Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #34 Jun 20 2022 10:12:27
%S 0,0,0,1,1,2,2,4,5,6,8,12,16,22,29,36,48,67,84,118,151,203,270,362,
%T 472,636,846,1142,1526,2024,2736,3666,4918,6550,8776,11796,15824
%N When A002487 is written as a triangle the n-th row has length 2^(n-1); a(n) is the maximal multiplicity of any entry in that row, considering the entries strictly between the initial 1 and the central 2.
%C The maximal entry is row n is Fibonacci(n+1), and the smallest missing number is A135510(n). The number of distinct numbers in each row is given by A293160.
%C It would be nice to have a formula for this sequence, or at least some bounds.
%e Rows 0 through 6 of A002487 are:
%e 0,
%e 1,
%e 1, 2,
%e 1, 3, 2, 3,
%e 1, 4, 3, 5, 2, 5, 3, 4,
%e 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5,
%e 1, 6, 5, 9, 4, 11, 7, 10, 3, 11, 8, 13, 5, 12, 7, 9, 2, 9, 7, 12, 5, 13, 8, 11, 3, 10, 7, 11, 4, 9, 5, 6,
%e To find a(5) we consider the entries 1, 5, 4, 7, 3, 8, 5, 7, 2 in row 5. Ignoring the initial 1 and the final 2, the maximal multiplicity is 2 (for example, 5 appears twice), so a(5) = 2.
%e From _Don Reble_, Nov 04 2017: (Start)
%e The initial values of a(n) for n >= 3 together with the terms that have the highest multiplicity are:
%e 3 1 [3]
%e 4 1 [3 4 5]
%e 5 2 [5 7]
%e 6 2 [5 7 9 11]
%e 7 4 [11]
%e 8 5 [13 17]
%e 9 6 [19 23 31 41]
%e 10 8 [23 37 43]
%e 11 12 [71]
%e 12 16 [71]
%e 13 22 [127]
%e 14 29 [109]
%e 15 36 [199 251]
%e 16 48 [263]
%e 17 67 [433]
%e 18 84 [701]
%e 19 118 [839]
%e 20 151 [1193]
%e 21 203 [1801]
%e 22 270 [2693]
%e 23 362 [4229]
%e 24 472 [4349]
%e 25 636 [7759]
%e 26 846 [11287]
%e 27 1142 [14627]
%e 28 1526 [20929]
%e 29 2024 [37243]
%e 30 2736 [43133]
%e 31 3666 [67231]
%e 32 4918 [90227]
%e 33 6550 [127819]
%e 34 8776 [181031]
%e 35 11796 [251071]
%e 36 15824 [394549]
%e (End)
%p A002487 := proc(n) option remember; if n <= 1 then n elif n mod 2 = 0 then procname(n/2); else procname((n-1)/2)+procname((n+1)/2); fi; end:
%p ans:=[];
%p for n from 3 to 18 do
%p b1:=2^(n-1); b2:=2^n-1; b3:=2^(n-2)-1; mx:=0;
%p ar:=Array(0..b1-1,0);
%p for k from 1 to b3 do
%p kk:=b1+k;
%p v:=A002487(kk);
%p ar[v]:=ar[v]+1;
%p od:
%p for k from 0 to b1-1 do if ar[k]>mx then mx:=ar[k]; fi; od:
%p ans:=[op(ans),mx];
%p od:
%p ans;
%o (Python)
%o from itertools import chain, product
%o from collections import Counter
%o from functools import reduce
%o def A293957(n): return 0 if n <= 2 else max(Counter(m for m in (sum(reduce(lambda x,y:(x[0],x[0]+x[1]) if y else (x[0]+x[1],x[1]),chain(k,(1,)),(1,0))) for k in product((False,True),repeat=n-2)) if m != 1 and m != 2).values()) # _Chai Wah Wu_, Jun 20 2022
%Y Cf. A002487, A049456, A135510, A293160.
%K nonn,more
%O 0,6
%A _N. J. A. Sloane_, Nov 03 2017
%E a(19)-a(36) from _Don Reble_, Nov 04 2017