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A355303
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a(n) is the smallest integer that has n normal undulating divisors.
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12
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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
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OFFSET
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1,2
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COMMENTS
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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).
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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