A113468 Least number k such that k, k+n, k+2*n and k+3*n have the same number of divisors.
242, 213, 3445, 111, 8718, 5, 2001, 69, 3526, 299, 1074, 5, 2222, 537, 9177, 129, 4114, 5, 8, 598, 7843, 111, 1235, 10, 2984, 303, 3538, 417, 987, 7, 1771, 91, 7659, 57, 9269, 10, 2264, 145, 1197, 219, 1606, 5, 1826, 115, 8897, 203, 618, 5, 8, 159, 2673, 183
Offset: 1
Keywords
Examples
a(19) = 8 because 8, 8 + 19 = 27, 8 + 2*19 = 46 and 8 + 3*19 = 65 each have 4 divisors.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A113465.
Programs
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Mathematica
a[n_] := Module[{k = 1, d}, While[(d = DivisorSigma[0, k]) != DivisorSigma[0, k+n] || DivisorSigma[0, k+2*n] != d || DivisorSigma[0, k+3*n] != d, k++]; k]; Array[a, 60] (* Amiram Eldar, Aug 04 2024 *) -
PARI
a(n) = {my(k = 1, d); while((d = numdiv(k)) != numdiv(k+n) || numdiv(k+2*n) != d || numdiv(k+3*n) != d, k++); k;} \\ Amiram Eldar, Aug 04 2024
Extensions
Name corrected by Amiram Eldar, Aug 04 2024
Comments