A358205 a(n) is the least number k such that 1 + 2*k + 3*k^2 has exactly n prime divisors, counted with multiplicity.
0, 2, 1, 13, 19, 7, 61, 331, 169, 1141, 6487, 898, 20581, 315826, 59947, 296143, 1890466, 6141994, 1359025, 49188715, 20490901, 264422320, 178328878, 1340590345, 9476420614, 5989636213, 72238539832, 103619599441, 668478672403, 794002910839, 417430195531
Offset: 0
Keywords
Examples
a(5) = 7 because 1 + 2*7 + 3*7^2 = 162 = 2*3^4 has 5 prime divisors, counted with multiplicity. From _Jon E. Schoenfield_, Nov 05 2022: (Start) Let m = 1 + 2*k + 3*k^2. Since no such number m is divisible by 2^2, 5, or 7, the smallest number m having a given number of prime factors counted with multiplicity will tend to have a large number of 3's among its prime factors: . n k = a(n) m = 1 + 2*k + 3*k^2 -- ------------ ----------------------------------------------------- 0 0 1 1 2 17 (prime) 2 1 6 = 2 * 3 3 13 534 = 2 * 3 * 89 4 19 1122 = 2 * 3 * 11 * 17 5 7 162 = 2 * 3^4 6 61 11286 = 2 * 3^3 * 11 * 19 7 331 329346 = 2 * 3^4 * 19 * 107 8 169 86022 = 2 * 3^6 * 59 9 1141 3907926 = 2 * 3^5 * 11 * 17 * 43 10 6487 126256482 = 2 * 3^5 * 11^2 * 19 * 113 11 898 2421009 = 3^10 * 41 12 20581 1270773846 = 2 * 3^9 * 19 * 1699 13 315826 299238818481 = 3^9 * 19 * 73 * ... 14 59947 10781048322 = 2 * 3^10 * 11 * 43 * 193 15 296143 263102621634 = 2 * 3^12 * 17 * 14561 16 1890466 10721588872401 = 3^12 * 11 * 19 * ... 17 6141994 113172283172097 = 3^16 * 2629057 18 1359025 5540849569926 = 2 * 3^14 * 11^2 * 4787 19 49188715 7258589148431106 = 2 * 3^17 * 28103531 20 20490901 1259631112357206 = 2 * 3^15 * 17 * 73 * ... 21 264422320 209757490471391841 = 3^16 * 11 * 17 * ... 22 178328878 95403566542874409 = 3^19 * 19 * 83 * ... 23 1340590345 5391547422002837766 = 2 * 3^19 * 11^2 * ... 24 9476420614 269407642979285252217 = 3^22 * 2617 * ... 25 5989636213 107627225904222216534 = 2 * 3^20 * 19 * 97 * ... 26 72238539832 15655219911322828844337 = 3^22 * 11 * 19 * ... 27 103619599441 32211064165147101736326 = 2 * 3^22 * 11 * 43 * ... 28 668478672403 1340591206374369138728034 = 2 * 3^22 * 19 * 331 * ... 29 794002910839 1891321867264002956873442 = 2 * 3^23 * 11 * 73 * ... 30 417430195531 522743904423981537506946 = 2 * 3^25 * 11 * 17 * ... . As a result, the last digits of the ternary representation of a(n) tend to fall into a pattern: . n a(n) a(n) in base 3 -- ------------ --------------------------- 0 0 0_3 1 2 2_3 2 1 1_3 3 13 111_3 4 19 201_3 5 7 21_3 6 61 2021_3 7 331 110021_3 8 169 20021_3 9 1141 1120021_3 10 6487 22220021_3 11 898 1020021_3 12 20581 1001020021_3 13 315826 121001020021_3 14 59947 10001020021_3 15 296143 120001020021_3 16 1890466 10120001020021_3 17 6141994 102120001020021_3 18 1359025 2120001020021_3 19 49188715 10102120001020021_3 20 20490901 1102120001020021_3 21 264422320 200102120001020021_3 22 178328878 110102120001020021_3 23 1340590345 10110102120001020021_3 24 9476420614 220110102120001020021_3 25 5989636213 120110102120001020021_3 26 72238539832 20220110102120001020021_3 27 103619599441 100220110102120001020021_3 28 668478672403 2100220110102120001020021_3 29 794002910839 2210220110102120001020021_3 30 417430195531 1110220110102120001020021_3 (End)
Links
- Gerry Martens, Table of n, a(n) for n = 0..40
Programs
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Maple
N:= 18: # for a(0)..a(N) V:= Array(0..N): count:= 0: for k from 0 while count < N+1 do v:= numtheory:-bigomega(1+2*k+3*k^2); if v <= N and V[v] = 0 then count:= count+1; V[v]:= k fi od: convert(V,list); -
Mathematica
a[n_] := Module[{i = 0},While[! PrimeOmega[1 + 2 i + 3 i^2] == n, i += 1]; i] Table[a[n], {n, 0, 14}] (* Gerry Martens, Nov 05 2022 *)
Extensions
a(21)-a(22) from Amiram Eldar, Nov 04 2022
a(23)-a(30) from Jon E. Schoenfield, Nov 05 2022
Comments