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.
Table[n + 1 + Sum[Floor[(n - k)/Sqrt[3]], {k, 0, n}], {n, 0, 200}] (* A022778 *)
(* Clark Kimberling, Mar 14 2015 *)
Take[ Transpose[ Sort[ Flatten[ Table[{i + j*Sqrt[2], i}, {i, 17}, {j, 15}], 1], #1[[1]] < #2[[1]] &]][[2]], 96] (* Robert G. Wilson v, Jul 24 2004 *)
Quiet[Block[{$ContextPath}, Needs["Combinatorica`"]], {General::compat}]
memos = <||>;
zeroBasedC[theta_, i_] := zeroBasedC[theta, i] = Module[{memo, depth},
memo = Lookup[memos, theta, {-1, 0}];
While[memo[[-1]] <= i, AppendTo[memo, memo[[-1]] + Ceiling[theta * (Length[memo] - 1)]]];
memos[i] = memo;
depth = Combinatorica`BinarySearch[memo, i] - 3/2;
If[IntegerQ[depth] && depth <= i, 1 + zeroBasedC[theta, i - depth], 0]
];
A007336[i_] := zeroBasedC[2^(1/2), i - 1] + 1;
Table[A007336[i], {i, 1, 100}] (* Brady J. Garvin, Aug 19 2024 *)
from bisect import bisect
from collections import defaultdict
from functools import cache
from math import ceil
memos = defaultdict(lambda: [-1, 0])
@cache
def zero_based_c(theta, i):
memo = memos[theta]
while memo[-1] <= i:
memo.append(memo[-1] + ceil(theta * (len(memo) - 1)))
depth = bisect(memo, i) - 1
return 0 if depth > i or memo[depth] == i else 1 + zero_based_c(theta, i - depth)
def A007336(i):
return zero_based_c(2 ** 0.5, i - 1) + 1
print([A007336(i) for i in range(1, 1001)]) # Brady J. Garvin, Aug 18 2024
For a(3), 1+7+7 == 3 (mod 12). For a(4), 1+7+7+7 == 10 (mod 12).
[1+7*(n-1) mod(12): n in [1..80]]; // Vincenzo Librandi, May 10 2015
PadRight[{}, 100, {1, 8, 3, 10, 5, 12, 7, 2, 9, 4, 11, 6}] (* Vincenzo Librandi, May 10 2015 *)
LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1},{1, 8, 3, 10, 5, 12, 7, 2, 9, 4, 11, 6},108] (* Ray Chandler, Aug 27 2015 *)
a(n)=7*(n-1)%12+1 \\ Charles R Greathouse IV, Jun 02 2015
Vec(x*(1 + 8*x + 3*x^2 + 10*x^3 + 5*x^4 + 12*x^5 + 7*x^6 + 2*x^7 + 9*x^8 + 4*x^9 + 11*x^10 + 6*x^11) / (1 - x^12) + O(x^80)) \\ Colin Barker, Nov 15 2019
Mathematica code based on that for A007337 by Robert G. Wilson v.: m = x /. Solve[x^( 10) + x^9 - x^7 - x^6 - x^5 - x^4 - x^3 + x + 1 == 0, x][[2]] Take[Transpose[Sort[Flatten[Table[{i + j*m, i}, {i, 25}, {j, 17}], 1], #1[[1]] < #2[[1]] &]][[2]], 95]
Northwest corner: 1 3 7 13 20 29 40 53 2 5 10 17 25 35 47 61 4 8 14 22 31 42 55 70 6 11 18 27 37 49 63 79 9 15 23 33 44 57 72 89 12 19 28 39 51 65 81 99 16 24 34 46 59 74 91 110 21 30 41 54 68 84 102 122
r = Sqrt[3]; 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}] (* A022778, col 1 of A283940 *)
v = Table[s[n], {n, 0, z}] (* A022777, row 1 of A283940*)
w[i_, j_] := u[[i]] + v[[j]] + (i - 1)*(j - 1) - 1;
Grid[Table[w[i, j], {i, 1, 10}, {j, 1, 10}]] (* A283940, array *)
Flatten[Table[w[k, n - k + 1], {n, 1, 20}, {k, 1, n}]] (* A283940, sequence *)
r = sqrt(3);
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 21 2017
r = 3 ** 0.5
def s(n): return 1 if n<1 else s(n - 1) + 1 + int(n*r)
def p(n): return n + 1 + sum([int((n - k)/r) 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 21 2017
import numpy as np r = np.sqrt(3) x = np.arange(11) u = np.cumsum(np.ceil(x / r)).astype(int) v = np.cumsum(np.ceil(x * r)).astype(int) print(*[1 + u[k] + v[n-k] + k*(n-k) for n in range(11) for k in range(n+1)], sep=', ') # David Radcliffe, May 10 2025
m = x /. Solve[x^3 - x - 1 == 0, x][[1]]
Take[Transpose[Sort[Flatten[Table[{i + j*m, i}, {i, 25}, {j, 17}], 1], #1[[1]] < #2[[1]] &]][[2]], 95]
m = x /. Solve[x^4 - x^3 - 1 == 0, x][[4]]
Take[Transpose[Sort[Flatten[Table[{i + j*m, i}, {i, 25}, {j, 17}], 1], #1[[1]] < #2[[1]] &]][[2]], 95]
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