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.
a(n)=1+sum(k=1,n-1,ceil(exp(1)*k)) \\ Benoit Cloitre, Jan 24 2009
Quiet[Block[{$ContextPath}, Needs["Combinatorica`"]], {General::compat}]
theta = 1 / E;
sums = {0};
cached = <||>;
A023124[i_] := Module[{term, path, base},
While[sums[[-1]] < i,
term = sums[[-1]] + Floor[theta * (Length[sums] - 1)] + 1;
AppendTo[sums, term];
cached[term] = 1
];
path = {i};
While[Not[KeyExistsQ[cached, path[[-1]]]],
AppendTo[path, path[[-1]] - Combinatorica`BinarySearch[sums, path[[-1]]] + 3/2];
];
base = cached[path[[-1]]];
MapIndexed[(cached[#1] = base + Length[path] - First[#2]) &, path];
cached[i]
];
Print[Table[A023124[i], {i, 1, 100}]]; (* Brady J. Garvin, Sep 13 2024 *)
from bisect import bisect
from sympy import floor, E
theta = 1 / E
sums = [0]
cached = {}
def A023124(i):
while sums[-1] < i:
term = sums[-1] + floor(theta * (len(sums) - 1)) + 1
sums.append(term)
cached[term] = 1
path = [i]
while path[-1] not in cached:
path.append(path[-1] - bisect(sums, path[-1]) + 1)
base = cached[path[-1]]
for offset, vertex in enumerate(reversed(path)):
cached[vertex] = base + offset
return cached[i]
print([A023124(i) for i in range(1, 1001)]) # Brady J. Garvin, Sep 13 2024
Northwest corner: 1 4 10 19 30 44 61 81 103 2 6 13 23 35 50 68 89 112 3 8 16 27 40 56 75 97 121 5 11 20 32 46 63 83 106 131 7 14 24 37 52 70 91 115 141 9 17 28 42 58 77 99 124 151
r = E; z = 100; s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*r];
u = Table[n + 1 + Sum[Floor[(n - k)/r], {k, 0, n}], {n, 0, z}] (* A022786, col 1 of A283943 *)
v = Table[s[n], {n, 0, z}] (* A022785, row 1 of A283943 *)
w[i_, j_] := u[[i]] + v[[j]] + (i - 1)*(j - 1) - 1;
Grid[Table[w[i, j], {i, 1, 10}, {j, 1, 10}]] (* A283943, array *)
Flatten[Table[w[k, n - k + 1], {n, 1, 20}, {k, 1, n}]] (* A283943, sequence *)
\\ Produces the triangle when the array is read by antidiagonals
r = exp(1);
z = 100;
s(n) = if(n<1, 1, s(n - 1) + 1 + floor(n*r));
p(n) = n + 1 + sum(k=0, n, floor((n - k)/r));
u = v = vector(z + 1);
for(n=1, 101, (v[n] = s(n - 1)));
for(n=1, 101, (u[n] = p(n - 1)));
w(i, j) = u[i] + v[j] + (i - 1) * (j - 1) - 1;
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(w(k, n - k + 1), ", "); ); print(); ); };
tabl(10) \\ Indranil Ghosh, Mar 26 2017
# Produces the triangle when the array is read by antidiagonals
import math
from mpmath import *
mp.dps = 100
def s(n): return 1 if n<1 else s(n - 1) + 1 + int(math.floor(n*e))
def p(n): return n + 1 + sum([int(math.floor((n - k)/e)) for k in range(0, n+1)])
v=[s(n) for n in range(0, 101)]
u=[p(n) for n in range(0, 101)]
def w(i, j): return u[i - 1] + v[j - 1] + (i - 1) * (j - 1) - 1
for n in range(1, 11):
print([w(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017
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