OFFSET
1,5
COMMENTS
For numbers > 1, iterate the map x -> A252463(x) which divides even numbers by 2, and shifts every prime in the prime factorization of odd n one index step towards smaller primes. a(n) counts the numbers of the form 4k+1 encountered until 1 has been reached, which is also included in the count. The count includes also n itself if it is of the form 4k+1 (A016813), thus a(1) = 1.
In other words, locate the node which contains n in binary tree A005940 and traverse from that node towards the root, counting all numbers of the form 4k+1 that occur on the path.
LINKS
FORMULA
a(1) = 1, a(2n) = a(n), and for odd numbers n > 1, a(n) = a(A064989(n)) + [n == 1 (mod 4)].
Other identities and observations. For n >= 1:
a(n) >= A292374(n).
From Antti Karttunen, Apr 22 2022: (Start)
(End)
MATHEMATICA
a[1] = 1; a[n_] := a[n] = a[Which[n == 1, 1, EvenQ@n, n/2, True, Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ n]] + Boole[Mod[n, 4] == 1]; Array[a, 105] (* Michael De Vlieger, Sep 17 2017 *)
PROG
(PARI)
A064989(n) = {my(f); f = factor(n); if((n>1 && f[1, 1]==2), f[1, 2] = 0); for (i=1, #f~, f[i, 1] = precprime(f[i, 1]-1)); factorback(f)};
(Scheme, with memoization-macro definec)
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Sep 17 2017
STATUS
approved