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

1,1

Comments

Solutions to related equation A000120(k) = A000120(k*n) are A340351.

Links

Pontus von Brömssen, Antidiagonals n = 1..100, flattened

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.

Example

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.

Crossrefs

Cf. A000120, A340351, A340069.

Cf. A263132 (superset of 1st row), A007583 (1st row), A299960 (2nd row).

Sequence in context: A071708 A136592 A351657 * A085416 A107802 A142351

Adjacent sequences: A340438 A340439 A340440 * A340442 A340443 A340444

Keyword

nonn,base,tabl

Author

Thomas Scheuerle, Jan 07 2021

Extensions

More terms from Pontus von Brömssen, Jan 08 2021

Status

approved