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 A172236 Array T(n,k) = n*T(n,k-1) + T(n,k-2) read by upward antidiagonals, starting T(n,0) = 0, T(n,1) = 1. 12
 0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 5, 3, 0, 1, 4, 10, 12, 5, 0, 1, 5, 17, 33, 29, 8, 0, 1, 6, 26, 72, 109, 70, 13, 0, 1, 7, 37, 135, 305, 360, 169, 21, 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 0, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 0, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89, 0, 1, 11, 101, 747, 4289, 18200, 53353, 98145, 98209, 42837, 5741, 144 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,9 COMMENTS Equals A073133 with an additional column T(.,0). If the first column and top row are deleted, antidiagonal reading yields A118243. Adding a top row of 1's and antidiagonal reading downwards yields A157103. Antidiagonal sums are 0, 1, 2, 5, 12, 32, 93, 297, 1035, 3911, 15917, 69350, .... From Jianing Song, Jul 14 2018: (Start) All rows have strong divisibility, that is, gcd(T(n,k_1),T(n,k_2)) = T(n,gcd(k_1,k_2)) holds for all k_1, k_2 >= 0. Let E(n,m) be the smallest number l such that m divides T(n,l), we have: for odd primes p that are not divisible by n^2 + 4, E(n,p) divides p - ((n^2+4)/p) if p == 3 (mod 4) and (p - ((n^2+4)/p))/2 if p == 1 (mod 4). E(n,p) = p for odd primes p that are divisible by n^2 + 4. E(n,2) = 2 for even n and 3 for odd n. Here ((n^2+4)/p) is the Legendre symbol. A prime p such that p^2 divides T(n,E(n,p)) is called an n-Wall-Sun-Sun prime. E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide T(n,E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by T(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1. Let pi(n,m) be the Pisano period of T(n,k) modulo m, i.e, the smallest number l such that T(n,k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p - 1 if ((n^2+4)/p) = 1 and 2(p+1) if ((n^2+4)/p) = -1. pi(n,p) = 4p for odd primes p that are divisible by n^2 + 4. pi(n,2) = 2 even n and 3 for odd n. pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide T(n,E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by T(n,p), so pi(n,p^e) = 4p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1),pi(n,m_2)) if gcd(m_1,m_2) = 1. If n != 2, the largest possible value of pi(n,m)/m is 4 for even n and 6 for odd n. For even n, pi(n,p^e) = 4p^e; for odd n, pi(n,2p^e) = 12p^e, where p is any odd prime factor of n^2 + 4. For n = 2 it is 8/3, obtained by m = 3^e. Let z(n,m) be the number of zeros in a period of T(n,k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: z(n,p) = 4 for odd primes p that are divisible by n^2 + 4. For other odd primes p, z(n,p) = 4 if E(n,p) is odd; 1 if E(n,p) is even but not divisible by 4; 2 if E(n,p) is divisible by 4; see the table below. z(n,2) = z(n,4) = 1. Among all values of z(n,p) when p runs through all odd primes that are not divisible by n^2 + 4, we have: ((n^2+4)/p)...p mod 8....proportion of 1.....proportion of 2.....proportion of 4 ......1..........1......1/6 (conjectured)...2/3 (conjectured)...1/6 (conjectured)* ......1..........5......1/2 (conjectured)...........0...........1/2 (conjectured)* ......1.........3,7.............1...................0...................0 .....-1.........1,5.............0...................0...................1 .....-1.........3,7.............0...................1...................0 * The result is that among all odd primes that are not divisible by n^2 + 4, 7/24 of them are with z(n,p) = 1, 5/12 are with z(n,p) = 2 and 7/24 are with z(n,p) = 4 if n^2 + 4 is a twice a square; 1/3 of them are with z(n,p) = 1, 1/3 are with z(n,p) = 2 and 1/3 are with z(n,p) = 4 otherwise. [Corrected by Jianing Song, Jul 06 2019] z(n,p^e) = z(n,p) for all odd primes p; z(n,2^e) = 1 for even n and 2 for odd n, e >= 3. (End) LINKS Jianing Song, Antidiagonals n = 1..100, flattened (the first 30 antidiagonals from Vincenzo Librandi) FORMULA T(n,k) = (((n + sqrt(n^2 + 4))/2)^k - ((n-sqrt(n^2 + 4))/2)^k)/sqrt(n^2 + 4), n >= 1, k >= 0. - Jianing Song, Jun 27 2018 For n >= 1, Sum_{i=1..k} 1/T(n,2^i) = ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/T(n,2^k)), where u = (n + sqrt(n^2 + 4))/2, v = (n - sqrt(n^2 + 4))/2 are the two roots of the polynomial x^2 - n*x - 1. As a result, Sum_{i>=1} 1/T(n,2^i) = (n^2 + 4 - n*sqrt(n^2 + 4))/(2*n). - Jianing Song, Apr 21 2019 EXAMPLE The array starts in row n = 1 with columns k >= 0 as   0      1      1      2      3      5      8   0      1      2      5     12     29     70   0      1      3     10     33    109    360   0      1      4     17     72    305   1292   0      1      5     26    135    701   3640   0      1      6     37    228   1405   8658   0      1      7     50    357   2549  18200   0      1      8     65    528   4289  34840   0      1      9     82    747   6805  61992   0      1     10    101   1020  10301 104030   0      1     11    122   1353  15005 166408 MATHEMATICA f[0, a_] := 0; f[1, a_] := 1; f[n_, a_] := f[n, a] = a*f[n - 1, a] + f[n - 2, a]; m1 = Table[f[n, a], {n, 0, 10}, {a, 1, 11}]; Table[Table[m1[[m, n - m + 1]], {m, 1, n}], {n, 1, 10}]; Flatten[%] PROG (PARI) T(n, k) = if (k==0, 0, if (k==1, 1, n*T(n, k-1) + T(n, k-2))); tabl(nn) = for(n=1, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Jianing Song, Jul 14 2018 (program from Michel Marcus; see also A316269) (PARI) T(n, k) = ([n, 1; 1, 0]^k)[2, 1] \\ Jianing Song, Nov 10 2018 CROSSREFS Cf. A000045 (first row), A000129 (2nd row), A006190 (3rd row), A001076 (4th row), A052918 (5th row), A005668 (6th row), A054413 (7th row), A041025 (8th row), A099371 (9th row), A041041 (10th row), A049666 (11th row), A041061 (12th row), A140455 (13th row), A041085 (14th row), A154597 (15th row), A041113 (16th row), A178765 (17th row), A041145 (18th row), A243399 (19th row), A041181 (20th row). (Note that there are offset shifts for rows n = 5, 7, 8, 10, 12, 14, 16..20.) Cf. A002522 (4th column), A054602 (5th column), A057721 (6th column), A124152 (7th column). Entry points for T(n,k) modulo m: A001177 (n=1), A214028 (n=2), A322907 (n=3). Pisano period for T(n,k) modulo m: A001175 (n=1), A175181 (n=2), A175182 (n=3), A175183 (n=4), A175184 (n=5), A175185 (n=6). Number of zeros in a period for T(n,k) modulo m: A001176 (n=1), A214027 (n=2), A322906 (n=3). Cf. A271782, A316269. Sequence in context: A317205 A296068 A144064 * A191646 A297321 A277938 Adjacent sequences:  A172233 A172234 A172235 * A172237 A172238 A172239 KEYWORD nonn,tabl,easy AUTHOR Roger L. Bagula, Jan 29 2010 EXTENSIONS More terms from Jianing Song, Jul 14 2018 STATUS approved

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Last modified October 13 18:57 EDT 2019. Contains 327981 sequences. (Running on oeis4.)