A352322 Starts of runs of 3 consecutive Pell-Niven numbers (A352320).

4, 28, 110, 168, 984, 1024, 3123, 3514, 5740, 6783, 6923, 8584, 12664, 16744, 18160, 19670, 23190, 23470, 24030, 34503, 34643, 36304, 40384, 45880, 47390, 50910, 51190, 51750, 57607, 61640, 68104, 73600, 78403, 78630, 78910, 79470, 86674, 89360, 95824, 101320

Offset: 1

Author

Amiram Eldar, Mar 12 2022

Keywords

• nonn
• base

Comments

Conjecture: There are no runs of 4 consecutive Pell-Niven numbers (checked up to 2*10^8).

Examples

			4 is a term since 4, 5 and 6 are all Pell-Niven numbers: the minimal Pell representation of 4, A317204(20) = 20, has the sum of digits 2+0 = 2 and 4 is divisible by 2, the minimal Pell representation of 5, A317204(5) = 100, has the sum of digits 1+0+0 = 1 and 5 is divisible by 1, and the minimal Pell representation of 6, A317204(6) = 101, has the sum of digits 1+0+1 = 2 and 6 is divisible by 2.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000129, A265744, A317204.

A182190 \ {0} is a subsequence.

Subsequence of A352320 and A352321.

Similar sequences: A154701, A328206, A328210, A328214, A330932, A331087, A333428, A334310, A331822, A342428, A344343, A351716, A351721, A352091, A352109.

Programs

Mathematica

pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; pellNivenQ[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]; Divisible[n, Plus @@ IntegerDigits[Total[3^(s - 1)], 3]]]; seq[count_, nConsec_] := Module[{pn = pellNivenQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ pn, c++; AppendTo[s, k - nConsec]]; pn = Join[Rest[pn], {pellNivenQ[k]}]; k++]; s]; seq[30, 3]