This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
(* A377090list is defined at A377090 *) Differences[A377090list[100]] (* Paolo Xausa, May 02 2025 *)
The initial terms are: n a(n) |a(n)-a(n-1)| -- ---- ------------- 0 0 N/A 1 1 1^2 2 2 1^2 3 -2 2^2 4 -1 1^2 5 3 2^2 6 4 1^2 7 5 1^2 8 -4 3^2 9 -3 1^2 10 6 3^2 11 7 1^2 12 8 1^2 13 -8 4^2 14 -7 1^2
A377091 = [0]; A377091.least_unused = 1; function a(n){ for(let i = A377091.length-1; i < n; ++i) { let k = A377091.least_unused; while(!Number.isInteger(Math.sqrt(Math.abs(A377091[i] - k))) || A377091.indexOf(k) > 0) k = (k<0)-k; A377091.push(k); if (k == A377091.least_unused) { do k = (k<0)-k; while ( A377091.indexOf( k ) > 0 ); A377091.least_unused = k; } }; return A377091[n]; } // M. F. Hasler, Jan 26 2025
h := proc(b, a, i) option remember; ifelse(issqr(abs(a[-1] - i)) and not is(i in a), ifelse(b < nops(a) + 1, a, h(b, [op(a), i], 1)), h(b, a, ifelse(i < 0, 1 - i, -i))) end: a_list := length -> h(length, [0], 1): a_list(62); # Peter Luschny, Jan 20 2025
A377091list[nmax_] := Module[{s, a, u = 1}, s[_] := False; s[0] = True; NestList[(While[s[u] && s[-u], u++]; a = u; While[s[a] || !IntegerQ[Sqrt[Abs[# - a]]], a = Boole[a < 0] - a]; s[a] = True; a) &, 0,nmax]]; A377091list[100] (* Paolo Xausa, Mar 18 2025 *)
\\ See Links section.
A377091_upto(n,S=[])={vector(n+1, k, S=setunion(S, [n=if(k>1, k=1; while(setsearch(S,k) || !issquare(abs(n-k)), k=(k<0)-k); k)]); n)} \\ M. F. Hasler, Jan 18 2025
from math import isqrt
from itertools import count, islice
def cond(n): return isqrt(n)**2 == n
def agen(): # generator of terms
an, aset, m = 0, {0}, 1
for n in count(0):
yield an
an = next(s for k in count(m) for s in [k, -k] if s not in aset and cond(abs(an-s)))
aset.add(an)
while m in aset and -m in aset: m += 1
print(list(islice(agen(), 62))) # Michael S. Branicky, Dec 25 2024
from math import sqrt
def a_list(b: int, a: list[int] = [0], i: int = 1) -> list[int]:
if sqrt(abs(a[-1] - i)).is_integer() and not (i in a):
a += [i]
if b < len(a):
return a
else:
return a_list(b, a)
else:
return a_list(b, a, int(i < 0) - i)
print(a_list(40)) # Peter Luschny, Jan 20 2025
class A377091: # A377091(n) gives a(n) terms = [0]; N = 1 # next candidate def _new_(A, n): A.extend(A, n-len(A.terms)+1); return A.terms[n] def extend(A, n): any((k:=A.N) in A.terms and setattr(A, 'N', k:=(k<0)-k) or A.terms.append(next(k for _ in range(9**9) if (abs(A.terms[-1]-k)**.5) .is_integer() and k not in A.terms or not(k:=(k<0)-k))) for _ in range(n)) # M. F. Hasler, Feb 08 2025
The first terms are: n a(n) |a(n)-a(n-1)| --- ----- --------------- 0 0 N/A 1 1 1 2 -1 2 3 2 3 4 3 1 5 -2 5 6 -3 1 7 -4 1 8 4 8 9 5 1 10 6 1 11 7 1 12 -6 13 13 -5 1 14 -7 2 The first terms are a(0) = 0, a(1) = 1 and a(2) = -1, clearly the |smallest| unused number so far, which yields |a(2)-a(1)| = 2, a Fibonacci number. - _M. F. Hasler_, Feb 21 2025
A377092list[nmax_] := Module[{s, a, u = 1, fibQ}, fibQ[n_] := fibQ[n] = (IntegerQ[Sqrt[# + 4]] || IntegerQ[Sqrt[# - 4]]) & [5*n^2]; s[_] := False; s[0] = True; NestList[(While[s[u] && s[-u], u++]; a = u; While[s[a] || !fibQ[Abs[# - a]], a = Boole[a < 0] - a]; s[a] = True; a) &, 0,nmax]]; A377092list[100] (* Paolo Xausa, Apr 19 2025 *)
\\ See Links section.
A377092_upto(N, U=[-1])={vector(N, n, if(n>1, for(k=U[1]+1,oo, A010056(k-N) && !setsearch(U, k) && [N=k, break]), N=0); U=setunion(U,[N]); while(#U>1&&U[1]+1==U[2],U=U[^1]); N)} \\ M. F. Hasler, Feb 21 2025
def A377092(n): if not getattr(A := A377092, 'N', 0): A.N = 1; A.terms = [0] while len(A.terms) <= n: while (k := A.N) in A.terms: A.N = (k<0)-k while not A010056(abs(k - A.terms[-1])) or k in A.terms: k = (k<0)-k A.terms.append(k) return A.terms[n] # M. F. Hasler, Feb 10 2025
(3 + sqrt(2) + sqrt(3))^3 = 14 + 6*sqrt(2) + 6*sqrt(3) + 2*sqrt(6), so a(3) = 6.
(* Program 1 generates sequences A377113-A377116. *) tbl = Table[Expand[(3 + Sqrt[2] + Sqrt[3])^n], {n, 0, 24}]; u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &, Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]]; {s1,s2,s3,s4}=Transpose[(PadRight[#1,4]&)/@Last/@u][[1;;4]]; s2 (* Peter J. C. Moses, Oct 16 2024 *) (* Program 2 generates this sequence. *) LinearRecurrence[{12, -44, 48, 8}, {0, 1, 6, 38}, 25]
(3 + sqrt(2) + sqrt(3))^3 = 14 + 6*sqrt(2) + 6*sqrt(3) + 2*sqrt(6), so a(3) = 6.
(* Program 1 generates sequences A377113-A377116. *) tbl = Table[Expand[(3 + Sqrt[2] + Sqrt[3])^n], {n, 0, 24}]; u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &, Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]]; {s1,s2,s3,s4}=Transpose[(PadRight[#1,4]&)/@Last/@u][[1;;4]]; s3 (* Peter J. C. Moses, Oct 16 2024 *) (* Program 2 generates this sequence. *) LinearRecurrence[{12, -44, 48, 8}, {0, 1, 6, 36}, 25]
Comments