A266577 Square array read by descending antidiagonals: T(n,k) = ((2^(n+1) + 1)^(k-1) + 1)/2.
1, 3, 1, 13, 5, 1, 63, 41, 9, 1, 313, 365, 145, 17, 1, 1563, 3281, 2457, 545, 33, 1, 7813, 29525, 41761, 17969, 2113, 65, 1, 39063, 265721, 709929, 592961, 137313, 8321, 129, 1, 195313, 2391485, 12068785, 19567697, 8925313, 1073345, 33025, 257, 1, 976563, 21523361, 205169337, 645733985, 580145313, 138461441, 8487297, 131585, 513, 1
Offset: 1
Examples
The array begins: 1 3 13 63 313 1 5 41 365 1 9 145 1 17 1 Example of the result concerning palindromic numbers: 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. 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.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..1275
- Ahmad J. Masad et al., Conjecture on palindromic numbers, MathOverflow, Apr 2018.
- Ahmad J. Masad et al., A more general conjecture about palindromic numbers, MathOverflow, 2025.
Crossrefs
Cf. A034478.
Programs
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Mathematica
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 *)
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PARI
tabl(n) = matrix(n, n, i, j, ((2^(i+1)+1)^(j-1)+1)/2); \\ Michel Marcus, Jan 02 2016
Extensions
a(31) corrected by Georg Fischer, Nov 07 2021
Comments