A352109 Starts of runs of 3 consecutive lazy-tribonacci-Niven numbers (A352107).
175, 1183, 2259, 5290, 12969, 21130, 51820, 70629, 78090, 79540, 81818, 129648, 160224, 169234, 180908, 228240, 238574, 249494, 278628, 332891, 376335, 383866, 398650, 399644, 454090, 550380, 565200, 683448, 683604, 694274, 728895, 754390, 782110, 809830, 837550
Offset: 1
Examples
175 is a term since 175, 176 and 177 are all divisible by the number of terms in their maximal tribonacci representation: k A352103(k) A352104(k) k/A352104(k) --- ---------- ---------- ------------ 175 11111110 7 25 176 11111111 8 22 177 100100100 3 59
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; trib[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; IntegerDigits[Total[2^(s - 1)], 2]]; lazyTriboNivenQ[n_] := Module[{v = trib[n]}, nv = Length[v]; i = 1; While[i <= nv - 3, If[v[[i ;; i + 3]] == {1, 0, 0, 0}, v[[i ;; i + 3]] = {0, 1, 1, 1}; If[i > 3, i -= 4]]; i++]; i = Position[v, ?(# > 0 &)]; If[i == {}, False, Divisible[n, Total[v[[i[[1, 1]] ;; -1]]]]]]; seq[count, nConsec_] := Module[{tri = lazyTriboNivenQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ tri, c++; AppendTo[s, k - nConsec]]; tri = Join[Rest[tri], {lazyTriboNivenQ[k]}]; k++]; s]; seq[30, 3]
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