A254606 -id:A254606 - cp's OEIS frontend

A254608 a(n) is the smallest number such that A254606(a(n)) = n.

1, 2, 9, 14, 27, 26, 34, 63, 64, 53, 89, 115, 88, 165, 101, 116, 132, 292, 149, 185, 166, 225, 271, 205, 270, 318, 247, 397, 294, 293, 319, 370, 371, 344, 399, 398, 427, 691, 489, 488, 553, 552, 520, 655, 658, 620, 622, 769, 621, 693, 656, 731, 851, 810, 732

Offset: 0

Lei Zhou, Feb 02 2015

nonn

			A254606(1)=0, so a(0)=1;
A254606(2)=1, so a(1)=2;
A254606(m)<2 for all m<9, and A254606(9)=2, so a(2)=9.

Lei Zhou, Table of n, a(n) for n = 0..10000

Cf. A254606.

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]]];
a = {1}; n = 0; p1 = 2; p2 = 1; lb = 0; While[lb < 100, n++; p2 = NextPrime[p2]; If[p2 > p1, p1 = p2; p2 = 2];
d = NumDiff[p1, p2]; While[l = Length[a]; l <= d, AppendTo[a, 0]]; If[a[[d + 1]] == 0, a[[d + 1]] = n; ps = Position[a, 0]; If[Length[ps] == 0, b = a, ps1 = ps[[1]][[1]]; b = Take[a, ps1 - 1]]; lb = Length[b]]]; b

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

Lei Zhou, Feb 02 2015

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.

			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.

Lei Zhou, Table of n, a(n) for n = 1..10000
Lei Zhou, Histogram for 1<=n<=60000
Index entries for sequences related to Goldbach conjecture

Cf. A045917, A254606

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}]

cp's OEIS Frontend

A254608 a(n) is the smallest number such that A254606(a(n)) = n.

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