu[n] - 1)); Array[a, 100] (* Amiram Eldar, Jun 09 2025 *)

a(n) = my(p=1); fordiv(n, d, if (gcd(d,n/d) == 1, p*=eulerphi(d))); p; \\ Michel Marcus, Jun 09 2025

from sympy import totient, divisors, gcd
def a(n):
   prod = 1
   for d in divisors(n):
      if gcd(d, n//d) == 1:
          prod *= totient(d)
   return prod
print([a(n) for n in range(1, 61)])

a(n) = Product_{d|n} phi(d) if gcd(n,floor(n/d)) = 1.
a(p) = p-1 for p prime.
a(p^k) = p^k-p^(k-1).
a(n) = phi(n)^(2^(omega(n)-1)) = A000010(n)^(A034444(n)/2). - Amiram Eldar, Jun 09 2025

A384666 Number of distinct values of the quadratic discriminant D=b^2-4ac, for a,b,c in the range [-n,n].

Original entry on oeis.org

1, 6, 17, 35, 56, 90, 125, 178, 223, 282, 344, 436, 499, 608, 701, 804, 904, 1062, 1164, 1339, 1450, 1604, 1765, 1988, 2114, 2335, 2525, 2735, 2909, 3194, 3366, 3679, 3887, 4137, 4389, 4661, 4840, 5237, 5536, 5835, 6068, 6507, 6759, 7195, 7473, 7773, 8148, 8645

Offset: 0

Darío Clavijo, Jun 06 2025

nonn

a(n) is lower bounded by n log n for n > 0.

The number of distinct a*c is 2*A027384(n)-1.

Cf. A000217, A027384.

def a(n):
    D, ac = {0}, {0}
    SQ = [i*i for i in range(0, n+1)]
    for i in range(1, n+1):
        ac.add(i)
        if (s:= SQ[i]) > n:
            ac.add(s)
    for a_ in range(2, n):
        for c in range(a_+ 1, n+1):
            ac.add(a_* c)
    for b in range(n + 1):
        b2 = SQ[b]
        for v in ac:
            ac4 = v << 2
            D.add(b2 + ac4)
            if b2 < ac4:
                D.add(b2 - ac4)
    return len(D)
print([a(n) for n in range(0, 48)])

A384543 Number of distinct values from the bitwise operation i XOR j for all integers i and j in the range [1, n].

Original entry on oeis.org

1, 2, 4, 7, 8, 8, 8, 15, 16, 16, 16, 16, 16, 16, 16, 31, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 63, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 127, 128, 128

Offset: 1

Darío Clavijo, Jun 02 2025

For any n, the maximum value of (i XOR j) is < 2^floor(log_2(n))+1.

			For n=3, a(3) = 4 because:
 i | j | i XOR j
---+---+-------
 1 | 1 | 0
 1 | 2 | 3
 1 | 3 | 2
 2 | 1 | 3
 2 | 2 | 0
 2 | 3 | 1
 3 | 1 | 2
 3 | 2 | 1
 3 | 3 | 0
In total there are 4 unique values for i XOR j.

Cf. A000079, A000225, A070939, A151821.

a[n_] := CountDistinct[Flatten[Table[BitXor[i, j], {i, 1, n}, {j, 1, i}]]]; Array[a, 100] (* Amiram Eldar, Jun 02 2025 *)

a(n) = #setbinop((x,y)->bitxor(x,y), [1..n]); \\ Michel Marcus, Jun 02 2025

def a(n):
    if n < 4: return [1,1,2,4][n]
    k2 = 1 << n.bit_length()
    if (n & (n - 1)) == 0: return k2 - 1
    return k2
print([a(n) for n in range(1, 68)])

a(2^k) = A000225(k+1) for k > 1.
a(2^k-1) = A151821(k).
a(2^k+1) = A000079(k+1).
a(n) = 2^k if 2^(k-1) < n < 2^k with k=2^floor(log_2(n))+1.

A384452 a(n) is the sum of squares of the unitary divisors of n!.

Original entry on oeis.org

1, 5, 50, 650, 16900, 547924, 27396200, 1746641000, 139773881000, 13460683752200, 1642203417768400, 236441876606410000, 40195119023089700000, 7723888546922636420000, 1735183690969722609168800, 444206919394766468845892000, 128820006624482275965308680000, 41737604550102658693597600532800

Offset: 1

Darío Clavijo, Jun 02 2025

nonn

Project Euler, Problem 429: Sum of Squares of Unitary Divisors.
Wikipedia, Legendre's formula.

Cf. A000142, A034676, A064028, A077610.

f[p_, e_] := p^(2*e)+1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 18] (* Amiram Eldar, Jun 02 2025 *)

row(n) = {my(d = divisors(n)); select(x->(gcd(x, n/x)==1), d); } \\ A077610
a(n) = norml2(row(n!)); \\ Michel Marcus, Jun 02 2025

from sympy import nextprime
def f(n,p):
  if n==0: return 0
  return f(n//p,p) + n//p
def a(n):
  s,p = 1, 2
  while p<=n:
    s *= p**(f(n,p)<<1)+1
    p = nextprime(p)
  return s
print([a(n) for n in range(1, 19)])

a(n) = Sum_{d|n!} (d^2 if gcd(d,n!//d) = 1).
a(n) = Product_{p <= n, p prime} (p^(2*f(n,p)))+1 with f(n,p) = f(floor(n/p)) + floor(n/p) and f(0,p) = 0 where f(n,p) is equivalent to the Legendre formula.
a(n) = A034676(n!).

A383752 Product of nonzero remainders n mod p, over all primes p < n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 6, 8, 3, 8, 10, 36, 24, 8, 30, 288, 420, 1920, 2268, 640, 270, 2880, 9240, 13824, 7560, 19200, 17820, 120960, 64064, 362880, 5054400, 1881600, 475200, 165888, 464100, 6386688, 4082400, 1228800, 2120580, 34836480, 23474880, 217728000

Offset: 1

Darío Clavijo, May 28 2025

nonn

Paolo Xausa, Table of n, a(n) for n = 1..1000

Cf. A000040, A024934, A013939, A102647, A309912.

A383752[n_] := Times @@ DeleteCases[Mod[n, Prime[Range[PrimePi[n - 2]]]], 0];
Array[A383752, 50] (* Paolo Xausa, Jun 05 2025 *)

a(n) = vecprod(select(x->(x!=0), apply(lift, apply(x->Mod(n, x), primes([2,n-1]))))); \\ Michel Marcus, May 28 2025

from sympy import primerange
def a(n):
    s = 1
    for p in primerange(0, n):
        if p > (n >> 1): s *= (n-p)
        elif (x:= n % p) > 0: s*= x
    return s
print([a(n) for n in range(1,41)])

a(p) = A102647(p) if p prime.

A382454 Number of solutions winning the Tchoukaillon game with n seeds and 2n pits.

Original entry on oeis.org

1, 2, 9, 49, 285, 1717, 10569, 66013, 416687, 2651355, 16976806, 109256095, 706071989, 4579020513, 29784426945, 194231327451, 1269457354069, 8313189986612, 54534379879411, 358298017624625, 2357331709694072, 15528887031395023, 102412421113465576, 676104332189192702

Offset: 0

Darío Clavijo, May 26 2025

nonn

a(n) is the number of permutations of [2n+1] with n inversions. a(2) = 9: 12453, 12534, 13254, 13425, 14235, 21354, 21435, 23145, 31245. - Alois P. Heinz, May 27 2025

Mancala World, Tchoukaillon

Cf. A000707, A008302.

a:= n-> coeff(series(mul((1-q^j)/(1-q), j=1..2*n+1), q, n+1), q, n):
seq(a(n), n=0..23);  # Alois P. Heinz, May 27 2025

a[n_]:=Coefficient[Series[Product[(1-q^j)/(1-q),{j,1,2*n+1}],{q,0,n+1}]//Normal,q,n];Array[a, 24, 0] (* Shenghui Yang, Jun 02 2025 *)

def a(n):
    if n == 0: return 1
    p = [1]
    for j in range(1, (n << 1) + 2):
        np = [0] * (len(p) + j - 1)
        for k in range(len(p)):
            for l in range(j):
                if (kl:=k+l) <= n:
                    np[kl] += p[k]
        p = np[:n+1]
    return p[n]
print([a(n) for n in range(1,24)])

a(n) = T(2n,n) with T(x,y) = Sum_{v=0..min(x,y)} T(x-1, y-v) and T(0,y) = 1 if y = 0 else 0.

a(n) = A008302(2n+1,n).

A384197 The Barret reducer reciprocal mod n.

Original entry on oeis.org

4, 8, 5, 16, 12, 10, 9, 32, 28, 25, 23, 21, 19, 18, 17, 64, 60, 56, 53, 51, 48, 46, 44, 42, 40, 39, 37, 36, 35, 34, 33, 128, 124, 120, 117, 113, 110, 107, 105, 102, 99, 97, 95, 93, 91, 89, 87, 85, 83, 81, 80, 78, 77, 75, 74, 73, 71, 70, 69, 68, 67, 66, 65, 256

Offset: 1

Darío Clavijo, May 21 2025

The aim of the Barret reducer is to compute x_i mod n for multiple i in an efficient manner in commodity hardware.
The reciprocal is a(n) = floor(4^k / n), with k = floor(log_2(n))+1 such that for any x_i where q_i = (x_i * a(n)) >> (2 * k) and x_i - q_i * k is congruent to x_i mod n.
This sequence contains no fixed points.

			For n = 97, a(97) = 168 because: k = floor(log_2(97))+1 = 6, so  a(97) = floor((4^6) / 97) = 168.
As an example application, let some x=65537 and q = (x * a(n)) >> (2 * 6) = (65537 * 168) >> (2 * 6) = 353, then 353 % 97 = 65537 % 97, where ">>" denotes right shift.

Paolo Xausa, Table of n, a(n) for n = 1..8191
Wikipedia, Barret reduction.

Cf. A070939, A143096.

A384197[n_] := Floor[4^BitLength[n]/n]; Array[A384197, 100] (* Paolo Xausa, Jun 05 2025 *)

a = lambda n: (1 << (n.bit_length() << 1)) // n
print([a(n) for n in range(1,65)])

a(n) = floor(4^k / n) where k = floor(log_2(n))+1.
a(2^j) = 2^(j+2).
a(2^j-1) = A143096(j+1).

A383036 The determinant of the matrix representing a totally anti-symmetric quasigroup of order 2*n+1.

Original entry on oeis.org

0, 9, 1250, 352947, 172186884, 129687123005, 139788510734886, 204350482177734375, 389289535005334947848, 937146152681201173795569, 2782184294469515486371964010, 9986310782535957929474146174619, 42632564145606011152267456054687500, 213501642487388555901009081409220318757

Offset: 0

Darío Clavijo, May 21 2025

A totally antisymmetric quasigroup of order 2*n+1 is constructed in a way such that M[i][j] != M[j][i] for i!=j with m = 2*n+1, k = 2 and M[j][i] = k*(j-i) mod m for 0 <= j,i < m.
For any k != 0 mod m the resulting matrix M has the same determinant for each n.
Also the resulting matrix M is circulant and a Latin square.

			For n = 1, a(1) = 9 because:
The resulting totally anti-symetric quasigroup has a matrix:
with k = 1:
  0, 1, 2,
  2, 0, 1,
  1, 2, 0
which has a determinant: 9.
with k = 2:
  0, 2, 1,
  1, 0, 2,
  2, 1, 0
has also the same determinant 9.

Paolo Xausa, Table of n, a(n) for n = 0..100
H. Michael Damm, Totally anti-symmetric quasigroups for all orders n not equal to 2 or 6, Discrete Math., 307:6 (2007), 715-729.
Wikipedia, Quasigroups

Cf. A005408, A081131, A375584.

A383036[n_] := n*(2*n+1)^(2*n); Array[A383036, 15, 0] (* Paolo Xausa, May 28 2025 *)

a(n) = n*(2*n+1)^(2*n) = A081131(2*n+1).

cp's OEIS Frontend

User: Darío Clavijo

A385029 a(n) = Sum_{-n <= a, b, c <= n} (b^2 - 4*a*c).

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A384716 The totient of the product of unitary divisors of n.

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A384763 Product of the Euler totients of the unitary divisors of n.

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A384666 Number of distinct values of the quadratic discriminant D=b^2-4*a*c, for a,b,c in the range [-n,n].

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A384543 Number of distinct values from the bitwise operation i XOR j for all integers i and j in the range [1, n].

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A384452 a(n) is the sum of squares of the unitary divisors of n!.

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A383752 Product of nonzero remainders n mod p, over all primes p < n.

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A382454 Number of solutions winning the Tchoukaillon game with n seeds and 2n pits.

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A385029 a(n) = Sum_{-n <= a, b, c <= n} (b^2 - 4ac).

A384666 Number of distinct values of the quadratic discriminant D=b^2-4ac, for a,b,c in the range [-n,n].