A385430 Least number k such that k and k + n! have the same number of divisors.
2, 3, 5, 5, 7, 7, 11, 23, 17, 11, 17, 29, 46, 19, 43, 23, 31, 37, 89, 29, 31, 31, 97, 62, 41, 59, 47, 67, 159, 107, 127, 79, 37, 97, 61, 131, 86, 43, 97, 53, 61, 97, 71, 47, 94, 101, 233, 53, 83, 61, 249, 53, 71, 158, 71, 149, 107, 134, 254, 206, 166, 131, 271
Offset: 1
Examples
a(1) = 2 since d(2) = d(3) = 2; a(5) = 7 since d(7) = d(7+5!) = 2; a(13) = 46 since d(46) = d(46+13!) = 4; etc.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..82
Programs
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Mathematica
a[n_] := Block[{k = 2}, While[ DivisorSigma[0, k] != DivisorSigma[0, k + n!], k++]; k]; Array[ a, 51]
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PARI
a(n) = my(k=1); while (numdiv(k) != numdiv(k+n!), k++); k; \\ Michel Marcus, Aug 02 2025
Extensions
More terms from Sean A. Irvine, Aug 08 2025
Comments