A340441 Square array, read by ascending antidiagonals, where row n gives all odd solutions k > 1 and n > 0 to A000120(2*n+1) = A000120((2*n+1)*k), A000120 is the Hamming weight.
3, 13, 11, 3, 205, 43, 57, 5, 3277, 171, 35, 3641, 7, 52429, 683, 21, 47, 233017, 19, 838861, 2731, 3, 79, 99, 14913081, 23, 13421773, 10923, 241, 5, 197, 187, 954437177, 37, 214748365, 43691, 7, 61681, 7, 325, 419, 61083979321, 39, 3435973837, 174763
Offset: 1
Examples
Five initial terms of rows 1-5 are listed below: 1: 3, 11, 43, 171, 683, ... 2: 13, 205, 3277, 52429, 838861, ... 3: 3, 5, 7, 19, 23, ... 4: 57, 3641, 233017, 14913081, 954437177, ... 5: 35, 47, 99, 187, 419, ... T(3,4) = 19 because: (3*2+1) in binary is 111 and (3*2+1)*19 = 133 in binary is 10000101, both have 3 bits set to 1.
Links
- Pontus von Brömssen, Antidiagonals n = 1..100, flattened
Crossrefs
Formula
If 2*n = 2^j, then T(n, m) = (1+2^(j+2*j*m))/(2*n+1) for m > 0. In particular:
T(1, m) = (1+2^(1+2*m))/3 = A007583(m),
T(2, m) = (1+2^(2+4*m))/5 = A299960(m),
T(4, m) = (1+2^(3+6*m))/9.
The third row consists of all numbers of the form (1+2^(1+b*3)+2^(2+c*3))/7, where b and c are natural numbers >= 0 and b+c > 0.
The seventh row consists of all numbers of the form (1+2^(1+b*2)+2^(2+c*2)+2^(3+d*2))/15 where b, c, and d are natural numbers >= 0 and b+c+d > 1.
Extensions
More terms from Pontus von Brömssen, Jan 08 2021
Comments