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.
Let s = A019565 and let t = A280864. a(0) = 1 since s(0) = 1 = t(1). a(1) = 2 since s(1) = 2 = t(2). a(2) = 4 since s(2) = 3 = t(4). a(3) = 5 since s(3) = 5 = t(5). Table relating this sequence to s and t. The last column shows Y if s(n) is divisible by the prime in the heading, otherwise ".": n s(n) a(n) 2357 ---------------------- 0 1 1 . 1 2 2 Y 2 3 4 .Y 3 6 5 YY 4 5 7 ..Y 5 10 8 Y.Y 6 15 17 .YY 7 30 26 YYY 8 7 11 ...Y 9 14 12 Y..Y 10 21 20 .Y.Y 11 42 37 YY.Y 12 35 36 .YYY 13 70 72 Y.YY 14 105 73 .YYY 15 210 207 YYYY ...
nn = 2^13; r = s = 1; c[_] := False;
rad[n_] := rad[n] = Times @@ FactorInteger[n][[All, 1]];
a = Monitor[Reap[Do[w = GCD[r, s]; k = m = r/w;
While[Or[c[k], ! CoprimeQ[w, k] ], k += m]; Sow[k]; c[k] = True;
s = r; r = rad[k], {i, nn}]][[-1, 1]], i];
Array[FirstPosition[a, Times @@ Prime@ Position[Reverse[IntegerDigits[#, 2]], 1][[All, 1]] ][[1]] &, 61, 0]
Let s = A019565 and let t = A280866. a(0) = 1 since s(0) = 1 = t(1). a(1) = 2 since s(1) = 2 = t(2). a(2) = 4 since s(2) = 3 = t(4). a(3) = 5 since s(3) = 5 = t(5). Table relating this sequence to s and t. The last column shows Y if s(n) is divisible by the prime in the heading, otherwise ".": n s(n) a(n) 2357 ---------------------- 0 1 1 . 1 2 2 Y 2 3 4 .Y 3 6 5 YY 4 5 7 ..Y 5 10 8 Y.Y 6 15 17 .YY 7 30 26 YYY 8 7 11 ...Y 9 14 12 Y..Y 10 21 20 .Y.Y 11 42 37 YY.Y 12 35 36 .YYY 13 70 67 Y.YY 14 105 68 .YYY 15 210 205 YYYY ...
nn = 2^14; c[] := False; m[] := 1; i = 1; j = m[1] = m[2] = 2; c[1] = c[2] = True; f[x_] := f[x] = Times @@ FactorInteger[x][[All, 1]]; s = Association[ Monitor[Reap[ Do[While[c[Set[k, # m[#]]], m[#]++] &[f[i * j]/f[i]]; If[SquareFreeQ[k], Sow[Total[2^(-1 + PrimePi[FactorInteger[k][[All, 1]]])] -> n] ]; Set[{c[k], i, j}, {True, j, k}], {n, 3, nn}] ][[-1, 1]], n]]; TakeWhile[{1, 2}~Join~Array[If[KeyExistsQ[s, #], Lookup[s, #], 0] &, Floor@ Sqrt[nn], 2], # > 0 &]
Table below shows rows j = 0..5: j\k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ------------------------------------------------------------------- 0: 1; 1: 2; 2: 3, 4; 3: 5, 6, 7, 8; 4: 9, 10, 12, 11, 13, 14, 15, 16; 5: 17, 18, 20, 19, 24, 21, 25, 22, 26, 23, 28, 27, 29, 30, 31, 32; . These correspond with values f(T(j,k)) as shown below: j\k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ------------------------------------------------------------------------------- 0: 1; 1: 2; 2: 3, 6; 3: 5, 10, 15, 30; 4: 7, 14, 21, 35, 42, 70, 105, 210; 5: 11, 22, 33, 55, 66, 77, 110, 154, 165, 231, 330, 385, 462, 770, 1155, 2310; . T(4,4) = a(11) = 35, while A019565(11) = 42, since 11_2 = "1011", f(11) = 2*3*7 = 42, but A019565(12) = 35 since 12_2 = "1100", f(12) = 5*7 is smaller than 42, therefore a(11) = 35, and a(12) = 42.
Flatten@ Table[
SortBy[Range[2^n, 2^(n + 1) - 1],
Times @@ Flatten@
MapIndexed[Prime[#2]^#1 &,
Reverse@ IntegerDigits[#, 2]] &], {n, 0, 8}]
A103790(2)=1*2*1-1; => a(1)=2 A103790(4)=3*2*2-1; => a(2)=4
A019565 = Function[tn, k1 = tn; o = 1; tt = 1; While[k1 > 0, k2 = Mod[k1, 2]; If[k2 == 1, tt = tt*Prime[o]]; k1 = (k1 - k2)/2; o = o + 1]; tt]; Array[fa, {1, 500}]; Do[fa[n] = 0, {n, 1, 500}]; n = 2; npd = Prime[n]; ct = 1; wt = 1; While[wt < 69, cr = (ct + 1)/2; If[fa[cr] == 0, fa[cr] = n; While[fa[wt] > 0, Print[fa[wt]]; wt = wt + 1]]; n = n + 1; npd = Prime[n]; ct = 1; tt = ct; cp = npd + A019565[tt]; While[ ! (PrimeQ[cp]), ct = ct + 1; tt = ct; cp = npd + A019565[tt]]]
A019565(0)=1, 1-2=-1 is not prime; A019565(1)=2, 2-2=0 is not prime; ... A019565(4)=5, 5-2=3 is prime, so a(1)=4;
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