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A355303
a(n) is the smallest integer that has n normal undulating divisors.
12
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
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
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Jun 29 2022
EXTENSIONS
Terms a(11) and beyond from Amiram Eldar, Jun 29 2022
STATUS
approved