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

Author

Bernard Schott, Jun 29 2022

Keywords

• nonn
• base

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).

Examples

			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.
		

Links

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

Crossrefs

Cf. A005179, A355301, A355302, A355304.

Programs

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 *)

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

Extensions

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