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A254610 Number of decompositions of 2n into sums of two primes p1 <= p2 such that the smallest |k*p1-p2| = 2^m+b, where |b|<=2. 1

%I #15 Mar 19 2015 07:13:08

%S 0,1,1,1,2,1,2,2,2,2,3,3,3,2,3,2,4,4,2,3,4,3,4,5,4,3,5,3,4,6,3,5,6,2,

%T 4,6,4,4,7,4,5,8,5,4,9,4,4,7,2,5,7,4,5,8,5,6,9,5,5,11,4,5,8,3,5,7,5,4,

%U 7,6,6,8,5,4,9,3,6,8,4,7,9,4,5,11,7,5,8

%N Number of decompositions of 2n into sums of two primes p1 <= p2 such that the smallest |k*p1-p2| = 2^m+b, where |b|<=2.

%C a(1)=0 is the only zero term up to n=200000.

%C It is hypothesized that a(1)=0 is the only zero term of this sequence.

%C The histogram for 1<=n<=60000 of this sequence shows the shape of a distribution with mode=10, and it has a regional maximum at 20.

%H Lei Zhou, <a href="/A254610/b254610.txt">Table of n, a(n) for n = 1..10000</a>

%H Lei Zhou, <a href="/A254610/a254610.jpg">Histogram for 1<=n<=60000 </a>

%H <a href="/index/Go#Goldbach">Index entries for sequences related to Goldbach conjecture</a>

%e For n=1, 2n=2, which cannot be decomposed into the sum of two primes, so a(1)=0.

%e For n=2, 2n = 4 = 2+2, and 2-2 = 0 = 2^0-1, so the difference from 2^0 is 1, which satisfies the condition. So a(2)=1;

%e ...

%e For n=5, 2n = 10 = 3+7 = 5+5. |3*2-7| = 1 = 2^0 and |5-5| = 0 = 2^0-1; both satisfy the condition, so a(5)=2.

%e ...

%e For n=35, 2n = 70 = 3+67 = 11+59 = 17+53 = 23+47 = 29+41. These five Goldbach decompositions make A045917(35)=5. Among these, |3*22-67| = 1 = 2^0; |11*5-59| = 4 = 2^2; |17*3-53| = 2 = 2^1; |23*2-47| = 1 satisfies the condition. However, |29-41| = 12 = 2^3+4 = 2^4-4 does not satisfy the condition. So, a(35)=4 < A045917(35). This is the first term where the two sequences differ.

%t NumDiff[n1_, n2_] := Module[{c1 = n1, c2 = n2}, If[c1 < c2, c1 = c1 + c2; c2 = c1 - c2; c1 = c1 - c2]; k = Floor[c1/c2]; a1 = c1 - k*c2; If[a1 == 0, a2 = 0, a2 = (k + 1) c2 - c1]; Return[Min[a1, a2]]];

%t Table[e = 2 n; p1 = 1; ct = 0; While[p1 = NextPrime[p1]; p1 <= n, p2 = e - p1; If[PrimeQ[p2], d = NumDiff[p1, p2]; k = Floor[Log[2, d]]; diff1 = d - 2^k; If[diff1 == 0, ct++, diff2 = 2^(k + 1) - d; If[(diff1 <= 2) || (diff2 <= 2), ct++]]]]; ct, {n, 1, 100}]

%Y Cf. A045917, A254606

%K nonn,easy

%O 1,5

%A _Lei Zhou_, Feb 02 2015

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Last modified March 28 08:22 EDT 2024. Contains 371236 sequences. (Running on oeis4.)