cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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, 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, 7, 6, 6, 8, 5, 4, 9, 3, 6, 8, 4, 7, 9, 4, 5, 11, 7, 5, 8
Offset: 1

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Author

Lei Zhou, Feb 02 2015

Keywords

Comments

a(1)=0 is the only zero term up to n=200000.
It is hypothesized that a(1)=0 is the only zero term of this sequence.
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.

Examples

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

Crossrefs

Programs

  • Mathematica
    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]]];
    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}]