A110441 Triangular array formed by the Mersenne numbers.
1, 3, 1, 7, 6, 1, 15, 23, 9, 1, 31, 72, 48, 12, 1, 63, 201, 198, 82, 15, 1, 127, 522, 699, 420, 125, 18, 1, 255, 1291, 2223, 1795, 765, 177, 21, 1, 511, 3084, 6562, 6768, 3840, 1260, 238, 24, 1, 1023, 7181, 18324, 23276, 16758, 7266, 1932, 308, 27, 1
Offset: 0
Examples
Triangle starts: 1; 3, 1; 7, 6, 1; 15, 23, 9, 1; 31, 72, 48, 12, 1; (0, 3, -2/3, 2/3, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, ...) begins: 1 0, 1 0, 3, 1 0, 7, 6, 1 0, 15, 23, 9, 1 0, 31, 72, 48, 12, 1. - _Philippe Deléham_, Mar 19 2012 With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins / 1 \/1 \/1 \ / 1 \ | 3 1 ||0 1 ||0 1 | | 3 1 | | 7 3 1 ||0 3 1 ||0 0 1 |... = | 7 6 1 | |15 7 3 1 ||0 7 3 1 ||0 0 3 1 | |15 23 9 1| |31 15 7 3 1 ||0 15 7 3 1||0 0 7 3 1| |... | |... ||... ||... | |... | - _Peter Bala_, Jul 22 2014
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150)
- Naiomi T. Cameron and Asamoah Nkwanta, On Some (Pseudo) Involutions in the Riordan Group, J. Int. Seq. 8 (2005), #05.3.7.
- Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.
Programs
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Maple
# Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left. PMatrix(10, n -> 2^n - 1); # Peter Luschny, Oct 09 2022
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Mathematica
With[{n = 9}, DeleteCases[#, 0] & /@ CoefficientList[Series[1/(1 - (3 + y) x + 2 x^2), {x, 0, n}, {y, 0, n}], {x, y}]] // Flatten (* Michael De Vlieger, Apr 25 2018 *)
Formula
Riordan array M(n, k): (1/(1-3z+2z^2), z/(1-3z+2z^2)). Leftmost column M(n, 0) is the Mersenne numbers A000225, first column is A045618, second column is A055582, row sum is A007070 and diagonal sum is even-indexed Fibonacci numbers A001906.
T(n,k) = Sum_{j=0..n} C(j+k,k)C(n-j,k)2^(n-j-k). - Paul Barry, Feb 13 2006
From Philippe Deléham, Mar 19 2012: (Start)
G.f.: 1/(1-(3+y)*x+2*x^2).
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) -2*T(n-2,k), T(0,0) = 1, T(n,k) = 0 if k<0 or if k>n.
Sum_{k, 0<=k<=n} T(n,k)*x^k = A000225(n+1), A007070(n), A107839(n), A154244(n), A186446(n), A190975(n+1), A190979(n+1), A190869(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7 respectively. (End)
Recurrence: T(n+1,k+1) = Sum_{i=0..n-k} (2^(i+1) - 1)*T(n-i,k). - Peter Bala, Jul 22 2014
From Peter Bala, Oct 07 2019: (Start)
Recurrence for row polynomials: R(n,x) = (3 + x)*R(n-1,x) - 2*R(n-2,x) with R(0,x) = 1 and R(1,x) = 3 + x.
The row reverse polynomial x^n*R(n,1/x) is equal to the numerator polynomial of the finite continued fraction 1 + x/(1 + 2*x/(1 + ... + x/(1 + 2*x/(1)))) (with 2*n partial numerators). Cf. A116414. (End)
Comments