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A273401
Numbers n such that n and n + 1 have exactly the same number of odd divisors.
4
1, 5, 6, 10, 11, 12, 13, 19, 22, 23, 28, 37, 40, 43, 46, 47, 49, 52, 54, 58, 61, 65, 67, 69, 73, 77, 79, 82, 84, 88, 96, 103, 106, 110, 112, 114, 119, 129, 132, 136, 140, 148, 151, 154, 155, 157, 163, 166, 172, 178, 182, 185, 186, 191, 192, 193, 203, 204, 211, 215, 216, 219, 220, 221
OFFSET
1,2
COMMENTS
If A001227(n) = A001227(n*2^m) for m >= 0 then:
1) A001227(n) is equal to number of ways to write 2n - 1 as (4*x + 2)*y + 4*x + 1 where x and y are nonnegative integers;
2) A001227(n) is equal to number of distinct values of k if k/(2n-1) + 1 divides (k/(2n - 1))^(k/(2n - 1)) + k, (k/(2n - 1))^k + k/(2n - 1) and k^(k/(2n - 1)) + k/(2n - 1).
LINKS
EXAMPLE
5 and 6 have both two odd divisors: (1 and 5) and (1 and 3) respectively; so 5 is a term in the sequence.
MAPLE
A001227:= n -> numtheory:-tau(n)/(1+padic:-ordp(n, 2)):
R:= map(A001227, [$1..1000]):
ListTools:-SearchAll(0, A001227[2..-1]-A001227[1..-2]); # Robert Israel, May 27 2016
MATHEMATICA
Select[Range@ 221, First@ Differences@ Map[Count[Divisors@ #, _?OddQ] &, {#, # + 1}] == 0 &] (* Michael De Vlieger, Jun 26 2016 *)
SequencePosition[Table[Count[Divisors[n], _?OddQ], {n, 250}], {x_, x_}] [[All, 1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Apr 06 2019 *)
PROG
(PARI) lista(nn) = for (n=1, nn, if (sumdiv(n, d, d%2) == sumdiv(n+1, d, d%2), print1(n, ", "))); \\ Michel Marcus, May 27 2016
(PARI) is(n)=numdiv(n>>valuation(n, 2))==numdiv((n+1)>>valuation(n+1, 2)) \\ Charles R Greathouse IV, Jul 15 2016
CROSSREFS
Cf. A001227, A206581 (primes in a(n)).
Sequence in context: A189056 A340325 A247561 * A042958 A265188 A163903
KEYWORD
nonn
AUTHOR
STATUS
approved