Shreyansh Jaiswal - cp's OEIS frontend

A385881 Algebraic rank of elliptic curve y^2 = x^3 - n*x - n.

0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 2, 1, 1, 0, 0, 0, 0, 0, 1, 2, 1, 1, 2, 2, 0, 2, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 2, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 2, 0, 1, 0, 1, 1, 1, 0, 1, 2, 1

Offset: 1

Shreyansh Jaiswal, Aug 20 2025

Terms from n = 43 onward are the analytic ranks (see PARI code) of the corresponding elliptic curves. By the BSD conjecture, these are expected to equal the algebraic ranks. Thus, the validity of these terms is conditional on BSD.

			a(1) = 0 because y^2 = x^3 - x - 1 has rank 0.

LMFDB, y^2 = x^3 - x - 1.

Cf. A060951, A060952.

a(n) = ellanalyticrank( ellinit([0, 0, 0, -n, -n]) )[1]; \\ Michel Marcus, Aug 20 2025

for k in range(1, 43):
    E = EllipticCurve([-k, -k])
    print(E.rank(), end=", ")

More terms from Michel Marcus, Aug 20 2025

A386928 Algebraic rank of elliptic curve y^2 = x^3 + n*x + n.

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 2, 0, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 2, 1, 2, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 2, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 2, 2, 1, 0, 1, 2, 1, 1, 1, 2, 0

Offset: 1

Shreyansh Jaiswal, Aug 08 2025

Terms from n = 29 onward are the analytic ranks (see PARI code) of the corresponding elliptic curves. By the BSD conjecture, these are expected to equal the algebraic ranks. Thus, the validity of these terms is conditional on BSD.

			a(1) = 1 because y^2 = x^3 + x + 1 has rank 1.

LMFDB, y^2 = x^3 + x + 1.

Cf. A060950, A060953.

a(n) = ellanalyticrank(ellinit([n, n]))[1]; \\ Jinyuan Wang, Aug 08 2025

for k in range(1,29):
    E = EllipticCurve([k,k])
    print(E.rank(),end=", ")

More terms from Jinyuan Wang, Aug 08 2025

A383649 Numbers k such that A206369(k) is prime.

Original entry on oeis.org

3, 4, 6, 8, 9, 16, 18, 49, 64, 81, 98, 162, 169, 338, 625, 729, 1024, 1250, 1458, 4096, 4489, 6241, 8978, 12482, 14641, 19321, 22801, 26569, 29282, 37249, 38642, 45602, 53138, 65536, 74498, 113569, 121801, 143641, 208849, 227138, 243602, 262144, 287282, 292681, 375769, 413449, 417698

Offset: 1

Shreyansh Jaiswal, May 04 2025

nonn

The corresponding primes are 2, 3, 2, 5, 7, 11, 7, 43, 43, 61, 43, 61, 157, 157, 521... Sorting and removing duplicates from this sequence of primes appears to give A127727.

All the terms are either prime powers or twice prime powers since A206369 is multiplicative and A206369(n) = 1 only for n = 1 and 2. 3 is the only prime term since A206369(p) = p-1 for a prime p. - Amiram Eldar, May 04 2025

			3 is a term since A206369(3) = 2, and 2 is a prime.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..122 from Shreyansh Jaiswal)

Cf. A127727, A206369.

seq[max_] := Module[{s = {3, 6}, e}, Do[e = Select[Range[2, Floor[Log[p, max]]], PrimeQ[Sum[(-1)^(# - k)*p^k, {k, 0, #}]] &]; s = Join[s, p^e]; If[p > 2, e = Select[e, # <= Floor[Log[p, max/2]] &]; s = Join[s, 2*p^e]], {p, Prime[Range[Floor[Sqrt[max]]]]}]; Sort[s]]; seq[500000] (* Amiram Eldar, May 04 2025 *)

isok(k) = ispseudoprime(sumdiv(k, d, eulerphi(k/d) * issquare(d))); \\ Michel Marcus, May 04 2025

from math import prod; from sympy import *
def ok(n): return isprime(prod((lambda x:x[0]+int((x[1]<<1)>=p+1))(divmod(p**(e+1), p+1)) for p, e in factorint(n).items())) # using code from Chai Wah Wu in A206369
print([k for k in range(1,10**5) if ok(k)])

A382545 a(n) = A071324(n) - A000010(n).

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 1, 1, 2, 0, 4, 0, 2, 4, 3, 0, 7, 0, 4, 4, 2, 0, 8, 1, 2, 2, 6, 0, 14, 0, 5, 4, 2, 8, 13, 0, 2, 4, 8, 0, 20, 0, 10, 12, 2, 0, 16, 1, 11, 4, 12, 0, 22, 8, 14, 4, 2, 0, 24, 0, 2, 10, 11, 8, 28, 0, 16, 4, 18, 0, 25, 0, 2, 22, 18, 12, 32, 0, 16, 7, 2, 0, 30, 8, 2, 4, 20, 0, 44, 12, 22, 4, 2, 8, 32, 0, 15, 10, 23

Offset: 1

Shreyansh Jaiswal, Apr 23 2025

nonn

a(n) >= 0, as A071324(n) >= A000010(n) for all n.

			a(100) = A071324(100) - A000010(100) = 63 - 40 = 23.

Shreyansh Jaiswal, Table of n, a(n) for n = 1..10000

Cf. A000010, A071324.

a[n_] := Plus @@ (-(d = Divisors[n])*(-1)^(Range[Length[d], 1, -1])) - EulerPhi[n]; Array[a, 100] (* Amiram Eldar, Apr 23 2025 *)

a(n) = my(f=factor(n), d=Vecrev(divisors(f))); sum(k=1, #d, (-1)^(k+1)*d[k]) - eulerphi(f); \\ Michel Marcus, Apr 23 2025

from sympy import divisors;from functools import lru_cache; from sympy import totient
cached_divisors = lru_cache()(divisors)
def c(n): return sum(d if i%2==0 else -d for i, d in enumerate(reversed(cached_divisors(n))))
for n in range(1, 101): print((c(n)-totient(n)),end=", ")

a(p) = 0 for prime p, as A071324(p) = p-1 = A000010(p).

A382837 Numbers k such that k - A071324(k) > A000010(k).

Original entry on oeis.org

60, 70, 84, 120, 140, 154, 168, 180, 200, 210, 220, 240, 252, 260, 264, 280, 286, 300, 312, 336, 340, 350, 360, 374, 390, 396, 408, 418, 420, 442, 456, 468, 480, 490, 494, 504, 510, 520, 528, 540, 560, 570, 588, 598, 600, 624, 630, 646, 660, 672, 680, 700

Offset: 1

Shreyansh Jaiswal, Apr 06 2025

nonn

This sequence lists the counterexamples to Atanassov's conjecture that n - A071324(n) <= A000010(n) for all n. (See Atanassov link, page 2).
Every positive integer multiple of 210 (4th primorial) is a term within the sequence. This can be proved by a slight modification of the proof of Theorem 2 (see Jaiswal link). - Ivan N. Ianakiev, Apr 13 2025
It appears that odd terms within the sequence are extremely uncommon.
It appears that this sequence probably has a nonzero natural density. Computations of the ratio f(n)/n, where f(n) is the number of terms <=n (see links) up to n = 10^6, suggest that the natural density of this sequence (if it exists) is around 0.08. Since every multiple of 210 is a term within the sequence, this value, if it exists, is >= 1/210, which is approximately 0.0047.
Conjecture: For each even natural number n, infinitely many pairs (k,k+n) exist, such that both k and k+n are terms in the sequence. (Example - n = 876 has pairs (84, 960), (180, 1056), (264, 1140), (300, 1176)...)

			60 - A071324(60) = 60 - 40 = 20 > A000010(60) = 16.

Shreyansh Jaiswal, Table of n, a(n) for n = 1..10000
Krassimir Atanassov, A remark on an arithmetic function. Part 2, Notes on Number Theory and Discrete Mathematics 15(3) (2009), 21-22.
Shreyansh Jaiswal, Logarithmic plot of the ratio f(n)/n, upto n = 10^6
Shreyansh Jaiswal, Plot of the ratio f(n)/n, from 10^5 <= n <= 10^6 [An enlarged version of the previous graph.]
Shreyansh Jaiswal, On two Conjectures by Atanassov on the Alternating sum-of-divisors function, Jun 14 2025.

Cf. A000010, A071324.

f[n_]:=Plus@@(-(d=Divisors[n])*(-1)^(Range[Length[d],1,-1]));
fQ[n_]:=n-f[n]>EulerPhi[n]; Select[Range[700], fQ[#]&] (* Ivan N. Ianakiev, Apr 13 2025, thanks to Amiram Eldar at A071324 *)

isok(k) = my(d=Vecrev(divisors(k))); k - sum(i=1, #d, (-1)^(i+1)*d[i]) > eulerphi(k); \\ Michel Marcus, Apr 12 2025

from sympy import divisors;  from functools import lru_cache
from sympy import totient
cached_divisors = lru_cache()(divisors)
def c(n):  return sum(d if i%2==0 else -d for i, d in enumerate(reversed(cached_divisors(n))))
for n in range(1, 701):
    if n - c(n) > totient(n):
        print(n, end=", ")

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User: Shreyansh Jaiswal

A385881 Algebraic rank of elliptic curve y^2 = x^3 - n*x - n.

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A386928 Algebraic rank of elliptic curve y^2 = x^3 + n*x + n.

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A383649 Numbers k such that A206369(k) is prime.

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A382545 a(n) = A071324(n) - A000010(n).

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A382837 Numbers k such that k - A071324(k) > A000010(k).

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