OFFSET
1,1
COMMENTS
Primes p > 5 are in the sequence. We may rewrite the k-th prime p instead as prime(k)^1, and thus convert it to a single run of 1 ones followed by (k-1) zeros, which becomes a binary number consisting of 1 followed by k-1 zeros. It is clear that for p > 5, prime(k) < 2^(k-1).
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..13175 (terms in A005940 for n < 2^21)
Michael De Vlieger, Fan style binary tree diagram of b(n) for n = 1..2^14-1, where b(n) = A005940(n), highlighting terms such that b(n+1) < n in red, b(n+1) = n in ochre, and b(n+1) > n in blue. Terms shown in red appear in this sequence, while b(A029747(n)+1) = A029747(n) appears in ochre.
EXAMPLE
7 is in the sequence since 7 = prime(3+1)^1, which we write as 1 following 3 zeros when approached from the least significant digit, i.e., "1000"_2 = 8, thus A005940(8) = 7; and 7 < 8.
5 is not in the sequence since 5 = prime(2+1)^1 -> "100"_2 = 4, and 5 > 4.
14 is in the sequence since 14 = prime(0+1)^1 * prime(3+1)^1, which we can express as a binary number with singleton 1s following 0 and 3 zeros, i.e., "10001"_2 = 17, hence A005940(17) = 14 and we see 14 < 17.
MATHEMATICA
nn = 2^10; a[0] = 1; Reap[Do[k = Prime[1 + DigitCount[n, 2, 0]]*a[n - 2^Floor@ Log2@ n]; Set[a[n], k]; If[k < n, Sow[k]], {n, nn}]][[-1, -1]]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael De Vlieger, Aug 07 2022
STATUS
approved