A241757 -id:A241757 - cp's OEIS frontend

A241758 Smallest prime in representation 2*A241757(n) by sum of two primes, the adding of which in binary requires only one carry.

2, 5, 13, 5, 17, 17, 5, 17, 5, 5, 13, 17, 5, 13, 5, 17, 37, 17, 5, 13, 17, 17, 29, 37, 41, 5, 5, 17, 13, 5, 13, 17, 5, 37, 41, 17, 5, 73, 5, 89, 13, 97, 5, 13, 17, 37, 41, 29, 137, 5, 5, 41, 5, 41, 13, 193, 5, 5, 17, 193, 17, 17, 37, 41, 37, 97, 53, 73, 53, 5

Offset: 1

Vladimir Shevelev, Apr 28 2014

			a(2)=5, since A241757(2)=22=5+17, and in binary in sum of 101+10001 involves only one carry.

Peter J. C. Moses, Table of n, a(n) for n = 1..10000
Aviezri Fraenkel and Alex Kontorovich, The Sierpiński Sieve of Nim-varieties and Binomial Coefficients, INTEGERS 7 (2)(2007), #A14.
E. E. Kummer, Über die Ergänzungssätze zu den allgemeinen Reciprocitätsgesetzen, J. Reine Angew. Math. 44 (1852), 93-146.

Cf. A241757, A241405.

2||Binomial(2*A241757(n), a(n)). Indeed, from the Kummer theorem (see reference) 2^t||Binomial(n,x) if and only if in adding x and n-x in binary we have exactly t carries. A proof of the Kummer theorem in arbitrary base one can find in [Fraenkel & Kontorovich].

More terms from Peter J. C. Moses, Apr 29 2014

A108421 Smallest number of ones needed to write in binary representation 2*n as sum of two primes.

Original entry on oeis.org

2, 4, 4, 4, 5, 5, 5, 5, 4, 4, 5, 6, 5, 5, 6, 4, 5, 6, 5, 5, 5, 5, 6, 6, 6, 5, 6, 5, 6, 7, 7, 7, 8, 5, 5, 6, 5, 5, 6, 6, 5, 6, 6, 5, 6, 6, 7, 8, 5, 5, 6, 6, 6, 6, 7, 5, 6, 6, 7, 8, 7, 7, 8, 6, 7, 5, 5, 6, 5, 5, 6, 6, 5, 6, 6, 5, 6, 7, 7, 7, 6, 6, 6, 6, 6, 6, 7, 6, 6, 7, 7, 7, 8, 7, 8, 6, 5, 5, 6, 6, 6, 6, 7, 5, 6

Offset: 2

Reinhard Zumkeller, Jun 03 2005

a(n) = Min{A000120(p)+A000120(q) : p,q prime and p+q=2*n}.
a(n) = A108422(n) - A108423(n).
a(n) >= A000120(n)+1, with equality for n in A241757. - Robert Israel, Mar 25 2018

			n=15: 2*15=30 and A002375(15)=3 with 30=7+23=11+19=13+17,
13+17 -> 1101+10001 needs a(15)=5 binary ones, whereas
7+23 -> 111+10111 and 11+19 -> 1011+10011 need more.

Cf. A000120, A004676, A005843, A007088, A108422, A108423, A241757.

N:= 200: # to get a(2)..a(N)
Primes:= select(isprime, [seq(i,i=3..2*N-3,2)]):
Ones:= map(t -> convert(convert(t,base,2),`+`), Primes):
V:= Vector(N): V[2]:= 2:
for i from 1 to nops(Primes) do
  p:= Primes[i];
  for j from 1 to i do
    k:= (p+Primes[j])/2;
    if k > N then break fi;
    t:= Ones[i]+Ones[j];
    if V[k] = 0 or t < V[k] then V[k]:= t fi
  od
od:
convert(V[2..N],list); # Robert Israel, Mar 25 2018

Min[#]&/@(Table[Total[Flatten[IntegerDigits[#,2]]]&/@Select[ IntegerPartitions[ 2*n,{2}],AllTrue[#,PrimeQ]&],{n,2,110}]) (* Harvey P. Dale, Jul 27 2020 *)

A241856 Smallest number of carries while adding two primes in binary, the sum of which is 2*n.

Original entry on oeis.org

1, 2, 3, 2, 3, 2, 4, 3, 2, 1, 3, 3, 2, 1, 5, 2, 3, 3, 3, 2, 2, 1, 4, 3, 3, 1, 3, 1, 2, 2, 6, 5, 6, 2, 3, 3, 2, 1, 4, 3, 2, 2, 3, 1, 2, 1, 5, 5, 2, 1, 3, 2, 2, 1, 4, 1, 2, 1, 3, 3, 2, 1, 7, 4, 5, 2, 3, 3, 2, 1, 4, 3, 2, 2, 3, 1, 2, 2, 5, 4, 3, 2, 3, 2, 2, 1, 4

Offset: 2

Vladimir Shevelev, Apr 30 2014

			Let n=5. We have 2*5=10=3+7=5+5. Adding 3+7 in binary requires 3 carries, while adding 5+5 in binary requires 2 carries. Thus a(5)=2.

Peter J. C. Moses, Table of n, a(n) for n = 2..1001

Cf. A241757, A241758.

More terms from Peter J. C. Moses, Apr 30 2014

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A241758 Smallest prime in representation 2*A241757(n) by sum of two primes, the adding of which in binary requires only one carry.

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A108421 Smallest number of ones needed to write in binary representation 2*n as sum of two primes.

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A241856 Smallest number of carries while adding two primes in binary, the sum of which is 2*n.

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