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A330927 revision #14

A330927
Numbers k such that both k and k + 1 are Niven numbers.
31
1, 2, 3, 4, 5, 6, 7, 8, 9, 20, 80, 110, 111, 132, 152, 200, 209, 224, 399, 407, 440, 480, 510, 511, 512, 629, 644, 735, 800, 803, 935, 999, 1010, 1011, 1014, 1015, 1016, 1100, 1140, 1160, 1232, 1274, 1304, 1386, 1416, 1455, 1520, 1547, 1651, 1679, 1728, 1853
OFFSET
1,2
COMMENTS
Cooper and Kennedy proved that there are infinitely many runs of 20 consecutive Niven numbers. Therefore this sequence is infinite.
REFERENCES
Jean-Marie De Koninck, Those Fascinating Numbers, American Mathematical Society, 2009, p. 36, entry 110.
LINKS
Curtis Cooper and Robert E. Kennedy, On consecutive Niven numbers, Fibonacci Quarterly, Vol. 21, No. 2 (1993), pp. 146-151.
Helen G. Grundman, Sequences of consecutive Niven numbers, Fibonacci Quarterly, Vol. 32, No. 2 (1994), pp. 174-175.
Wikipedia, Harshad number.
Brad Wilson Construction of 2n consecutive n-Niven numbers, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.
EXAMPLE
1 is a term since 1 and 1 + 1 = 2 are both Niven numbers.
MATHEMATICA
nivenQ[n_] := Divisible[n, Total @ IntegerDigits[n]]; nq1 = nivenQ[1]; seq = {}; Do[nq2 = nivenQ[k]; If[nq1 && nq2, AppendTo[seq, k - 1]]; nq1 = nq2, {k, 2, 2000}]; seq
PROG
(Magma) f:=func<n|n mod &+Intseq(n) eq 0>; a:=[]; for k in [1..2000] do if forall{m:m in [0..1]|f(k+m)} then Append(~a, k); end if; end for; a; // Marius A. Burtea, Jan 03 2020
KEYWORD
nonn,base
AUTHOR
Amiram Eldar, Jan 03 2020
STATUS
approved