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%I #18 Mar 02 2021 02:11:53
%S 3,1,0,2,2,2,4,0,0,3,1,0,3,1,2,1,4,1,2,1,2,0
%N Number of palindromic hexagonal numbers with exactly n digits.
%C Number of terms in A054969 with exactly n digits.
%H G. J. Simmons, <a href="/A002778/a002778_2.pdf">Palindromic powers</a>, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy] See page 95.
%e There are only two 4-digit hexagonal numbers that are palindromic, 3003 and 5995. Thus, a(4)=2.
%t A054969 = {0, 1, 6, 66, 3003, 5995, 15051, 66066, 617716, 828828, 1269621, 1680861, 5073705, 5676765, 1264114621, 5289009825, 6172882716, 13953435931, 1313207023131, 5250178710525, 6874200024786, 61728399382716, 602224464422206, 636188414881636, 1250444114440521, 16588189498188561, 58183932923938185, 66056806460865066, 67898244444289876, 514816979979618415, 3075488771778845703, 6364000440440004636, 15199896744769899151}; Table[Length[ Select[A054969, IntegerLength[#] == n || (n == 1 && # == 0) &]], {n, 19}]
%o (Python)
%o def afind(terms):
%o m, n, c = 0, 1, 0
%o while n <= terms:
%o p = m*(2*m-1)
%o s = str(p)
%o if len(s) == n:
%o if s == s[::-1]: c += 1
%o else:
%o print(c, end=", ")
%o n, c = n+1, int(s == s[::-1])
%o m += 1
%o afind(14) # _Michael S. Branicky_, Mar 01 2021
%Y Cf. A000384, A054969, A054970, A082721, A263618, A307752.
%K nonn,base,more
%O 1,1
%A _Robert Price_, Apr 27 2019
%E a(20)-a(22) from _Michael S. Branicky_, Mar 01 2021
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