A352344 Starts of runs of 3 consecutive lazy-Pell-Niven numbers (A352342).
2196, 2650, 5784, 17459, 28950, 57134, 112878, 124506, 147078, 162809, 169694, 191538, 210494, 218654, 223344, 223459, 230894, 239360, 258740, 277455, 278900, 285615, 289695, 291328, 291858, 295408, 311524, 314658, 324734, 332894, 335179, 341900, 347718, 362880
Offset: 1
Examples
2196 is a term since 2196, 2197 and 2198 are all divisible by the sum of the digits in their maximal Pell representation: k A352339(k) A352340(k) k/A352340(k) ---- ---------- ---------- ------------ 2196 121222020 12 183 2197 121222021 13 169 2198 121222022 14 157
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; pellp[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[pell[k] <= m, k++]; k--; AppendTo[s, k]; m -= pell[k]; k = 1]; IntegerDigits[Total[3^(s - 1)], 3]]; lazyPellNivenQ[n_] := Module[{v = pellp[n]}, nv = Length[v]; i = 1; While[i <= nv - 2, If[v[[i]] > 0 && v[[i + 1]] == 0 && v[[i + 2]] < 2, v[[i ;; i + 2]] += {-1, 2, 1}; If[i > 2, i -= 3]]; i++]; i = Position[v, ?(# > 0 &)]; Divisible[n, Plus @@ v[[i[[1, 1]] ;; -1]]]]; seq[count, nConsec_] := Module[{lpn = lazyPellNivenQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ lpn, c++; AppendTo[s, k - nConsec]]; lpn = Join[Rest[lpn], {lazyPellNivenQ[k]}]; k++]; s]; seq[30, 3]