Greg Knowles - cp's OEIS frontend

User: Greg Knowles

Greg Knowles has authored 2 sequences.

A194627 a(1)=1, a(n+1) = p(n)^2 + q(n)^2 + 1, where p(n) and q(n) are the number of prime and nonprime numbers respectively in the sequence so far.

1, 2, 3, 6, 9, 14, 21, 30, 41, 46, 59, 66, 81, 98, 117, 138, 161, 186, 213, 242, 273, 306, 341, 378, 417, 458, 501, 546, 593, 602, 651, 702, 755, 810, 867, 926, 987, 1050, 1115, 1182, 1251, 1322, 1395, 1470, 1547, 1626, 1707, 1790, 1875, 1962, 2051, 2142

Offset: 1

Greg Knowles, Sep 15 2011

			For n=1, we have no primes and one nonprime (a(1)=1), so a(2)=0^2+1^2+1=2. Now we have one prime (a(2)=2) and one nonprime, so a(3)=1^2+1^2+1=3.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A079545, A377791, A377882.

R:= 1: v:= 1: p:= 0: q:= 0:
for i from 2 to 100 do
  if isprime(v) then p:= p+1 else q:= q+1 fi;
  v:= p^2 + q^2 + 1;
  R:= R,v
od:
R; # Robert Israel, Nov 10 2024

t = {1}; Do[ps = Count[t, ?(PrimeQ[#] &)]; AppendTo[t, ps^2 + (n - ps - 1)^2 + 1], {n, 2, 100}]; t (* _T. D. Noe, Sep 15 2011 *)

p=q=0;for(n=1,50,print1(k=p^2+q^2+1", ");if(isprime(k),p++,q++)) \\ Charles R Greathouse IV, Sep 16 2011

A191598 a(1)=1. Then, if the first n terms contain all the numbers 1 to n, a(n+1)=2*a(n), otherwise a(n+1) is the smallest number not already in the sequence.

Original entry on oeis.org

1, 2, 4, 3, 6, 5, 10, 7, 8, 9, 18, 11, 12, 13, 14, 15, 16, 17, 34, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 66, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 130

Offset: 1

Greg Knowles, Jun 08 2011

			There are no gaps with just one term, so we double to get a(2)=2. Still no gaps, so we double again to get a(3)=4. Now 3 has not appeared yet, so a(4)=3. The sequence is compact again, so we double to get a(5)=6.

Michael S. Branicky, Table of n, a(n) for n = 1..10000

from itertools import count, islice
def agen():  # generator of terms
    an, aset, m = 1, {1}, 2
    for n in count(1):
        yield an
        an = 2*an if m > n else m
        aset.add(an)
        while m in aset: m += 1
print(list(islice(agen(), 67))) # Michael S. Branicky, Nov 13 2024

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User: Greg Knowles

A194627 a(1)=1, a(n+1) = p(n)^2 + q(n)^2 + 1, where p(n) and q(n) are the number of prime and nonprime numbers respectively in the sequence so far.

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