cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A372313 Table read by antidiagonals: A(n,1) = 2n-1, and for k > 1, A(n,k) = A372289(A(n,k-1)+A(n,1)).

Original entry on oeis.org

1, 5, 3, 13, 13, 5, 29, 341, 21, 7, 61, 2773, 53, 29, 9, 125, 22229, 117, 149, 37, 11, 253, 177877, 245, 629, 93, 45, 13, 509, 1423061, 501, 2549, 205, 469, 53, 15, 1021, 11384533, 1013, 10229, 429, 15701, 133, 61, 17, 2045, 91076309, 2037, 40949, 877, 503125, 293, 309, 69, 19
Offset: 1

Views

Author

Ali Sada, Apr 26 2024

Keywords

Comments

Conjecture: if A(n,1) is congruent to 33 (mod 100), then all terms on the n-th row are congruent to 33 (mod 100).

Examples

			A(3,1) = 5. A(3,2) = A372289(5+5) = 21. A(3,3) = A372289(21+5) = 53. A(3,4) = A372289(53+5) = 117.
Table begins:
  1,  5,   13,   29,     61,        125,          253,           509, ...
  3,  13,  341,  2773,   22229,     177877,       1423061,       11384533, ...
  5,  21,  53,   117,    245,       501,          1013,          2037, ...
  7,  29,  149,  629,    2549,      10229,        40949,         163829, ...
  9,  37,  93,   205,    429,       877,          1773,          3565, ...
  11, 45,  469,  15701,  503125,    16100693,     515222869,     16487132501, ...
  13, 53,  133,  293,    613,       1253,         2533,          5093, ...
  15, 61,  309,  1301,   5269,      21141,        84629,         338581, ...
  17, 69,  173,  381,    797,       1629,         3293,          6621, ...
  19, 77,  3413, 27477,  219989,    1760085,      14080853,      112646997, ...
		

Crossrefs

A371094 a(n) = m*(2^e) + ((4^e)-1)/3, where m = 3n+1, and e is the 2-adic valuation of m.

Original entry on oeis.org

1, 21, 7, 21, 13, 341, 19, 45, 25, 117, 31, 69, 37, 341, 43, 93, 49, 213, 55, 117, 61, 5461, 67, 141, 73, 309, 79, 165, 85, 725, 91, 189, 97, 405, 103, 213, 109, 1877, 115, 237, 121, 501, 127, 261, 133, 1109, 139, 285, 145, 597, 151, 309, 157, 5461, 163, 333, 169, 693, 175, 357, 181, 1493, 187, 381, 193, 789, 199
Offset: 0

Views

Author

Antti Karttunen (proposed by Ali Sada), Apr 19 2024

Keywords

Comments

Construction: take the binary expansion of 3n+1 (A016777(n)), and substitute "01" for all trailing 0-bits that follow after its odd part (= A067745(1+n)), of which there are A371093(n) in total. See the examples.

Examples

			For n=1, 3*n+1 = 4, "100" in binary, when we substitute 01's for the two trailing 0's, we obtain 21, "10101" in binary, therefore a(1) = 21.
For n=6, 3*6+1 = 19, "10011" in binary, and there are no trailing 0's, and no changes, therefore a(6) = 19.
For n=7, 3*7+1 = 22, "10110" in binary, with one trailing 0, which when replaced with 01 gives us 45, "101101" in binary, therefore a(7) = 45.
For n=229, there are e=4 trailing bit expansions 0 -> 01,
  3n+1 = binary  101011  0 0 0 0
  a(n) = binary  101011 01010101
		

Crossrefs

Cf. A016921, A372351 (even and odd bisection), A372290 (numbers occurring in the latter).
Cf. also A302338.

Programs

  • Mathematica
    Array[#2*(2^#3) + ((4^#3) - 1)/3 & @@ {#1, #2, IntegerExponent[#2, 2]} & @@ {#, 3 #1 + 1} &, 67, 0] (* Michael De Vlieger, Apr 19 2024 *)
  • PARI
    A371094(n) = { my(m=1+3*n, e=valuation(m,2)); ((m*(2^e)) + (((4^e)-1)/3)); };
    
  • Python
    def A371094(n): return ((m:=3*n+1)<<(e:=(~m & m-1).bit_length()))+((1<<(e<<1))-1)//3 # Chai Wah Wu, Apr 28 2024

Formula

a(n) = A372289(A016777(n)).
a(2n) = A016777(2n) = A016921(n).
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