David Rabahy - cp's OEIS frontend

A378628 SHA-256 hash of binary strings ordered by length then lexicographically, interpreting the bits of the resulting hash as a number in binary.

102987336249554097029535212322581322789799900648198034993379397001115665086549, 85627803894273957621009139791088467240572419248799306945317937552975942820725, 84071438648566403273885734069370854758478523414599514048347910236850479717025, 91949452561369181223012808967964775320445203057864513807052896586217122700773

Offset: 1

David Rabahy, Dec 02 2024

nonn

The n-th binary string is the bits of the binary expansion of n after its most significant 1 bit, so beginning from n=1: [empty], 0, 1, 00, 01, 10, 11, 000, 001, 010, ... .
The SHA-256 hash function takes as input a vector of bits, possibly empty, and the resulting hash is 256 bits so that a(n) ranges 0 to 2^256-1 inclusive.
The length of the input bit vector is part of the hash, so initial 0 bits will, in general, result in a different hash.

			For n = 1, there are no bits after the most significant 1 bit, so a(1) is the SHA-256 hash of the empty string.
For n = 6 = 110_2, the bits after the most significant 1 are 10, so a(5) is the SHA-256 hash of the bits 10.

David Rabahy, Table of n, a(n) for n = 1..1000
David Rabahy, Python program
NIST, Secure Hash Standard (SHS), FIPS PUB 180-4.
Wikipedia, SHA-2

Cf. A007931, A076478, A365749, A366061.

Python
```
# See links
```

from sha256bit import Sha256bit
def a(n):
    s = bin(n)[3:]
    t = bytearray(int(s[i*8:i*8+8].ljust(8, "0"), 2) for i in range((len(s)+7)//8)) if n > 1 else ""
    return int(Sha256bit(t, bitlen=len(s)).hexdigest(), 16)
print([a(n) for n in range(1, 5)]) # Michael S. Branicky, Dec 08 2024

A288493 First differences of A006878 (record new trajectory lengths of Collatz function) (Hailstone sequence).

Original entry on oeis.org

1, 6, 1, 8, 3, 1, 3, 88, 1, 3, 3, 3, 3, 3, 3, 13, 1, 26, 8, 3, 1, 26, 8, 21, 24, 6, 8, 3, 3, 26, 3, 13, 16, 11, 3, 21, 8, 3, 57, 6, 21, 39, 16, 3, 3, 26, 3, 3, 21, 13, 16, 52, 21, 3, 3, 13, 1, 39, 205, 1, 3, 3, 8, 1, 21, 1, 13, 8, 42, 37, 44, 1, 21, 31, 26, 3, 6, 1, 8, 6, 8, 13, 52, 1, 13, 3, 8, 3, 13, 8, 52, 3, 26, 3, 3, 106, 1, 13, 3, 3, 16, 3, 13, 16, 21, 13, 8

Offset: 1

David Rabahy, Jun 13 2017

nonn

The sequence appears to return to 1 again and again forever when the minimal possible new record of just (previous record + 1) is reached at the latest possible value of 2X. Through 129 there are only 2 entries, a(17) and a(21), that are a minimal new record but aren't 2X.

			For n = 3 the difference between A006878(4) = 8 and A006878(3) = 7 is 1.

Hugo Pfoertner, Table of n, a(n) for n = 1..147 (from Eric Rosendaal's 3x+1 Delay Records, terms 1..129 from David Rabahy)
Eric Roosendaal, 3x+1 Delay Records

Cf. A006877, A006878, A375094.

(* This script is not suitable to compute a large number of terms. *)
terms = 40; steps[x0_] := steps[x0] = Block[{x = x0, nos = 0}, While[x != 1, If[Mod[x, 2] == 0, x = x/2, x = 3*x + 1]; nos++]; nos]; b[1] = 1; b[n_] := b[n] = Block[{x = b[n - 1] + 1}, record = steps[x - 1]; While[steps[x] <= record, x++]; x];
A006877 = Table[Print[b[n]]; b[n], {n, 1, terms+1}];
A006878 = steps /@ A006877;
Differences[A006878] (* Jean-François Alcover, Jun 15 2017 *)

A259753 For increasing z > 0, integers, y - x, where x^3 + y^3 = z^3 + 1, with y > x > 1.

Original entry on oeis.org

1, 30, 71, 100, 104, 485, 1169, 705, 887, 1727, 421, 775, 4499, 4260, 3180, 5748, 9719, 307, 6092, 18521, 20304, 18825, 32255, 6174, 33082, 7601, 29400, 17607, 13457, 52487, 1727, 44794, 41772, 76328, 1801, 29707, 80999, 119789, 111226, 132105, 122730, 171071, 123117, 237275

Offset: 1

David Rabahy, Jul 21 2015

nonn

It seems to me the sequence can never drop all the way to 1 again.
From Robert Israel, Oct 13 2015: (Start)
There are only finitely many n with a(n) = 1.
Such n correspond to solutions of the Diophantine equation x (2 x^2 + 3 x + 3) = z^3.
Since gcd(x, 2 x^2 + 3 x + 3) = 1 or 3, we get two cases:
  if x is not divisible by 3, x = s^3, z = s^3 t^3 where 2 s^6 + 3 s^3 + 3 = t^3,
  otherwise x = 9  s^3, z = 3 s t, where 54 s^6 + 9 s^3 + 1 = t^3.
The algebraic curves 2 s^6 + 3 s^3 + 3 = t^3 and 54 s^6 + 9 s^3 + 1 = t^3 both have genus 4, so by Faltings's theorem they have only finitely many rational solutions.
(End)

			10 - 9 = 1 is the first number in the sequence because 10^3 + 9^3 = 12^3 + 1^3 and no other lower z produces a result.

Robert G. Wilson v, Table of n, a(n) for n = 1..368

Cf. A001235.

Cf. x = A050792, y = A050793, z = A050791 , x^3+y^3 = A050794.

Cubes:= {seq(x^3, x=2..10^4)}:
count:= 0:
for z from 1 to 10^4 do
  s:= z^3+1;
  M:= map(t -> s-t, select(`<`,Cubes,floor(s/2))) intersect Cubes;
  for m in M do
    count:= count+1;
    y:= simplify(m^(1/3));
    x:= simplify((s-m)^(1/3));
    A[count]:= y-x;
  od
od:
seq(A[i],i=1..count); # Robert Israel, Oct 13 2015

y = 3; lst = {}; While[y < 100001, x = 2; While[x < y, z = (y^3 + x^3 - 1)^(1/3); If[IntegerQ[z], AppendTo[lst, {z, y, x, y - x}]; Print[{z, y, x, y - x}]]; x++]; y++]; Last@ Transpose@ Sort@ lst (* Robert G. Wilson v, Jul 21 2015 and modified Oct 14 2015 *)

a(n) = A050793(n) - A050792(n). - Robert G. Wilson v, Jul 21 2015

A259836 Integers n where n^3 + (n+1)^3 is a Taxicab number A001235.

Original entry on oeis.org

9, 121, 235, 301, 1090, 1293, 1524, 3152, 8010, 15556, 15934, 19247, 20244, 21498, 24015, 25363, 25556, 45462, 57872, 63758, 80016, 93349, 94701, 101929, 113098, 119942, 132414, 143653, 167147, 186540, 192629, 229508, 246122, 247318, 292154, 307534, 322870

Offset: 1

David Rabahy, Jul 06 2015

nonn

			9^3 + 10^3 = 1729 = A001235(1), so 9 is in the sequence.

David Rabahy and Alois P. Heinz and Chai Wah Wu, Table of n, a(n) for n = 1..90 (first 38 terms from David Rabahy, next 12 terms from Alois P. Heinz)

Cf. A001235, A005898.

filter:= proc(n)
  local D, b, a, Q;
  D:= numtheory:-divisors(n);
  for b in D do
    a:= n/b;
    Q:= 12*b - 3*a^2;
    if Q > 9 and issqr(Q) and Q < 9*a^2 then return true fi
  od;
  false
end proc:
select(x -> filter(x^3 +(x+1)^3), [$1..100000]); # Robert Israel, Jul 07 2015

Select[Range[10000], Length[PowersRepresentations[#^3 + (# + 1)^3, 2, 3]]==2 &] (* Vincenzo Librandi, Jul 10 2015 *)

from _future_ import division
from gmpy2 import is_square
from sympy import divisors
A259836_list = []
for n in range(10000):
    m = n**3+(n+1)**3
    for x in divisors(m):
        x2 = x**2
        if x2 > m:
            break
        if x != (2*n+1) and m < x*x2 and is_square(12*m//x-3*x2):
            A259836_list.append(n)
            break # Chai Wah Wu, Jan 10 2016

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User: David Rabahy

A378628 SHA-256 hash of binary strings ordered by length then lexicographically, interpreting the bits of the resulting hash as a number in binary.

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A288493 First differences of A006878 (record new trajectory lengths of Collatz function) (Hailstone sequence).

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A259753 For increasing z > 0, integers, y - x, where x^3 + y^3 = z^3 + 1, with y > x > 1.

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A259836 Integers n where n^3 + (n+1)^3 is a Taxicab number A001235.

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Author

Keywords

Examples

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Programs

Maple

Mathematica

Python