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A211869 a(n) = Sum_{j=1..n-1} j*(n-j)*b^(j-1) with b = floor(n^2/4)+1.

%I A211869 #44 Mar 03 2018 11:07:28
%S A211869 0,1,8,98,1712,58985,2541896,187337236,15687030920,2014736789165,
%T A211869 280434300560320,55591630021883014,11642487182670742552,
%U A211869 3294318202343411333713,969986091740868071844464,371055858906757952457992360
%N A211869 a(n) = Sum_{j=1..n-1} j*(n-j)*b^(j-1) with b = floor(n^2/4)+1.
%C A211869 Equivalently, a(n) is the number having the digits (j*(n-j); j=1..n-1), in base b = floor(n^2/4)+1.
%C A211869 From _R. J. Cano_, Mar 03 2018: (Start)
%C A211869 If a(n) were converted to the base 1+floor(n^2/4)=A033638(n) then a palindrome would be obtained. Such palindrome is related to A215940(n!);
%C A211869 a(7)=2541896 and A033638(7)=13, giving the palindrome "6ACCA6". Such palindrome cannot be converted directly to decimal, but it might be defined instead from these digits the polynomial f(t)= 6*t^5 +10*t^4 +12*t^3 +12*t^2+10*t^1+6*t^0, then evaluating for t=10, we get f(10)=713306=A215940(7!). 713306 clearly looks distinct than "6ACCA6". f(11) and f(12) respectively are 1130256 with "7021A6", and 1722942 with "6B10A6". Now evaluating f(14) we get 3646530 and if converted to base 14 it yields "6ACCA6". The same happens with f(15) converted to base 15, f(16) converted to Hexadecimal, and also in general for f(y) converted to base y, if it were provided that y>=13.
%C A211869 Here A033638(n) gives the lower bound for the infinite set of bases where this behavior can be observed. For simplicity it is chosen the base A033638(n) when defining this sequence, although what we actually want is to keep the pattern generated by the products j*(n-j). (End)
%C A211869 This sequence together with A033638 and A215940 demonstrates the connection among permutation sets and palindromes obtained by symmetric products. - _Alexander R. Povolotsky_, Feb 08 2013
%H A211869 R. J. Cano, <a href="/A211869/b211869.txt">Table of n, a(n) for n = 1..57</a>
%H A211869 R. J. Cano, <a href="/w/images/b/be/Another_Illustration_of_more_symmetries_in_A215940_.txt">Additional information on this sequence.</a>
%H A211869 R. J. Cano, <a href="/w/images/a/a6/Tribute_to_Gauss.pdf">A211869: Gauss taught me that.</a>
%F A211869 a(n) = Sum_{j=1..n-1} j*(n-j)*A033638(n)^(n-1-j).
%e A211869 For n=5, the four products are 1*4 = 4, 2*3 = 6, 3*2 = 6, 4*1 = 4, giving the base-7 concatenation 4664. In base 10, this is a(5) = 1712.
%e A211869 For a(6) we have that 1+floor(6^2/4) = 10 so there is no need of converting the concatenation to decimal. By definition the products are j*(n-j) for j in 1..5: 1*(6-1) = 5 = 5*(6-5), 2*(6-2) = 8 = 4*(6-2), 3*(6-3) = 9 so the result is a(6)=58985.
%o A211869 (PARI) a(n,base=1+n^2\4)=sum(j=1, n-1, j*(n-j)*base^(n-1-j));
%Y A211869 Cf. A033638, A215940, A083449, A019566, A002113.
%K A211869 nonn,easy,base
%O A211869 1,3
%A A211869 _R. J. Cano_, Feb 02 2013