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A266577 Square array read by descending antidiagonals: T(n,k) = ((2^(n+1) + 1)^(k-1) + 1)/2.

%I A266577 #95 May 18 2025 02:29:01
%S A266577 1,3,1,13,5,1,63,41,9,1,313,365,145,17,1,1563,3281,2457,545,33,1,7813,
%T A266577 29525,41761,17969,2113,65,1,39063,265721,709929,592961,137313,8321,
%U A266577 129,1,195313,2391485,12068785,19567697,8925313,1073345,33025,257,1,976563,21523361,205169337,645733985,580145313,138461441,8487297,131585,513,1
%N A266577 Square array read by descending antidiagonals: T(n,k) = ((2^(n+1) + 1)^(k-1) + 1)/2.
%C A266577 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
%C A266577 Conjecture: For integers n and m > 1, let b(n,m)=n^m+1, S(n,m) = set of numbers of the form (b(n,m)^k+...+b(n,m)^((n-1)*k)+1)/n, where k is any nonnegative integer. Then for each positive integer s, (b(n,m)^s-1)*(product of any m not necessarily distinct terms of S(n,m)) is palindromic in base b(n,m). - _Ahmad J. Masad_, Jan 11 2025
%C A266577 The conjecture is true. See my proof in MathOverflow (2025) link. - _Max Alekseyev_, May 15 2025
%H A266577 Amiram Eldar, <a href="/A266577/b266577.txt">Table of n, a(n) for n = 1..1275</a>
%H A266577 Ahmad J. Masad et al., <a href="https://mathoverflow.net/q/298313">Conjecture on palindromic numbers</a>, MathOverflow, Apr 2018.
%H A266577 Ahmad J. Masad et al., <a href="https://mathoverflow.net/q/492658">A more general conjecture about palindromic numbers</a>, MathOverflow, 2025.
%e A266577 The array begins:
%e A266577   1   3  13  63 313
%e A266577   1   5  41 365
%e A266577   1   9 145
%e A266577   1  17
%e A266577   1
%e A266577 Example of the result concerning palindromic numbers:
%e A266577 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.
%e A266577 Example of the conjecture: assume n=5 and m=3, then b(5,3)=5^3+1=126. Assume k1=1 and k2=1 and k3=2 (they are three values since m=3). Assume s=3; then we have the calculation ((126+126^2+126^3+126^4+1)/5)^2*(126^2+126^4+126^6+126^8+1)/5*(126^3-1) which is equal to: 32807046133985032885720309126001 and this number has the base-126 expansion  (1,3,7,12,19,25,31,34,37,37,37,34,31,25,19,12,7,3,1)_126 which is a palindromic number in base 126.
%t A266577 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 A266577 (PARI) tabl(n) = matrix(n, n, i, j, ((2^(i+1)+1)^(j-1)+1)/2); \\ _Michel Marcus_, Jan 02 2016
%Y A266577 Cf. A034478.
%K A266577 nonn,tabl
%O A266577 1,2
%A A266577 _Ahmad J. Masad_, Jan 01 2016
%E A266577 a(31) corrected by _Georg Fischer_, Nov 07 2021