A355303 a(n) is the smallest integer that has n normal undulating divisors.

1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 126, 60, 320, 144, 168, 120, 252, 180, 560, 240, 630, 420, 780, 360, 1890, 960, 1920, 720, 1560, 1080, 1260, 1440, 1680, 4368, 2160, 3240, 3120, 3360, 4320, 2520, 6300, 6120, 8640, 6240, 13104, 5040, 12480, 9360, 12240, 7560

Offset

1,2

Comments

Normal undulating numbers are in A355301.

The first ten terms are the same as A005179, then A005179(11) = 1024 while a(11) = 126 (see example); also, a(n) = A005179(n) for n = 12, 16, 18, 20, 24 (up to n = 50).

Links

David A. Corneth, Table of n, a(n) for n = 1..402

Example

16 has 5 divisors: {1, 2, 4, 8, 16}, all of which are normal undulating integers; no positive integer smaller than 16 has five normal undulating divisors, hence a(5) = 16.
126 has 12 divisors: {1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126}; only 126 is not normal undulating; no positive integer smaller than 126 has eleven normal undulating divisors, hence a(11) = 126.

Mathematica

nuQ[n_] := AllTrue[(s = Sign[Differences[IntegerDigits[n]]]), # != 0 &] && AllTrue[Differences[s], # != 0 &]; f[n_] := DivisorSum[n, 1 &, nuQ[#] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n]; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[50, 10^5] (* Amiram Eldar, Jun 29 2022 *)

Prog

(PARI) isok(m) = if (m<10, return(1)); my(d=digits(m), dd = vector(#d-1, k, sign(d[k+1]-d[k]))); if (#select(x->(x==0), dd), return(0)); my(pdd = vector(#dd-1, k, dd[k+1]*dd[k])); #select(x->(x>0), pdd) == 0; \\ A355301
a(n) = my(k=1); while (sumdiv(k, d, isok(d)) != n, k++); k; \\ Michel Marcus, Jun 30 2022

Crossrefs

Cf. A005179, A355301, A355302, A355304.

Sequence in context: A343019 A355594 A357172 * A099315 A005179 A037019

Adjacent sequences: A355300 A355301 A355302 * A355304 A355305 A355306

Keyword

nonn,base

Author

Bernard Schott, Jun 29 2022

Extensions

Terms a(11) and beyond from Amiram Eldar, Jun 29 2022

Status

approved