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 A230762 List of commonest number of decompositions of 2k into an unordered sum of two odd primes in range 3 <= k <= m, integer m >= 3, where m is explained below. 2
 1, 2, 3, 4, 5, 7, 8, 9, 11, 18, 27, 44, 48, 52, 58, 61, 75, 77, 98, 141, 165, 200, 231, 337, 360, 378, 384, 466, 517, 697, 880, 1061, 1400, 1503, 1615, 1700, 1896, 2082, 2163, 3242, 3929, 4232, 5373 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS If making a statistical bar chart with x-axis denoting the number of decompositions of an even number, and y-axis denoting the number of hits of an x-axis value for all 3 <= k <= m, there are one or more commonest x value presenting with highest y value.  Such commonest x values increase when m increases, and fall on the x values listed in this sequence. Hypothesis: With the increase of m, the commonest number of decompositions of 2n into an unordered sum of two odd primes in the range of 3 <= k <= m ascends. This hypothesis derives that the corresponding smallest m to the terms of this sequence makes an ascending sequence.  Or say, when testing with m ascending, once a number a(n) enters this sequence, no number smaller than a(n) will be able to enter this sequence if they had not enter previous. LINKS Lei Zhou, Table of n, a(n) for n = 1..45 EXAMPLE When m=3, k has only one value 3, 2k=6=3+3.  Only one possible decomposition, making a decomposition statistics {{x,y}}={{1,1}}.  So a(1)=1; When m=4, k gets another value 4, 2k=8=3+5. The decomposition statistics {{x,y}}={{1,2}};... Thereafter, k=5 makes 2k=10=5+5=3+7, {{x,y}}={{1,2},{2,1}}, the commonest value is still 1. k=6, 2k=12=5+7, {{x,y}}={{1,3},{2,1}}, commonest x is still 1. k=7, 2k=14=3+11=7+7, {{x,y}}={{1,3},{2,2}}, commonest x is still 1. k=8, 2k=16=3+13=5+11, {{x,y}}={{1,3},{2,3}}, except for 1, 2 is now eligible to be the new possible commonest x, so a(2)=2. ... Counting up to k=28, the decomposition statistics is {{1,3},{2,8},{3,8},{4,5},{5,2}}, 2 and 3 are now the commonest decompositions. It is the first time for 3 to appear.  So a(3)=3. MATHEMATICA check=0; posts={}; mpos=0; res={}; sres=0; s={}; size=0; k=2; While[k++; k2=2*k; p2=k-1; ct=0;   While[p2=NextPrime[p2]; p2size, Do[AppendTo[s, 0], {i, size+1, ct}]; size=ct];   (*and construct statistics in array s*)   s[[ct]]++; m=Max[s]; aa=Position[s, m]; la=Length[aa];   Do[a=aa[[pos, 1]];     If[a>sres,       While[sres

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Last modified November 21 16:04 EST 2019. Contains 329371 sequences. (Running on oeis4.)