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%I #51 Apr 23 2023 22:20:40
%S 1,3,1,13,5,1,63,41,9,1,313,365,145,17,1,1563,3281,2457,545,33,1,7813,
%T 29525,41761,17969,2113,65,1,39063,265721,709929,592961,137313,8321,
%U 129,1,195313,2391485,12068785,19567697,8925313,1073345,33025,257,1,976563,21523361,205169337,645733985,580145313,138461441,8487297,131585,513,1
%N Square array read by descending antidiagonals: T(n,k) = ((2^(n+1) + 1)^(k-1) + 1)/2.
%C The matrix M in the definition of A292625 is given by this sequence, also, for each natural number m and each natural number c, ((2^(m+1)+1)^c-1)*(the product of any (m+1) not necessarily distinct terms of the m-th row) is palindromic in base (2^(m+1)+1), see the MathOverflow link. - _Ahmad J. Masad_, Apr 19 2023
%H Amiram Eldar, <a href="/A266577/b266577.txt">Table of n, a(n) for n = 1..1275</a>
%H Ahmad J. Masad, <a href="https://mathoverflow.net/questions/298313/conjecture-on-palindromic-numbers">Conjecture on palindromic numbers</a>, MathOverflow, Apr 2018.
%e The array begins:
%e 1 3 13 63 313
%e 1 5 41 365
%e 1 9 145
%e 1 17
%e 1
%e Example of the result concerning palindromic numbers:
%e Take m=2, c=4, 2^(m+1) + 1 = 2^3 + 1 = 9, we choose 3 not necessarily distinct terms from the second row. Let them be 41, 365, 365; then we get 41*365*365*(9^4 - 1) = 35832196000 = 112435534211_9, which is a palindromic number in base 9.
%t T[n_, k_] := ((2^(n + 1) + 1)^(k - 1) + 1)/2; Table[T[k, n - k + 1], {n, 1, 10}, {k, 1, n}] // Flatten (* _Amiram Eldar_, Sep 14 2022 *)
%o (PARI) tabl(n) = matrix(n, n, i, j, ((2^(i+1)+1)^(j-1)+1)/2); \\ _Michel Marcus_, Jan 02 2016
%Y Cf. A034478.
%K nonn,tabl
%O 1,2
%A _Ahmad J. Masad_, Jan 01 2016
%E a(31) corrected by _Georg Fischer_, Nov 07 2021
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